Mathematics First Year

Ministry of Higher Education and Scientific Research University of Technology Chemical Engineering Department

Mathematics First Year BY

Dr. Asawer A.Alwasiti 2008-2009

Mathematics Contents - Chapter one : Revision lines, equation of straight line, functions and graphs, even and odd functions, function in pieces, the absolute function, how to shift graph, sum difference product and quantitative of functions, composition of functions, inverse functions.

- Chapter two : Quadratic Functions Circles, parabolas, ellipses, hyperbolas, eccentricity, rotating the coordinate axes, discriminate.

- Chapter three: Transcendental Functions Power functions, exponential functions, logarithmic functions, natural logarithmic functions.

- Chapter four : Trigonometric Functions Graphs of trigonometric functions, periodicity, even and odd trigonometric functions, Inverse trigonometric functions, hyperbolic trigonometric functions, inverse hyperbolic trigonometric functions.

- Chapter Five: Limits and Continuity Properties, limits of trigonometric functions, limits involving infinity, limits of exponential functions, continuity.

- Chapter Six: Derivatives Definition, differentiation by rules, second and higher order derivative, application, implicit functions, the chain rule, derivative of trigonometric functions, derivative of hyperbolic functions, derivative of inverse hyperbolic functions, derivative of exponential and logarithmic functions, hospitals rule, partial derivative.

- Chapter seven: Integration Indefinite integration, integration of trigonometric functions, integration of inverse trigonometric functions, integration of logarithmic and exponential functions, integration of hyperbolic functions, integration of inverse hyperbolic functions, integration

methods; (substitution, by part, power trigonometric functions, trigonometric substitution, by part fraction), definite integration, applications; (area between two curves, length of curves, surface area, volumes). - Chapter eight: Determinants Definition, properties, solution of systems (Cramers rule).

- Chapter nine: Polar Coordinates Definition, Cartesian versus polar coordinates, graphing in polar coordinate.

- Chapter ten : Complex Number Definition, Graphing in complex number, Polar form of complex number, power of complex number, roots of complex number

University of Technology_______________ Mathematics_____________Chemical Engineering Department

Coordinates and Graphs in the Plain

y

II (-) negative quarter

I (+) positive quarter

x III (-) negative quarter

IV (+) positive quarter

Chapter One_________________________ Revision____________________Dr. Asawer A. Alwasiti

1


Lines Increment When particle moves from one point (x1, y1) to (x2, y2), the increments are: ∆x = x2 – x1 and ∆y =y2 – y1 Example Find the net changes in coordinates when particle moves from (4, -3) to (2, 5). Solution ∆x = 2-4 = -2 and ∆y = 5- (-3) = 8

(2, 5)

∆y

∆x

(4, -3)


2


Slope of a line The slope of the straight line is the ratio of rise to sum, so when point p1 (x1, y1) and p2 (x2, y2) are points on a nonverticle line L, the slope of L is m=

∆y y 2 − y1 = ∆x x 2 − x1

P2 (x2, y2)

∆y P1 (x1, y1) ∆x

Q (x, y)

Parallel and Perpendicular Lines

L1

L2

L2

L1 m2

m1 θ1

m1

m2 θ2

If L1//L2 then θ1=θ2 and m1=m2

θ1

θ2

If L1┴L2 then m1 =-1/m2


3


Equations of Straight Lines - Horizontal Lines The standard form of equation of horizontal lines is y=b

- Vertical lines The standard form of equation of vertical lines is x=a

y

y L (x, y) (b, 0)

x

L (x, y)

x

(a, 0)

- Neither horizontal nor vertical Point – slope equation The general form of point –slope equation of the point (x1, y1) with slope m is : y = m ( x – x1) +y1 Example Write an equation for the line thought (-2, -1) to (3, 4). Solution m=

−1− 4 − 5 = =1 −2−3 −5

Using (x2, y2) = (-2, -1) y = 1( x – (-2)) +(-1) y=x+1


4


Slope – intercept equation The general form of slope- intercept equation of line L with slope m and yintercept b is y = mx + b Example Find the slope and y-intercept of the line 8x + 5y = 20 Solution 8x + 5y = 20 20 8 − x 5 5 −8 y= x+4 5 y=

Slope = -8/5 and intercept is b=4 - General linear equation The general linear equation is Ax +By = C where both A and B are not zero Example Find the formula relating Fahrenheit and Celsius temperature, then find the Celsius equivalent of 90ºF and the Fahrenheit equivalent of -5 ºC. F = m ºC + b

(linear relationship)

The freezing point of water is TF =32 or TC= 0 and boiling point is TF=212 or TC=100. Solution F = m ºC + b 32 = m(0) +b

→ b = 32

and 212 = m (100) +32 F = 9/5 C +32

→ m = 9/5 or

C = 5/9 (F-32)


5


When F = 90 ºF then C = 5/9 (90 – 32) ≈ 32.2 º and when C = -5 ºC then F = 9/5 (-5) +32 = 23 º

H.W Q1) Find an equation for the line throught the point (p) perpendicular to L and then graph the equation a-

p (1, 2),

b-

p (-2, 4) ,

L: x +2y =3 L: x = 5

Q2) For what value of k are the two lines 2x + ky =3 and x + y = 1 if they are (a)parallel (b) perpendicular

Q3) The pressure p experienced by a driver under water is related to the drivers depth d by an equation of the form p = kd + 1 (k is a constant). When d=0 meters, the pressure is 1 atmospheres. The pressure at 100meter is about 10.94 atmospheres. Find the pressure at 50 meters.


6


Functions and Graphs Functions A function is like a machine that assigns a unique output to every allowable input.

x input (domain)

f

y = f(x) output (range)

y= f(x) in which : y = dependent variable x = independent variable Example Does the equation y2 + x = 1 represents a function y in terms of x? Solution Solve the above equation for y y 2= 1 - x y = + SQRT(1 - x) or y = - SQRT(1 - x) •

For one value of x we have two values of y and this is not a function.


7


Domain and Range Domain : is the set of real numbers over which x may vary and makes the value of y true ( input to the function). Rang : is the set of all values of y that makes the value of x true ( output of the function). Example: Find the domain and range of the following : a) y = x2 b) y = SQRT( 1 - x2) c) y = SQRT (4 - x) d) y = 1/x e) y = SQRT (x – 2) f) g(x) = SQRT ( - x 2 + 9) + 1 / (x - 1) Solution a) y = x2 domain : -∞0 → 2x < 3x < 10x bfor x 3-x >10-x Example Graph the function f(x)=log2x Solution f ( x) = log 2 x =

ln x ln 2

y = log2x

(1,0)

Example First order chemical reaction CA=CA0e-kt ,when t=t1/2 (half life) CA=0.5CAo . Prove that half life depend only on k. Solution CAoe-kt = 0.5 CAo So t 0.5 =

− ln 2 ln 2 = −k k

Chapter Three____________________ Transcendental Fnctions_________Dr. Asawer A. Alwasiti 32

University of Technology____________ Mathematics_____________Chemical Engineering Department

Example Solve for x a-ln x =3t +5 b- e2x = 10 c- 3log 7 − 4 log 2 = 5 (log a

4

5

x − log 5 x 2

)

solution a- x = e3t+5 b- x = 1.15 c- x = 1/5 H.W 1- Graph the functions below and state their domain and range. a- y = -2x +3

b- y = 3e-x – 2

2- Simplify a- 0.5ln( 4t4) – ln2

b- ln (e2lnx)

Chapter Three____________________ Transcendental Fnctions_________Dr. Asawer A. Alwasiti 33


Chapter Four Trigonometric Functions

Trigonometric Functions When an angle of measure θ is placed in standard position at the center of a circle of radius r, the six trigonometric functions of θ are: y r r cos θ = Cosine function: y y Tangent function: tan θ = x r 1 = Cosecant function: cscθ = sin θ y 1 r Secant function: secθ = = cos θ x 1 x = Cotangent function: cot θ = tan θ y

Sine function:

sin θ =

Conversion formula: 1 degree = π / 180 ( ≈ 0.02) radian 1 radian = 180 / π ( ≈ 57 ) degree

Identities: sin 2 θ + cos 2 θ = 1 sec 2 θ = 1 + tan 2 θ sin 2θ = 2 sin θ cosθ cos 2θ = cos 2 θ − sin 2 θ cos 2 θ =

1 + cos 2θ 2

Chapter Four____________________ Trigonometric Fnctions_________Dr. Asawer A. Alwasiti

34


sin 2 θ =

1 − cos 2θ 2

sin( A ± B ) = sin A cos B ± cos A sin B cos( A + B) = cos A cos B − sin A sin B

Periodicity A function f is periodic if there is a positive number p such that f (x+p) = f (x). the smallest such value of p is the period of f. cos (θ + 2π) = cos θ

,

sin (θ + 2π) = sin θ

tan (θ + 2π) = tan θ

,

sec (θ + 2π) = sec θ

sec (θ + 2π) = csc θ

,

cot (θ + 2π) = cot θ

similarly cos (θ - 2π) = cos θ

,

sin (θ - 2π) = sin θ and so on….

Graphs of trigonometric functions When we graph trigonometric functions in the coordinate plane, we denote the independent variable (radians) by x instead of θ.


35


Even and odd trigonometric functions The graph in the above figures suggest that cos θ and sin θ are even functions because their graphs are symmetric about the y-axis. The other four basic trigonometric functions odd. cos (-θ) = cos θ sin (-θ) = -sin θ sec(-θ)=1/cos (-θ)= 1/cos θ = sec θ Example Modeling temperature is ⎡ 2π ⎤ f ( x) = A sin ⎢ ( x − c)⎥ + D ⎣B ⎦

Where f is temperature and x is the number of the day. Graph this model.

Solution ‫׀‬A‫ ׀‬is the amplitude, ‫׀‬B‫ ׀‬is the period, C is the horizontal shift, and D is the vertical shift. Chapter Four____________________ Trigonometric Fnctions_________Dr. Asawer A. Alwasiti

36


Inverse trigonometric functions and their graphs


37


Important notes csc-1 x = sin-1(1/x) sec-1 x = cos-1(1/x) cot-1 x = π/2-tan-1(x) Example Prove that cos(A-B)=cosA cosB + sinA sinB Solution Since cos( A + B) = cos A cos B − sin A sin B So

cos( A − B ) = cos A cos( − B ) − sin A sin( − B )

= cosA cosB + sinA sinB


38


Hyperbolic Functions

Identifies


39


Inverse Hyperbolic Functions y = sinh −1 x = ln( x + x 2 + 1)

-∞ < x < ∞

y = cosh −1 x = ln( x + x 2 − 1) 1 1+ x y = tanh −1 x = ln 2 1− x 1+ 1− x2 ⎛1⎞ −1 y = sec h x = ln = cosh −1 ⎜ ⎟ x ⎝ x⎠ ⎛1 1 + x 2 ⎞⎟ ⎛1⎞ y = csc h −1 x = ln⎜ + = sinh −1 ⎜ ⎟ ⎜x x ⎟⎠ ⎝ x⎠ ⎝ 1 x +1 y = coth −1 x = ln 2 x −1

x >= 1 ‫׀‬x‫ < ׀‬1 0 < x ׀‬1

Graphing of Inverse Hyperbolic Functions


40


Important notes

Example Prove that cosh-1x – sinh-1x = 1 Solution ⎛ e x + e −x ⎜⎜ 2 ⎝

2

⎞ ⎛ e x − e −x ⎟⎟ − ⎜⎜ 2 ⎠ ⎝

2

⎞ ⎟⎟ = 1 ⎠

Simplifying the above equation yield 1

Example Prove that sinh(x+y) = sinhx cushy + coshx sinhy Solution ⎛ e x − e−x ⎞⎛ e y + e− y ⎞ ⎛ e x + e−x ⎞⎛ e y − e− y ⎞ ⎟⎟⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎜⎜ ⎟⎟ = ⎜⎜ 2 2 2 2 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ =

e x+ y − e − x− y = sinh( x + y ) 2

Example 1 2

Prove that y = tanh −1 x = ln

1+ x 1− x

Solution y = tanh-1x x = tanh y =

sinh y cosh y


41


x=

e2y −1 e2y + 1

1 2

Simplifying yield y = tanh −1 x = ln

1+ x 1− x

H.W 1 2

1- Prove that sin x cos y = {sin( x + y ) + sin( x − y )} 2- If t = tanh(x/2), show that tanh x =

2t 1+ t2

3- Prove that cosh 3x = 4 cosh 3 x − 3 cosh x 1 2

4- Show that cosh = m

1 (cosh x + 1) 2


42


Chapter Five Limits and Continuity Limits Definition: If the values of a function f of x approach the value L as x approaches c, we say f has limit L as x approach c and we write: lim f ( x) = L x →c

and read " the limit of f of x approach c equals L"

Properties of Limits Theorem 1 If L, M, c, and k are real numbers and lim f ( x) = L

and

x →c

1- Sum and Difference Rule:

lim g ( x) = M , then: x →c

lim( f ( x) + g ( x)) = L + M x →c

lim( f ( x).g ( x)) = L.M

2- Product Rule

x →c

( 3- Quotient Rule lim x →c

f ( x) L )= g ( x) M

(k . f ( x)) = k .L 4- Constant Multiple Rule lim x →c

5- Power Rule

lim( f ( x)) r / s = Lr / s x →c

where r and s are integers and s≠0

Theorem 2 Limits of Polynomial Function If P(x)= a n xn + a n-1 xn-1 + ……+a o , then lim P ( x) = P (c) = a n c n + a n −1c n −1 + ..... + a o x →c

Theorem 3 Limits of Rational functions If P(x) and Q(x) are two polynomials and Q(c) ≠ 0 them lim( x →c

P( x) P (c ) )= Q( x) Q (c )

Chapter Five____________________ Limits and Continuity _________Dr. Asawer A. Alwasiti

43


Example Find lim t →3

t3 −8 t2 − 4

Solution t3 −8 lim 2 =3 t →3 t − 4 Example Find lim x →3

4 − x2 + 7 x−3

Solution 3 4 − x2 + 7 4 + x2 + 7 =− * lim x →3 4 x−3 4 + x2 + 7 H.W Find the limits of 2+h − 2 1- lim h →0 h x −3 2- lim x →9 x − 9 5+ x − 5− x 3- lim x →0 2+ x − 2− x

Limits of Trigonometric Functions sin θ = 0 1- lim θ →0

cosθ = 1 2- lim θ →0

3- lim tan θ = θ →0

lim sin θ θ →0

lim cosθ θ →0

4- lim θ →0

sin θ

5- θlim → ±∞

θ sin θ

θ

=

0 =0 1

=1 =0


44


Example Find lim θ →0

sin( −θ ) −θ

Solution − sin(θ ) =1 = lim θ →0 −θ Example Find lim θ →0

tan θ

θ

Solution 1 sin θ = lim . = 1 ∗1 = 1 θ →0 θ cos θ Example Show that lim h →0 Solution 2 Since sin θ =

cosh − 1 =0 h

1 − cos 2θ 2

− So we can rewrite the limits as lim h →0 = − lim

sin θ

h →0

θ

sin 2 (h / 2) h

. sin θ = 0

Example Show that lim x →0

sin 2 x 2 = 5x 3

Solution sin 2 x sin 2 x 2 / 5 2 = lim . lim x →0 5 x 2 / 5 5 x →0 2 x = (2/5) H.W Find 1- lim x →0

sin 3 x x

2- lim x →0

sin x 2x 2 − x


45


Limits Involving Infinity If f ( x) =

1 x

It could be seen from the graph: 1- as x approaches 0 from right then 1/x tends to +∞ 2- as x approached 0 from the left then 1/x tends to -∞ 3- as x approached ∞ from the right (+∞), then 1/x tends to 0 4- as x approached ∞ from the right (-∞), then 1/x tends to 0 so 1 =0 x → +∞ x lim

And

lim+

x →0

1 = +∞ x

1 =0 x → −∞ x 1 lim− = −∞ x →0 x lim

, ,

Example 1 1 lim(5 + ) = lim 5+ lim = 5 x →∞ x →∞ x x →∞ x Example π 3 lim 2 = π 3 ∗ 0 ∗ 0 = 0 x → −∞ x Example 1 sin( ) Find lim x →∞ x Solution Let t = 1/x

So t → 0+ as x → ∞ 1 sin( ) = 0 So lim x →∞ x


46


Vertical and Horizontal asymptotes - A line y = b is a horizontal asymptote of the graph function y= f(x) if either; lim f ( x) = b x →∞

f ( x) = b or xlim → −∞

- A line x = a is a vertical asymptote of the graph of the function y = f(x) if either;

lim f ( x) = ±∞

x→a +

Example x+3 lim x →∞ x + 2 Solution 1+ 3/ x lim =1 x →∞ 1 + 2 / x asymptote) x+3 =∞ lim x → −2 x + 2 asymptote)

or lim f ( x) = ±∞ x →a −

i.e y = 1 ( horizontal

i.e. x = -2 ( vertical

Limits of Rational Functions 1- Degree of Denominator and Numerator of same degree. Example 5x 2 + 8x − 3 lim x →∞ 3x 2 + 2

y

Solution 8 3 − 2 x x =5 = lim x →∞ 2 3 3+ 2 x 2- Degree of Denominator greater than Numerator 5+

Example 11x + 2 lim 3 x → −∞ 2 x − 1


47


Solution 11 2 + 3 2 x x =0 lim x → −∞ 1 2− 3 x 3- Degree of Denominator less than Numerator Example 2x 2 − 3 x → −∞ 7 x + 4 lim

Solution 3 x 2 = −∞ lim x →− −∞ 4 7+ 2 x 2x −

H.W 1- find − 4x3 + 7x a- x→−∞ 2 2 x − 3 x − 10 lim

b- xlim → ±∞

3x 4 − 2 x 3 + 3x 2 − 5 x + 6 3x 4

2- Find the vertical and horizontal asymptotes of: x+3 x lim a- lim bx →∞ x − 1 x →∞ 7 x + 4

Limits of Exponential Functions The curve y = ex The line y = 0 is a horizontal asymptote of the graph so: ex = 0 1- xlim → −∞

2- lim x →∞

e−x =0 x

3- xlim → −∞

x =0 e−x


48


Example Find lim e

1 x

x →0 −

Solution Let t = 1/x so t→-∞

as x →0-

1 x

So lim e = 0 x →0 −

Example lim( x − x 2 + 1) x →∞

Solution lim( x − x 2 + 1) * x →∞

−1

= lim

x + x2 +1 Example 3 x −5 5 lim 3 x →∞ x +5 5 x →∞

x + x2 +1 x + x2 +1

=0

Solution x1 / 5 x1 / 3 = 1 = lim x →∞ x1 / 5 1 + 1/ 3 x 1−

H.W Find the limits and vertical and horizontal asymptote if exists: 1 3x 3 + 1 x lim sin lim 2- x→∞ 1- x→∞ x 2x + 5 2 3t 2 + 1 s s ( )( ) 3- lim 4- lim (try to graph this) x →∞ t 2 − 1 s →∞ s + 1 5 + s 2


49


Continuity A function f(x) is continuous at x =c if and only if it meets the following three conditions: 1- f(c) exists ( c lies in the domain of f) f ( x) exists ( f has a limit as x approaches c) 2- lim x→c f ( x) i.e lim f ( x) = lim f ( x) = lim x →c x →c +

x →c −

f ( x) =f(c) ( the limit equals the function value) 3- lim x→c

In other meaning: a function y = f(x) that can be graphed over each interval of its domain with one continuous motion of the pen is a continuous function.

Example The graph y = 1/x is continuous at every value of x except x=0. It has a point of discontinuity at x=0 The following types of functions are continuous at every point in their domain: 1- polynomials 2- rational functions: they have points of discontinuity at the zero of their denominator. 3- Root functions ( y = n x , n positive integer great than 1) 4- Trigonometric functions 5- Inverse trigonometric functions 6- Exponential functions 7- Logarithmic functions * the inverse function of any continuous function is continuous.


50


Chapter Six Derivatives Definition: The function f- defined by the formula f − ( x) = lim

∆x →0

f ( x + ∆x) − f ( x) ∆x

is called the derivative with respect to x of the function f. The domain of fconsists of all x for which the limit exists.

Geometric meaning of the derivative: f- is the function value at x and it’s the slope of the line to y=f(x) at x. Common notification for derivatives: y- ( y prime) dy dx

( dy by dx)

d [] dx

(the derivative with respect to (w.r.t))

Differentiation by Rules Let u =f(x)

and

v = g(x)

1-

d [c ] = 0 dx

2-

d [ax] = a a= constant dx d n [ x ] = nx n −1 dx d d d [u ± v] = [u ] ± [v] dx dx dx d [v ] d [u ] d [u.v] = v +u dx dx dx d d v [u ] − u [v] d u dx [ ] = dx 2 dx v v

3456-

c = constant

Chapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

51


du d n u = nu n−1 dx dx du d −n u = −nu −n −1 8dx dx

7-

Example s=(2t+3)2

Find the derivative of Solution s− =

ds = 2(2t + 3) * 2 = 8t + 12 dt

Example Find the derivative of r =

( x − 1)( x − 2 x) x4

Solution r− =

dr − x + 16 x − 6 = dx x4

Example Does the curve y=x4-2x2+2 have any horizontal tangent ? If so, where? Solution y- = 4x3 -4x solving the above equation when x=0 for x we get x=0, x=1, x=-1 at x=0 → (0,2) at x=1 → (1,1) at x=-1 → (-1,1) Example Find the derivative for y= (x2 + 1)3(x3 -1)2 Solution Let u=(x2 +1)3

and

v= (x3 -1)2

so y- = uv- +vuChapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

52


y- = 6x ( x2 +1)2 (x3 -1)(2x3+x-1) H.W Find the derivative of a- y =

1 ( x − 1) 5

b- y = (

2

2x −1 6 ) ( x 2 + 7)

Second and Higher Order Derivative dy dx d dy d2y y= = ( ) = 2 dx dx dx 2 d d y d3y y≡ = ( 2 ) = 3 dx dx dx d y n = ( y ) n −1 dx y− =

(first order) (second order) (third order) (nth order)

Application Velocity is the derivative of the distance. Speed is the absolute value of velocity. Speed = ‫׀‬velocity‫׀‬ Speed = ‫׀‬v(t)‫׀ = ׀‬ds/dt‫׀‬ Acceleration is the derivative of velocity with respect to time. a (t ) =

dv d 2 s = dt dt 2

Example The distance of a ball falls freely from rest is proportional with time as s=4.9t2 P

a- How long did it take the ball bearing to fall the first 14.7m? b- What is the velocity, speed and acceleration after 2 second? Solution a-

s = 4.9t2 P

14.7=4.9t2 P

P

so t = ±√3 second

t= √3 ( time increase from t=0 so we ignore the negative root) b-

velocity at any time


53

University of Technology____________ Mathematics_____________Chemical Engineering Department v(t ) =

ds = 9.8t dt

Velocity after 2 second = v(2) = 19.6 m/s Speed = ‫׀‬19.6‫ = ׀‬19.6 m/s Acceleration at any time d 2s a (t ) = 2 = 9.8m / s 2 dt

Acceleration after 2 second = 9.8 m/s2 P

Example A dynamite blast blows heavy rock straight up with a lunch velocity of 160 ft/sec. It reaches a height of s=160t -16t2 after time (sec) P

P

a- How height does the rock ago? b- what are thw velocity and speed of the rock when its 256 ft above the ground on the way up ? on the way down? c- what is the acceleration of the rock at any time t during its flight (after the blast)? d- when does the rock hit the ground again? Solution a- v =

ds = 160 − 32t ft/sec dt

at v=0 t=5 sec S max = s(5) = 400ft R

R

b- s(t) = 160 – 16 t2 P

P

at s = 256 ft then the time will be 2 and 8 second v(2) = 160 – 32 (2) = 96 ft/s v(8) = 160 - 32 (8) = - 96 ft /s At both instants, the rocks speed is 96 ft/s


54


c- a =

dv = −32 ft / sec 2 dt

( the acceleration is always downword)

d- at s=0 then time will be 10 sec

The Chain Rule If

y = f(t) and x = g(t), then dy dy dx = / dx dt dt

or

dy dy dt = . dx dt dx

And d 2 y dy − dx = / dx dt dx 2

or

d 2 y dy − dt = . dx dx dx 2

Example Find dy/dx

if

y = SQRT(1-t2)

and

x=t

Solution y− =

dy dy / dt = = −t (1 − t 2 ) dx dx / dt

Example Find d2y / dx2 if x = t – t2

and

y = t – t3

Solution y− =

dy dy / dt 1 − 3t 2 = = dx dx / dt 1 − 2t

dy − 2 − 6t + 6t 2 = dt (1 − 2t ) 2

d 2 y dy − / dt 2 − 6t + 6t 2 = = dx / dt dx 2 (1 − 2t ) 3

H.W Find d2y / dx2 for a- y = 2t , x = 3 - 3t

b- y = SQRT(t) , x = SQRT(t+1)


55


Implicite Function Différentiation We can find the derivative of implicit functions in two steps Step 1: Differentiate both sides of the equation with respect to x, treating y as a differentiable function of x Step 2: Solve for dy/dx Example Find dy/dx for y2 +5xy – 6x2 = 0 Solution dy dy + 5 x + 5 y − 12 x = 0 dx dx dy 12 x − 5 y = dx 2 y + 5 x

2y

Example Find the second derivative of the function 2x3 – 3y2 = 0 Solution 6x2 − 6 y

dy =0 dx

dy x 2 = dx y d 2 y 2x x4 = − dx 2 y y3

where y ≠ 0 Example Show that the point (2 , 4) lies on the curve x3 + y3 -9xy = 0. Then find the tangent and normal to the curve there (see figure) Solution Point (2,4) lies on the curve because 23 + 43 -9*2*4=0 To find the slope of the curve at (2,4)


56


dy dy − 9( x + y ) = 0 dx dx 2 dy 3 y − x = dx y 2 − 3 x dy = 4/5 dx ( 2, 4 )

3x 2 + 3 y 2

the tangent at (2,4) is the line thought (2,4) with slope 4/5 The equation of the tangent will be y =

4 12 x+ 5 5

The normal to the curve at (2,4) is the line perpendicular to the tangent, so the slope will be (-5/4) 5 4

And the equation is y = − x +

13 2

Example Assume that the radius r and the height of a cone are differentiable functions of t and let V be the volume of the cone. Find an equation that relates dV/dt, dr/dt and dh/dt. Solution V =

π 3

r 2h

dV π 2 = (r 3 dt dV π 2 = (r 3 dt

dh dr + 2r h dt dt dh dr + 2rh ) dt dt


57


Derivative of Trigonometric Functions If u = f(x) then y = sin u

y- = cos u .du/dx

y = cos u

y- = -sin u .du/dx

y = tan u

y- = sec2 u .du/dx

y = cot u

y- = -csc2 u .du/dx

y = sec u

y- = sec u . tan u.du/dx

y = csc u

y- = -csc u .cot u. du/dx

Example xy + sin y = 0 dy dy + y ) + cos y. = 0 dx dx dy −y = dx x + cos y

(x

Example Find the slope of the line tangent to the curve y = sin5x at the point where x=π/3. Solution y- = 5sin4x . cosx the tangent line has slope dy dx

x=

π 4

=

45 32

Example A body hanging from a spring is starched 5 units beyond its rest position and released at time t=0 to bob up and down. Its position at any later time is s = 5 cost What are its velocity and acceleration at time t? Solution Position s = 5 cost Velocity v(t) = -5sint Chapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

58


Acceleration a(t) = -5cost H.W Find the derivative for sin x x 1 y= sin x

y=

y 2 = x 2 + sin xy

2- Find the second derivative of y = sec25x

Derivative of Inverse Trigonometric Functions

Example y = cos −1

x 2 −1

dy 1 1 = . =− 2 dx 1− x / 4 2 4 − x2

Example

1 + 1− x2 x −1 dy − 2x 1 1 = x( = csc −1 ) + csc −1 + 2 dx x 2 1− x x 1 1 −1 x x y = x csc −1


59


Example

x −1 ) x +1  ( x + 1)(1) − ( x − 1)(−1)  1 =  x − 1 2  ( x + 1) 2  1+ ( ) x +1 2 = 2 ( x + 1) + ( x − 1) 2

y = tan −1 ( dy dx dy dx

H.W Find the derivative of a- y sin x + y = tan x 2

−1

b- y =

x2 − 4 1 x + sec −1 2 2 2 x

Derivative of Hyperbolic Function

Example

y = tanh(1 + x 2 ) dy = 2 x(sec h 2 (1 + x 2 ) dx

Example

y = x sec hx 2 dy = −2 x 2 sec hx 2 tanh x 2 + sec hx 2 dx


60


Dérivative of Inverse Hyperbolic Function

Example Find the derivative of y = cosh-1(secx) Solution 1 dy = . sec x. tan x dx sec 2 − 1 dy = sec x dx

Example Proof that

d 1 sinh −1 x = dx 1+ x2

Solution Let y = sinh-1 x

so

x = sinh y

1 dx dy = cosh y ⇒ = dy dx cosh y  cosh 2 − sinh 2 = 1 ⇒ cosh y = 1 + sinh 2 y ∴

1 dy = dx 1 + sinh 2 y

1 dy = dx 1+ x2 1 d ∴ sinh −1 x = dx 1+ x2 Chapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

61


Example If y = tanh-1u

Proof that

1 du dy = . dx 1 − u 2 dx

Solution y = tanh-1u 𝑑𝑑𝑑𝑑

→ u = tanhy

𝑑𝑑𝑑𝑑

= 𝑠𝑠𝑠𝑠𝑠𝑠ℎ2 𝑦𝑦

∴

1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = . 𝑑𝑑𝑑𝑑 1 − 𝑢𝑢2 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 ∴ = . 𝑑𝑑𝑑𝑑 𝑠𝑠𝑠𝑠𝑠𝑠ℎ2 𝑦𝑦 𝑑𝑑𝑑𝑑

∴

Derivative of Exponential and Logarithmic Function If u = f(x) 1- y = b u

dy du = b u . ln b. dx dx

2- y = e u

du dy = eu . dx dx

3- y = log b u

dy 1 1 du = . . dx u ln b dx

4- y = ln u

dy 1 du = . dx u dx

Example y = x sin x ⇒ ln y = sin x. ln x 1 dy 1 . = cos x. ln x + sin x. y dx x dy sin x = x sin x ( + cos x. ln x) dx x


62


Example 1+ x 1 ⇒ ln y = (ln(1 + x) − ln(1 − x)) 1− x 2 1 dy = dx 1 − x 2 y = ln

Example tan y = e x + ln x 1 dy = ex + dx x x 1  xe + 1  dy   = dx sec 2 y  x 

sec 2 y.

Example y = 3− x dy = −3 − x. ln x dx

Example y = π sin x dy = π sin x . ln π . cos x dx

H.W Find the derivative for a) y = x x

2

b) y = ln( x 2 + 3) c) y = e sin x

Example How rabidly will the fluid level inside a vertical cylindrical tank drop if we pump the fluid out at the rate of 300 L/min? Solution dv/dt = -3000 L/min = -3 m3/min the rate is negative because the volume is decreasing


63


V = πr 2 h dV dh = πr 2 dt dt 3 dh =− 2 dt πr

The fluid level will drop at a rate of 3/πr2 m/min Example A hot air balloon rising straight up from a level field is tracked by a range finder 500ft from the lift off point. At the moment the range finders elevation angle isπ/4, the angle is increasing at the rate of 0.14 rad/min. How fast is the balloon rising at that moment? Solution θ = f(t) = the angle in radius that range finder makes with the ground y = f(t) = the height in feet of the balloon y ⇒ y = 500. tan θ 500 dy dθ = 500 sec 2 θ . dt dt dy = 140 ft / min dt

tan θ =

The balloon is rising at the rate of 140 ft/min Example A police cruiser, approaching a right- angled intersection from the north, is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6m north of the intersection and the car is 0.8m to the east. The police determined with radar that the distance between them and the car is


64


increasing at 20 mph. If the cruiser is moving at 60 mph at the instance of the measurement, What is the speed of the car? Solution x = position of the car at time t y = position of the cruiser at time t s = distance between car and the cruiser at time t at x = 0.8m and y = 0.6m , ds/dt = 20 mph

and dy/dt = -60 mph ( because y is decreasing)

s2 = x2 + y2 ⇒ s = x2 + y2 ds dx dy = 2x + 2 y dt dt dt ds 1 dx dy = (x + y ) dt s dt dt ds dx dy 1 = (x + y ) dt dt dt x2 + y2

2s

∴

dx = 70mph dt

So the car speed is 70 mph Example Water runs into a conical tank at the rate of 9 ft3/min. The tank stands point down and has a height of 10 ft and a base radius of 5ft. How fast is the water level rising when the water is 6ft deep? Solution V = volume of the water in the tank at time t x = radius of the surface of the water at time t y = depth of the water in the tank at time t


65

University of Technology____________ Mathematics_____________Chemical Engineering Department x 5 1 ⇒x= y = y 10 2 V = V =

π 3

x2 y

π  y

2

  y 32 dV π 2 dy = y 4 dt dt dy = 0.32 ft / min dt

Hence, the water level is rising at about 0.32 ft/min.

Application of Derivatives on Limits: Hopitals Rule Intermediate forms of L' Hopitals, Rule: - Intermediate form 0/0 lim x→a

f ( x) f − (a) = g ( x) g − (a)

where f(a) = g(a) = 0

Example Find lim x →0

3 x − sin x x

Solution 3 − cos x =2 x →0 1

= lim

Example Find lim x →0

x − sin x x3

Solution 1 − cos x x →0 3x 2 sin x = lim x →0 6 x cos x 1 = lim = x →0 6 6 = lim


66


H.W Find 1- lim x →0

1− x −1− x2

x 2

2- lim x →π

sin 2 x 1 + cos x

- Other intermediate form ∞/∞, 0/∞, ∞/-∞ lim x→a

f ( x) f − (a) = g ( x) g − (a)

where f(a) = g(a) = ∞

Example tan x x →π 1 + tan x sec 2 x = lim =1 x →π sec 2 x

lim

Example 1 1 − ) x →0 sin x x  x − sin x  = lim  x →0  x sin x  1 − cos x 0 = lim = =0 x →0 x cos x + sin x 2

lim(

Example e 2 x + ln x lim( ) x →0 x + x 2  xe 2 x = lim x →0 x + x 2 

   2x 2e . x + e 2 x 1 = lim = =1 x →0 1 + 2x 1

Theorem ln f ( x) = L If lim x→a f ( x) = lim e ln f ( x ) = e L Then lim x→a x→a Chapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

67


Example lim(1 − x 2 )

−

1 x2

x →o

2

= f ( x) = (1 − x )

−

1 x2

1 ln(1 − x 2 ) 2 x ln(1 − x 2 ) ∴ lim ln f ( x) = − lim x →0 x →0 x2 − 2x 2 = − lim 1 − x x →0 2x 1 ∴ lim ln f ( x) = + lim =1 x →0 x →0 1 − x 2 ∴ lim f ( x) = e1 ln f ( x) = −

x →0

H.W Find 1- xlim →π / 3

cos x − 0.5 x −π /3

x − x2 + x 2- lim x →∞


68


Partial Derivative Let f(x,y) be a function of two variables, then we define the partial derivative as: 𝑓𝑓𝑥𝑥 =

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝑓𝑓𝑦𝑦 =

= limℎ→𝑜𝑜 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕

𝑓𝑓(𝑥𝑥+ℎ,𝑦𝑦)−𝑓𝑓(𝑥𝑥,𝑦𝑦)

= limℎ→𝑜𝑜

ℎ

𝑓𝑓(𝑥𝑥,𝑦𝑦+ℎ)−𝑓𝑓(𝑥𝑥,𝑦𝑦) ℎ

Algebraically, we can think of partial derivative of a function with respect to x as the derivative of the function with y held constant. Example 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 3𝑥𝑥𝑦𝑦 2 − 2𝑦𝑦𝑥𝑥 2

𝜕𝜕𝜕𝜕 = 3𝑦𝑦 2 − 4𝑥𝑥𝑥𝑥 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑓𝑓𝑦𝑦 = = 6𝑥𝑥𝑥𝑥 − 2𝑥𝑥 2 𝜕𝜕𝜕𝜕

𝑓𝑓𝑥𝑥 =

Example

Find f x and f y if 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = Solution 𝑓𝑓𝑥𝑥 =

𝑓𝑓𝑦𝑦 =

H.W

2𝑦𝑦

𝑦𝑦=𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

𝜕𝜕𝜕𝜕 2𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 = 𝜕𝜕𝜕𝜕 (𝑦𝑦 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)2

𝜕𝜕𝜕𝜕 2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝜕𝜕𝜕𝜕 (𝑦𝑦 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)2

Find f x and f y for

1) 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥

2) 𝑓𝑓(𝑥𝑥, 𝑦𝑦) =

𝑥𝑥 + 𝑦𝑦 𝑥𝑥 − 𝑦𝑦

Higher Order Partial Derivative: Just as with function of one variable, we can define second derivatives for functions of two variables. For function of two variables we have four types: Chapter six ____________________ Derivative ___________________Dr. Asawer A. Alwasiti

69


𝜕𝜕 2 𝑓𝑓 𝑜𝑜𝑜𝑜 𝜕𝜕𝜕𝜕 2

𝑓𝑓𝑥𝑥𝑥𝑥

𝑜𝑜𝑜𝑜

𝑓𝑓𝑦𝑦𝑦𝑦

𝜕𝜕 2 𝑓𝑓

,

,

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕


𝜕𝜕 2 𝑓𝑓 𝑜𝑜𝑜𝑜 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝑓𝑓𝑦𝑦𝑦𝑦 𝑜𝑜𝑜𝑜

𝜕𝜕 2 𝑓𝑓

𝜕𝜕𝜕𝜕 2

Theorem Let f(x,y) be a function with continues second order derivative, then f xy = f yx Example let f(x,y) = yex then f x = yex , f xx = yex ,

f y = ex

f xy = ex

, f yy = 0

, f yx = ex

so f xy = f yx Example 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + 𝑦𝑦𝑒𝑒 𝑥𝑥 𝑓𝑓𝑥𝑥 =

𝜕𝜕𝜕𝜕 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑦𝑦𝑒𝑒 𝑥𝑥 𝜕𝜕𝜕𝜕

𝑓𝑓𝑥𝑥𝑥𝑥 =

𝜕𝜕 2 𝑓𝑓 = 𝑦𝑦𝑒𝑒 𝑥𝑥 2 𝜕𝜕𝑥𝑥


𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑓𝑓 = � �= = −𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑒𝑒 𝑥𝑥 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕


𝜕𝜕 2 𝑓𝑓 = 2 = −𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝜕𝜕𝑦𝑦

𝑓𝑓𝑦𝑦 =


𝜕𝜕𝜕𝜕 = −𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + 𝑒𝑒 𝑥𝑥 𝜕𝜕𝜕𝜕

𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑓𝑓 = � �= = −𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑒𝑒 𝑥𝑥 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕𝜕𝜕𝜕𝜕


70


Functions of More Than Two Variables Suppose that f(x,y,z) = xy -2yz is a function of three variables, then we can define the partial derivatives in such the same way as we defined the partial derivatives, so we have: 𝑓𝑓𝑥𝑥 =

𝜕𝜕𝜕𝜕

𝑓𝑓𝑧𝑧 =

𝜕𝜕𝜕𝜕

𝑓𝑓𝑦𝑦 =

H.W

𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕

= 𝑦𝑦

= 𝑥𝑥 − 2𝑧𝑧

= −2𝑦𝑦

Find f x , f y , f z for f(x,y,z) = xsin(y+3z)

The Chain Rule for Functions of Two and Three Variable: If w = f(x,y) ∴

and x = f(t), y = f(t)

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 = . + . 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑

In similar way if w=f(x, y, z) And x = f(t), y = f(t), z=f(t) ∴

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 = . + . + . 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑

Example

Use the Chain Rule to find the derivative of w = xy with respect to t along the path x = cost, y = sint. What is the derivatives value when t=π/2? Solution 𝜕𝜕𝜕𝜕 = 𝑦𝑦 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 = −𝑠𝑠𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑

,

,

𝜕𝜕𝜕𝜕 = 𝑥𝑥 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑


71


∴

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 = . + . 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑

𝜕𝜕𝜕𝜕 = 𝑐𝑐𝑐𝑐𝑐𝑐2𝑡𝑡 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑙𝑙 = −1 𝜕𝜕𝜕𝜕 𝑡𝑡=𝜋𝜋/2

Example

Find the partial derivative of w if w =xy+z , x = cost, y = sint, z =t What the derivatives value at t=0? Solution ∴

= 𝑦𝑦(−𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠) + 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + (1)(1)

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 𝑙𝑙 =2 𝜕𝜕𝜕𝜕 𝑡𝑡=0

Implicit Differentiation U

Suppose f(x,y) and its partial derivatives f x and f y are continues and the equation R

R

R

R

f(x,y)=0 defines y as a differentiable function of x. Then at any point where f y ≠0: R

R

𝑓𝑓𝑥𝑥 𝑑𝑑𝑑𝑑 =− 𝑓𝑓𝑦𝑦 𝑑𝑑𝑑𝑑 Example Find dy/dx if x2 +siny =2y P

PR

R

Solution f(x,y) = x2 +siny -2y P

PR

R

𝑓𝑓𝑥𝑥 𝑑𝑑𝑑𝑑 =− 𝑓𝑓𝑦𝑦 𝑑𝑑𝑑𝑑


72


Chapter Seven Integration It’s a mathematics we use to find lengths, area and volumes of irregular shapes; to calculate the average values of functions; and to predict future population sizes and future costs of living. Two kinds of integral: 1- Identified integral

2- Definite Integral

1-Indefinite Integral:

If f-(x) is a derivative then the ant derivative of f-(x) is called the indefinite integral of f-(x) with respect to x and is denoted by: � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝑓𝑓(𝑥𝑥) + 𝑐𝑐 Integration formula: 𝑑𝑑𝑑𝑑 1- ∫ 𝑑𝑑𝑑𝑑 = 𝑢𝑢(𝑥𝑥) + 𝑐𝑐 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑

2- ∫ 𝑎𝑎𝑎𝑎(𝑥𝑥) = 𝑎𝑎 ∫ 𝑢𝑢(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

3- ∫�𝑢𝑢(𝑥𝑥) ± 𝑣𝑣 (𝑥𝑥)�𝑑𝑑𝑑𝑑 = ∫ 𝑢𝑢(𝑥𝑥) 𝑑𝑑𝑑𝑑 ± ∫ 𝑣𝑣 (𝑥𝑥)𝑑𝑑𝑑𝑑 4- ∫

𝑢𝑢𝑛𝑛 (𝑥𝑥)

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 =

∫ 𝑥𝑥 𝑛𝑛 𝑑𝑑𝑑𝑑 =

𝑢𝑢(𝑥𝑥)𝑛𝑛 +1 𝑛𝑛+1

𝑥𝑥 𝑛𝑛 +1 𝑛𝑛+1

Example 𝑑𝑑𝑑𝑑 Find y if = 𝑥𝑥 2 �𝑦𝑦 𝑑𝑑𝑑𝑑

√𝑦𝑦

𝑑𝑑𝑑𝑑

+ 𝑐𝑐

+ 𝑐𝑐

y" > 0

= 𝑥𝑥 2 𝑑𝑑𝑑𝑑

2�𝑦𝑦 = 𝑥𝑥 3 + 𝑐𝑐

Example 2𝑥𝑥 2 + 1 �3 𝑑𝑑𝑑𝑑 √2𝑥𝑥 3 + 3𝑥𝑥 + 1

1

= �(2𝑥𝑥 3 + 3𝑥𝑥 + 1)3 (2𝑥𝑥 2 + 1)𝑑𝑑𝑑𝑑

∗

3 3

Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

73

University of Technology____________ Mathematics_____________Chemical Engineering Department 2 1 = (2x 3 + 3x + 1)3 + c 2

Example

At any point (x,y) of curve

𝑑𝑑 2 𝑦𝑦

𝑑𝑑𝑥𝑥 2

= 12𝑥𝑥 2 − 6 when x =-1, y= -11 and when x=1, y=

-3. Find the equation of the curve and the equation of the tangent and normal at (1,-3). Solution 𝑑𝑑2 𝑦𝑦 � 2 = �(12𝑥𝑥 2 − 6)𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥 �

𝑑𝑑𝑑𝑑 = �(4𝑥𝑥 3 − 6𝑥𝑥 + 𝑐𝑐)𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

y=x4-3x2+cx+d

at (-1.-11) and (1,-3) c = 4 and d=-5

Hence, the equation of the curve is y=x4 -3x +4x -5 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

= 4𝑥𝑥 3 − 6𝑥𝑥 + 4

at (1,-3)

The slope of the tangent is 2 and the slope of the normal is -1/2 The equation of the tangent is y=2x+5 The slope of the normal is y = -x-5 Example As a function of elapsed time t, the velocity v(t) of a body falling from rest in a vacuum near the surface of the earth satisfies the differential equation with initial condition v=0 at t=0 find v as a function of t.


= 9.8,

Solution dv =9.8dt v = 9.8t +c

(0,0)

v=9.8 t m/s Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

74


H.W 1- Speed of the body movement straight line is given by v=6(2t-t2), find the distance of the body from a fixed point after two second. 2- Find the equation of the curve whose slope is (x – 1/x2) and passing through (2,5) 3- Find y if 4-Find r if


𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑

=

𝑥𝑥+1

𝑦𝑦−1

y≠1

= (2𝑧𝑧 + 1)3

Integration of Trigonometric Functions 𝑑𝑑𝑑𝑑

1- ∫ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

2- ∫ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

3- ∫ 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑢𝑢 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

4- ∫ 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑢𝑢 = −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

5- ∫ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

6- ∫ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐 𝑑𝑑𝑑𝑑

Example

� 𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥. 𝑑𝑑𝑑𝑑

Example 𝑐𝑐𝑐𝑐𝑐𝑐 2𝑥𝑥

∫ 𝑠𝑠𝑠𝑠𝑠𝑠 3 2𝑥𝑥 𝑑𝑑𝑑𝑑 = ∫ 𝑠𝑠𝑠𝑠𝑠𝑠−3 2𝑥𝑥. 𝑐𝑐𝑐𝑐𝑐𝑐2𝑥𝑥. 𝑑𝑑𝑑𝑑 Let u= sin2x

(power function)

so du=cos2x.2dx

= � 𝑢𝑢−3 . Example �


𝑑𝑑𝑑𝑑 2

1 𝑐𝑐𝑐𝑐𝑐𝑐 2 (7𝑥𝑥

= � 𝑠𝑠𝑠𝑠𝑠𝑠 2 (7 75


1 = tan(7𝑥𝑥 7

Example

� 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥. 𝑑𝑑𝑑𝑑

1 + 𝑐𝑐𝑐𝑐 2 1 1 = 𝑥𝑥 + 𝑠𝑠 2 4 =�

Example

� 𝑠𝑠𝑠𝑠𝑠𝑠3 2𝑥𝑥. 𝑑𝑑

= � 𝑠𝑠𝑠𝑠𝑠𝑠2 2𝑥𝑥

= �(1 − 𝑐𝑐𝑐𝑐

= � 𝑠𝑠𝑠𝑠𝑠𝑠2𝑥𝑥.

1 = 𝑐𝑐𝑐𝑐𝑐𝑐 3 2𝑥𝑥 6

Example

� 𝑡𝑡𝑡𝑡𝑡𝑡4 2𝑥𝑥. 𝑠𝑠

= � 𝑡𝑡𝑡𝑡𝑡𝑡4 2𝑥𝑥

= � 𝑡𝑡𝑡𝑡𝑡𝑡4 2𝑥𝑥

𝑡𝑡𝑡𝑡𝑡𝑡7 2𝑥𝑥 =� 14

Integration of Inverse Trigonometric Functions U

𝑑𝑑𝑑𝑑

1

𝑢𝑢

1- ∫ 2 2 = 𝑠𝑠𝑠𝑠𝑠𝑠−1 + 𝑐𝑐 𝑎𝑎 𝑎𝑎 √𝑎𝑎 +𝑢𝑢 2- ∫

3- ∫

𝑑𝑑𝑑𝑑

𝑎𝑎 2 +𝑢𝑢 2 𝑑𝑑𝑑𝑑

1

𝑢𝑢

= 𝑡𝑡𝑡𝑡𝑡𝑡−1 + 𝑐𝑐

𝑢𝑢√𝑢𝑢 2 −𝑎𝑎 2

𝑎𝑎

1

𝑎𝑎

𝑢𝑢

= 𝑠𝑠𝑠𝑠𝑠𝑠 −1 � � + 𝑐𝑐 𝑎𝑎

𝑎𝑎

𝑢𝑢2 < 𝑎𝑎2 𝑢𝑢2 > 𝑎𝑎2


76


Example

u=x

3 P

2

so du=3x .dx P

P

P

2

→ du/3 = x .dx P

P

Example

�

𝑥𝑥 2 𝑑𝑑𝑑𝑑

√1 + 𝑥𝑥 6

1 𝑑𝑑𝑑𝑑 � 3 √1 − 𝑢𝑢

�

3

(𝑥𝑥 + 2)√

=� =�

(𝑥𝑥 + 2

(𝑥𝑥 + 2

Example

= 3𝑠𝑠𝑠𝑠𝑠𝑠 −1 |𝑥𝑥

Let u =2x

�

du=2dx

and a=5

𝑑𝑑𝑑𝑑 4𝑥𝑥 2 + 25

1 2 = � 2 (2𝑥𝑥)

=

H.W Find: 1- ∫

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 .𝑑𝑑𝑑𝑑

√1−𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥

2- ∫

2(𝑥𝑥−1).𝑑𝑑𝑑𝑑

�1−(𝑥𝑥−1)4

3- ∫ 4𝑐𝑐𝑐𝑐𝑐𝑐3𝑦𝑦. 𝑑𝑑𝑑𝑑 𝜋𝜋

𝜋𝜋

4- ∫ csc �𝑥𝑥 + � . cot �𝑥𝑥 + � . 𝑑𝑑𝑑𝑑 5- ∫

𝑑𝑑𝑑𝑑

√9−𝑥𝑥 2

4

4


77

1 𝑡𝑡𝑡𝑡𝑡𝑡−1 10


Integration of Basic Logarithmic and Exponential Functions U

1- ∫

𝑑𝑑𝑑𝑑

= 𝑙𝑙𝑙𝑙|𝑢𝑢| + 𝑐𝑐

𝑢𝑢

2- ∫ 𝑒𝑒 𝑢𝑢 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑢𝑢 + 𝑐𝑐 1

. 𝑎𝑎𝑢𝑢 + 𝑐𝑐 3- ∫ 𝑎𝑎𝑢𝑢 𝑑𝑑𝑑𝑑 = 𝑙𝑙𝑙𝑙𝑙𝑙

𝑎𝑎 ≠ 1, 𝑎𝑎 > 0

Example 𝑑𝑑𝑑𝑑 .𝑑𝑑𝑑𝑑 ∫ 3𝑥𝑥 2 +4 = 𝑙𝑙𝑙𝑙|3𝑥𝑥 2 + 4| + 𝑐𝑐 Since |3𝑥𝑥 2 + 4| > 0

= ln (3x2+4)+c P

P

Example 2

Example

� 𝑥𝑥𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 =

4𝑒𝑒 3𝑥𝑥 − 3 � 𝑒𝑒 𝑥𝑥+1 𝑒𝑒 3𝑥𝑥 = 4 � 𝑥𝑥+1 𝑒𝑒 = 4 � 𝑒𝑒 2𝑥𝑥−

Example

Example

= 2𝑒𝑒 2𝑥𝑥−1 −

� 2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 �

𝑑𝑑𝑑𝑑

√𝑥𝑥 + (1 +

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 1 + = 2� Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

78

𝑑𝑑𝑑𝑑 = 𝑢𝑢


Example

Example

𝑒𝑒 𝑥𝑥 − 𝑒𝑒 −𝑥𝑥 � 𝑥𝑥 𝑒𝑒 + 𝑒𝑒 −𝑥𝑥

= 𝑙𝑙𝑙𝑙|𝑒𝑒 𝑥𝑥 + 𝑒𝑒 � 𝑒𝑒 𝑥𝑥

2 +𝑒𝑒 𝑥𝑥 2 +𝑙𝑙 2

= � 𝑒𝑒 𝑥𝑥 . 𝑒𝑒 𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑒𝑒 𝑥𝑥 = � 𝑒𝑒 𝑢𝑢 .

2

𝑑𝑑𝑑𝑑 2

Integration of Hyperbolic Functions U

Example

� 𝑐𝑐𝑐𝑐𝑐𝑐ℎ5𝑥𝑥. 𝑑𝑑

=�


79

𝑐𝑐𝑐𝑐𝑐𝑐ℎ5𝑥𝑥 𝑠𝑠𝑠𝑠𝑠𝑠ℎ5𝑥𝑥


Example �

𝑠𝑠𝑠𝑠𝑠𝑠ℎ√𝑥𝑥 √𝑥𝑥

𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = √𝑥𝑥 Example

= 2𝑐𝑐𝑐𝑐𝑐𝑐ℎ√𝑥𝑥

� 𝑠𝑠𝑠𝑠𝑠𝑠ℎ𝑥𝑥. 𝑑𝑑𝑑𝑑

=�

Integration of Inverse Hyperbolic Functions

= 2�

U

Example �


2𝑑𝑑𝑑𝑑 𝑒𝑒 𝑥𝑥 + 𝑒𝑒

𝑒𝑒 𝑥𝑥 (𝑒𝑒 𝑥𝑥 )2

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

√1 + 𝑠𝑠𝑠𝑠𝑠𝑠

80


Example �

𝑑𝑑𝑑𝑑

√𝑥𝑥 2 + 6𝑥𝑥

=�

=�

Example �

√𝑥𝑥 2 + 𝑑𝑑

�(𝑥𝑥 +

1 (𝑥𝑥 + )� 2

=� =�

Integration Methods

(𝑥𝑥 + (𝑥𝑥 +

U

1 2

1 2

1- Integration by Substitution U

Example

Let u= 4-x2 P

so du = -2xdx →xdx=-du/2 P

�

√4 − 𝑥𝑥 2

=� Example Let u=sin(x2+1) P

P

(du=cos(2x2+1).4x.dx)*4 P

16xdx.cos(2x2+1) =4du P


𝑑𝑑

−𝑑𝑑𝑑𝑑/2 √𝑢𝑢

� 𝑠𝑠𝑠𝑠𝑠𝑠3 (2𝑥𝑥 2

P

P

𝑥𝑥

81


� 𝑢𝑢3 . 4𝑑𝑑𝑑𝑑 = Example �

sin(𝑙𝑙𝑙𝑙𝑙𝑙 ) 𝑥𝑥

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑙𝑙𝑙𝑙𝑙𝑙

= − cos(𝑙𝑙𝑙𝑙𝑙𝑙

2- Integration by Parts U

u =f(x)

,

v=g(x)

𝑑𝑑 (𝑢𝑢. 𝑣𝑣 ) = 𝑑𝑑𝑑𝑑

� 𝑢𝑢𝑢𝑢𝑢𝑢 = 𝑢𝑢.

Example

Let u =lnx du=dx/x

dv=dx

� 𝑙𝑙𝑙𝑙𝑙𝑙. 𝑑𝑑𝑑𝑑

v=x ∴ � 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙

� 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = Example

� 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑥𝑥 =

𝑡𝑡𝑡𝑡𝑡𝑡−1 � . 𝑑𝑑 1 + 𝑥𝑥 2


82


Example

� 𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥. 𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑡𝑡𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑 =

1

� 𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥. 𝑑𝑑 Example � 𝑥𝑥𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑑𝑑

� 𝑥𝑥𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 =

Example

� 𝑒𝑒 𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑

Take ∫ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑒𝑒 𝑥𝑥 . 𝑑𝑑𝑑𝑑

2 � 𝑒𝑒 𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑒𝑒 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

� 𝑒𝑒 𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑 = Example

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒

� 𝑒𝑒 𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑒𝑒

∴ � 𝑒𝑒 𝑥𝑥 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

1 𝑥𝑥 (𝑒𝑒 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑒𝑒 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) + 𝑐𝑐 2

� 𝑥𝑥 2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑥𝑥 2 83


𝑑𝑑𝑑𝑑 = 2

� 𝑥𝑥 2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 = −𝑥𝑥 2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 2 � 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥. 𝑑𝑑𝑑𝑑

Take ∫ 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥. 𝑑𝑑𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑑𝑑

∴ � 𝑥𝑥 2 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 = −𝑥𝑥 2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 2𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 + 2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐 Example

� ln(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) . 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 ∴ � ln(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) . 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 = −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. ln(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) — 𝑐𝑐𝑐𝑐𝑠𝑠𝑠𝑠 � � ln(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ) . 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 = (1 − ln(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐))𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � 𝑑𝑑𝑑𝑑 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = ln⁡ − 𝑑𝑑𝑑𝑑 =

H.W. 1- ∫ ln(𝑥𝑥 + 1) . 𝑑𝑑𝑑𝑑

2- ∫ 𝑒𝑒 2𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐3𝑥𝑥. 𝑑𝑑𝑑𝑑

3- Power Trigonometric Functions Example � tan4 𝑥𝑥. 𝑑𝑑𝑑𝑑

= � tan2 𝑥𝑥. tan2 𝑥𝑥. 𝑑𝑑𝑑𝑑

= � tan2 𝑥𝑥. (sec 2 𝑥𝑥 − 1). 𝑑𝑑𝑑𝑑

= � tan2 𝑥𝑥. sec 2 𝑥𝑥. 𝑑𝑑𝑑𝑑 − �(sec 2 𝑥𝑥 − 1). 𝑑𝑑𝑑𝑑 1 = 𝑡𝑡𝑡𝑡𝑡𝑡3 𝑥𝑥 − 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑥𝑥 + 𝑐𝑐 3


84


Example � tan2 𝑥𝑥. sec 3 𝑥𝑥. 𝑑𝑑𝑑𝑑

= � sec 3 𝑥𝑥. (sec 2 𝑥𝑥 − 1). 𝑑𝑑𝑑𝑑

= � sec 5 𝑥𝑥. dx − � sec 3 𝑥𝑥 . 𝑑𝑑𝑑𝑑

For ∫ sec 3 𝑥𝑥 . 𝑑𝑑𝑑𝑑 = ∫ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥. 𝑑𝑑𝑑𝑑 ∴ � sec 3 𝑥𝑥 . 𝑑𝑑𝑑𝑑 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − � 𝑡𝑡𝑡𝑡𝑡𝑡2 𝑥𝑥. 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = sec 𝑑𝑑𝑑𝑑 = 𝑠𝑠

= 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − � 𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑥𝑥. 𝑑𝑑𝑑𝑑 + � 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑

∴ � sec 3 𝑥𝑥 . 𝑑𝑑𝑑𝑑 =

1 1 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑙𝑙𝑙𝑙|𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 | + 𝑐𝑐 2 2

For ∫ sec 5 𝑥𝑥 . 𝑑𝑑𝑑𝑑 = ∫ 𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑥𝑥. 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥. 𝑑𝑑𝑑𝑑

� sec 5 𝑥𝑥 . 𝑑𝑑𝑑𝑑 = sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − � 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 3𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥. 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 𝑑𝑑𝑑𝑑

𝑙𝑙𝑙𝑙𝑙𝑙 𝑢𝑢 = sec 𝑑𝑑𝑑𝑑 = 3

= sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 3 � 𝑡𝑡𝑡𝑡𝑡𝑡2 𝑥𝑥. 𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑥𝑥. 𝑑𝑑𝑑𝑑

= sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 3 �(𝑠𝑠𝑠𝑠𝑠𝑠 2 𝑥𝑥 − 1). 𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑥𝑥. 𝑑𝑑𝑑𝑑

= sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 3 � 𝑠𝑠𝑠𝑠𝑠𝑠 5 𝑥𝑥. 𝑑𝑑𝑑𝑑 + 3 � 𝑠𝑠𝑠𝑠𝑠𝑠 3 𝑥𝑥. 𝑑𝑑𝑑𝑑

1 1 ∴ 4 � sec 5 𝑥𝑥 . 𝑑𝑑𝑑𝑑 = sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 3[ 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝑙𝑙𝑙𝑙|𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 |] 2 2

1 1 1 ∴ � tan2 𝑥𝑥. sec 3 𝑥𝑥. 𝑑𝑑𝑑𝑑 = sec 3 x. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 − 𝑙𝑙𝑙𝑙|𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 | + 𝑐𝑐 4 8 8

Example

� 𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑥𝑥. 𝑐𝑐𝑐𝑐𝑐𝑐 5 𝑥𝑥. 𝑑𝑑𝑑𝑑 = � 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥. 𝑐𝑐𝑐𝑐𝑐𝑐 4 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑

� 𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑥𝑥. 𝑐𝑐𝑐𝑐𝑐𝑐 5 𝑥𝑥. 𝑑𝑑𝑑𝑑 = �(𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑥𝑥 − 1). 𝑐𝑐𝑐𝑐𝑐𝑐 4 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑 Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

85


� 𝑐𝑐𝑐𝑐𝑐𝑐 3 𝑥𝑥. 𝑐𝑐𝑐𝑐𝑐𝑐 5 𝑥𝑥. 𝑑𝑑𝑑𝑑 = � 𝑐𝑐𝑐𝑐𝑐𝑐 6 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑 − � 𝑐𝑐𝑐𝑐𝑐𝑐 4 𝑥𝑥 . 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐. 𝑑𝑑𝑑𝑑 1

H.W

� 𝑡𝑡𝑡𝑡𝑡𝑡3 2𝑥𝑥. 𝑠𝑠𝑠𝑠𝑠𝑠 2 2𝑥𝑥. 𝑑𝑑𝑑𝑑

1

= 𝑐𝑐𝑐𝑐𝑐𝑐 7 𝑥𝑥 + 𝑐𝑐𝑐𝑐𝑐𝑐 5 𝑥𝑥 + 𝑐𝑐 7

5

5- Trigonometric Substitution This method enables us to replace the binomials: 𝑎𝑎

1- 𝑎𝑎2 + 𝑏𝑏 2 𝑢𝑢2

𝑏𝑏𝑏𝑏 𝑢𝑢 = 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

3- 𝑏𝑏 2 𝑢𝑢2 − 𝑎𝑎2

𝑏𝑏𝑏𝑏 𝑢𝑢 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

2- 𝑎𝑎2 − 𝑏𝑏 2 𝑢𝑢2

𝑏𝑏

𝑎𝑎

𝑏𝑏𝑏𝑏 𝑢𝑢 = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑏𝑏

𝑎𝑎 𝑏𝑏

1- If the binomials in the form of 𝑎𝑎 2 + 𝑏𝑏 2 𝑢𝑢 2

Example

𝑑𝑑𝑑𝑑 (4 + 𝑥𝑥 2 )2 Let x = 2tanθ so dx = 2sec2θ.dθ 𝑑𝑑𝑑𝑑 2sec 2 θ. dθ ∴� =� (4 + 4𝑡𝑡𝑡𝑡𝑡𝑡2 𝜃𝜃)2 (4 + 𝑥𝑥 2 )2 2sec 2 θ. dθ =� (4 + 4𝑡𝑡𝑡𝑡𝑡𝑡2 𝜃𝜃)2 2sec 2 θ. dθ =� 16(1 + 𝑡𝑡𝑡𝑡𝑡𝑡2 𝜃𝜃)2

�

dθ 1 = � 𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑑𝑑𝑑𝑑 = � 8 𝑠𝑠𝑠𝑠𝑠𝑠 2 𝜃𝜃

1 1 + 𝑐𝑐𝑐𝑐𝑐𝑐2𝜃𝜃 𝑑𝑑𝑑𝑑 = � 8 2 1 1 = 𝜃𝜃 + 𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃 + 𝑐𝑐 16 32

Since x= 2tanθ tanθ=x/2

→ θ=tan-1(x/2)

𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃 = 2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐


86


𝑥𝑥

=2

2

.

√4 + 𝑥𝑥 2 √4 + 𝑥𝑥 2 1 𝑥𝑥 1 𝑥𝑥 𝑑𝑑𝑑𝑑 −1 ∴� = 𝑡𝑡𝑡𝑡𝑡𝑡 + + 𝑐𝑐 (4 + 𝑥𝑥 2 )2 16 2 8 4 + 𝑥𝑥 2

2- If the binomial in the form of 𝑎𝑎2 − 𝑏𝑏 2 𝑢𝑢2

Example 𝑢𝑢2 � . 𝑑𝑑𝑑𝑑 √16 − 𝑢𝑢2

Let u = 4sinθ

du = 4cosθ.dθ

�

𝑢𝑢2

√16 −

=�

𝑢𝑢2

. 𝑑𝑑𝑑𝑑 = �

16𝑠𝑠𝑠𝑠𝑠𝑠2 𝜃𝜃. 4𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 √16 −

16𝑠𝑠𝑠𝑠𝑠𝑠2 𝜃𝜃

. 𝑑𝑑𝑑𝑑 = � 𝑠𝑠𝑠𝑠𝑠𝑠2 𝜃𝜃. 𝑑𝑑𝑑𝑑

1 1 − 𝑐𝑐𝑐𝑐𝑐𝑐2𝜃𝜃 . 𝑑𝑑𝑑𝑑 = 8 �𝜃𝜃 − 𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃� + 𝑐𝑐 2 2

Since u = 4sinθ → θ=sin-1(u/4)

𝑠𝑠𝑠𝑠𝑠𝑠2𝜃𝜃 = 2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ∴�

1 = 𝑢𝑢�16 − 𝑢𝑢2 8 𝑢𝑢2

√16 − 𝑢𝑢2

. 𝑑𝑑𝑑𝑑 = 8 �𝑠𝑠𝑠𝑠𝑠𝑠−1

𝑢𝑢 1 − 𝑢𝑢�16 − 𝑢𝑢2 � + 𝑐𝑐 4 16

3- If the binomial in the form of 𝑏𝑏 2 𝑢𝑢2 − 𝑎𝑎2

Example 𝑑𝑑𝑑𝑑 � √𝑥𝑥 2 − 25

Let x = 5secθ dx = 5secθ.tanθ.dθ

�

𝑑𝑑𝑑𝑑

√𝑥𝑥 2 − 25

=�

5𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 𝑑𝑑𝑑𝑑

√25𝑠𝑠𝑠𝑠𝑠𝑠 2 𝜃𝜃 − 25


87


=�

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡. 𝑑𝑑𝑑𝑑 |𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 |

= ∓ � 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑑𝑑𝑑𝑑 = ∓𝑙𝑙𝑙𝑙|𝑠𝑠𝑠𝑠𝑐𝑐𝑐𝑐 + 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 | + 𝑐𝑐

Since secθ =x/5 → θ = sec-1(x/5) √𝑥𝑥 2 − 25 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = 5

𝑥𝑥 √𝑥𝑥 2 − 25 = 𝑙𝑙𝑙𝑙 � + � + 𝑐𝑐 ∴� 5 5 √𝑥𝑥 2 − 25 𝑑𝑑𝑑𝑑

4- Integration by Part Fraction

a- If the denominator has two linear factor Example 5𝑥𝑥 + 3 � 𝑑𝑑𝑑𝑑 (𝑥𝑥 + 1)(𝑥𝑥 − 3)

𝐴𝐴 𝐵𝐵 5𝑥𝑥 − 3 = + (𝑥𝑥 + 1)(𝑥𝑥 − 3) 𝑥𝑥 − 1 𝑥𝑥 − 3 5𝑥𝑥 − 3 = 𝐴𝐴𝐴𝐴 − 3𝐴𝐴 + 𝐵𝐵𝐵𝐵 + 𝐵𝐵 Hence, A=2 and B=3

�

�

5𝑥𝑥 + 3 2 3 𝑑𝑑𝑑𝑑 = � 𝑑𝑑𝑑𝑑 + � 𝑑𝑑𝑑𝑑 (𝑥𝑥 + 1)(𝑥𝑥 − 3) 𝑥𝑥 + 1 𝑥𝑥 − 3

5𝑥𝑥 + 3 𝑑𝑑𝑑𝑑 = 2𝑙𝑙𝑙𝑙|𝑥𝑥 + 1| + 3𝑙𝑙𝑙𝑙|𝑥𝑥 − 3| + 𝑐𝑐 (𝑥𝑥 + 1)(𝑥𝑥 − 3)

b- If the denominator has a repeated linear factor Example 6𝑥𝑥 + 7 � 𝑑𝑑𝑑𝑑 (𝑥𝑥 + 2)2

6𝑥𝑥 + 7 𝐴𝐴 𝐵𝐵 = + (𝑥𝑥 + 2)2 𝑥𝑥 + 2 (𝑥𝑥 + 2)2 6𝑥𝑥 + 7 = 𝐴𝐴𝐴𝐴 + (2𝐴𝐴 + 𝐵𝐵)

Hence, A=6 and B=-5


88


�

�

6𝑥𝑥 + 7 6 −5𝐵𝐵 � � 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑑𝑑 (𝑥𝑥 + 2)2 𝑥𝑥 + 2 (𝑥𝑥 + 2)2 5 6𝑥𝑥 + 7 | | 𝑑𝑑𝑑𝑑 = 6𝑙𝑙𝑙𝑙 + 𝑐𝑐 𝑥𝑥 + 2 + 𝑥𝑥 + 2 (𝑥𝑥 + 2)2

c- If the denominator has a quadratic factor Example −2𝑥𝑥 + 4 � 2 𝑑𝑑𝑑𝑑 (𝑥𝑥 + 1)(𝑥𝑥 − 1)2

−2𝑥𝑥 + 4 𝐴𝐴𝐴𝐴 + 𝐵𝐵 𝐶𝐶 𝐷𝐷 = + + (𝑥𝑥 2 + 1)(𝑥𝑥 − 1)2 (𝑥𝑥 2 + 1) 𝑥𝑥 − 1 (𝑥𝑥 − 1)2

−2𝑥𝑥 + 4 = (𝐴𝐴𝐴𝐴 + 𝐵𝐵)(𝑥𝑥 − 1)2 + 𝑐𝑐 (𝑥𝑥 2 + 1)(𝑥𝑥 − 1) + 𝐷𝐷 (𝑥𝑥 2 + 1) 𝐶𝐶 = −2, D=1

𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻, 𝐴𝐴 = 2, 𝐵𝐵 = 1, �

�

−2𝑥𝑥 + 4 2𝑥𝑥 + 1 −2 1 � � � 𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝑑𝑑 (𝑥𝑥 2 + 1)(𝑥𝑥 − 1)2 (𝑥𝑥 2 + 1) (𝑥𝑥 − 1)2 𝑥𝑥 − 1

1 −2𝑥𝑥 + 4 2 −1 | | | | 𝑑𝑑𝑑𝑑 = 𝑙𝑙𝑙𝑙 + 1 + 𝑡𝑡𝑡𝑡𝑡𝑡 𝑥𝑥 − 2𝑙𝑙𝑙𝑙 + 𝑐𝑐 𝑥𝑥 𝑥𝑥 − 1 − (𝑥𝑥 2 + 1)(𝑥𝑥 − 1)2 𝑥𝑥 − 1

d- If the numerator is higher than the degree of denominator Example 𝑥𝑥 5 − 𝑥𝑥 4 − 3𝑥𝑥 + 5 � 4 𝑑𝑑𝑑𝑑 𝑥𝑥 − 2𝑥𝑥 3 + 2𝑥𝑥 2 − 2𝑥𝑥 + 1

First we have to divide this function and this results:

𝑥𝑥 5 − 𝑥𝑥 4 − 3𝑥𝑥 + 5 � 4 𝑑𝑑𝑑𝑑 𝑥𝑥 − 2𝑥𝑥 3 + 2𝑥𝑥 2 − 2𝑥𝑥 + 1 = � �𝑥𝑥 + 1 +

Take :

= � �𝑥𝑥 + 1 +

−2𝑥𝑥 + 4 � 𝑑𝑑𝑑𝑑 𝑥𝑥 4 − 2𝑥𝑥 3 + 2𝑥𝑥 2 − 2𝑥𝑥 + 1

−2𝑥𝑥 + 4 � 𝑑𝑑𝑑𝑑 (𝑥𝑥 2 + 1)(𝑥𝑥 − 1)2

𝐴𝐴𝐴𝐴 + 𝐵𝐵 𝐶𝐶 𝐷𝐷 −2𝑥𝑥 + 4 = + + (𝑥𝑥 2 + 1)(𝑥𝑥 − 1)2 (𝑥𝑥 2 + 1) 𝑥𝑥 − 1 (𝑥𝑥 − 1)2


89


From the previous example we get: A=2, B=1, C=-1, D=1 𝑥𝑥 5 − 𝑥𝑥 4 − 3𝑥𝑥 + 5 ∴� 4 𝑑𝑑𝑑𝑑 𝑥𝑥 − 2𝑥𝑥 3 + 2𝑥𝑥 2 − 2𝑥𝑥 + 1 =

H.W Find: 1- ∫ 23-

𝑥𝑥 4 −𝑥𝑥 3 −𝑥𝑥−1

𝑥𝑥 2 1 + 𝑥𝑥 + 𝑙𝑙𝑙𝑙|𝑥𝑥 2 + 1| + 𝑡𝑡𝑡𝑡𝑡𝑡−1 𝑥𝑥 − 2𝑙𝑙𝑙𝑙|𝑥𝑥 − 1| − + 𝑐𝑐 𝑥𝑥 − 1 2

𝑑𝑑𝑑𝑑

𝑥𝑥 3 −𝑥𝑥 2 𝑥𝑥 2 ∫ (𝑥𝑥−1)(𝑥𝑥−2)(𝑥𝑥−3) 𝑑𝑑𝑑𝑑 𝑥𝑥 2 ∫ (𝑥𝑥−1)(𝑥𝑥 2 +2𝑥𝑥+1) 𝑑𝑑𝑑𝑑

2- Definite Integral

Properties of Definite Integral: 𝑏𝑏 𝑏𝑏 1- ∫𝑎𝑎 𝑘𝑘. 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 = 𝑘𝑘 ∫𝑎𝑎 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 𝑏𝑏

𝑏𝑏

𝑏𝑏

𝑘𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

2- ∫𝑎𝑎 �𝑓𝑓(𝑥𝑥) ∓ 𝑔𝑔(𝑥𝑥)�𝑑𝑑𝑑𝑑 = ∫𝑎𝑎 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 ∓ ∫𝑎𝑎 𝑔𝑔(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑏𝑏

𝑎𝑎

3- ∫𝑎𝑎 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 = − ∫𝑏𝑏 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 𝑏𝑏

4- 𝐼𝐼𝐼𝐼 𝑎𝑎 = 𝑏𝑏 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 ∫𝑎𝑎 𝑓𝑓(𝑥𝑥). 𝑑𝑑𝑑𝑑 =0 𝑏𝑏

𝑏𝑏𝑏𝑏

𝑐𝑐

5- ∫𝑎𝑎 𝑓𝑓(𝑥𝑥). 𝑑𝑑𝑑𝑑 + ∫𝑏𝑏 𝑓𝑓(𝑥𝑥). 𝑑𝑑𝑑𝑑 = ∫𝑎𝑎 𝑓𝑓(𝑥𝑥). 𝑑𝑑𝑑𝑑 Example

Find the area bounded by the x-axis and the parabola y = 6 - x - x2 Solution 0 = 6 – x –x2 So x = -3 and x= 2 2

2

Area = A = ∫−3 𝑓𝑓 (𝑥𝑥). 𝑑𝑑𝑑𝑑 = ∫−3(6 − 𝑥𝑥 − 𝑥𝑥 2 )𝑑𝑑𝑑𝑑

Area = 205/6


90


Example Find the area bounded by the x-axis and the parabola y = x2 -4x , x=3, x=1 Solution 3

2

Area = A = ∫1 𝑓𝑓(𝑥𝑥). 𝑑𝑑𝑑𝑑 = ∫−3(𝑥𝑥 2 − 4𝑥𝑥 )𝑑𝑑𝑑𝑑 A = -22/7

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = �−

Example

22 22 �= 3 3

Find the total area and net area bounded by the curve y = x3 -4 and the x-axis. Solution 0 = x3 - 4x So x=0 and x = ±2 0

𝐴𝐴1 = � (𝑥𝑥 3 − 4𝑥𝑥). 𝑑𝑑𝑑𝑑 = 4 −2 2

𝐴𝐴2 = � (𝑥𝑥 3 − 4𝑥𝑥). 𝑑𝑑𝑑𝑑 = − 4 0

Total area = 𝐴𝐴1 + 𝐴𝐴2 = 4 + |−4| = 8

Net area = 𝐴𝐴1 + 𝐴𝐴2 = 4 − 4 = 0

H.W

Find the area bounded by y = tanx from x=0 and x=π/4e

Application of Definite Integral. 1- Area between two curves If y 1 = f 1 (x) and y 2 = f 2 (x) are continuous for a≤x≤b and f 1 (x) ≥ f 2 (x) for same interval then the area between y 1 and y 2 𝑏𝑏

𝑏𝑏

= ∫𝑎𝑎 𝑓𝑓1 (𝑥𝑥). 𝑑𝑑𝑑𝑑 − ∫𝑎𝑎 𝑓𝑓2 (𝑥𝑥). 𝑑𝑑𝑑𝑑 𝑏𝑏

= � [𝑓𝑓1 (𝑥𝑥) − 𝑓𝑓2 (𝑥𝑥)]𝑑𝑑𝑑𝑑 𝑎𝑎


91


Example Find the area bounded by the parabola y = 6x – x2 and y = x2 -2x Solution 6x – x2 = x2 -2x So x = 0 → y = 0 And x = 4 → y = 8 4

∴ 𝐴𝐴 = � [𝑓𝑓1 (𝑥𝑥) − 𝑓𝑓2 (𝑥𝑥)]𝑑𝑑𝑑𝑑 4

0

= � [(6𝑥𝑥 − 𝑥𝑥 2 ) − (𝑥𝑥 2 − 2𝑥𝑥)]𝑑𝑑𝑑𝑑 0

𝐴𝐴 =

64 3

Example Find the area of the region enclosed by the parabola y = 2 – x2 and the curve y = x. Solution Let [𝑓𝑓1 (𝑥𝑥) = 𝑓𝑓2 (𝑥𝑥)] 2 – x2 = -x

Then x= -1 and x =2 2

∴ 𝐴𝐴 = � [𝑓𝑓1 (𝑥𝑥) − 𝑓𝑓2 (𝑥𝑥)]𝑑𝑑𝑑𝑑 2

−1

= � [2 − 𝑥𝑥 2 + 𝑥𝑥]𝑑𝑑𝑑𝑑 −1

𝐴𝐴 =

9 2

Example Find the area bounded on the right by the line y = x-2, on the left by the parabola x = y2 and from below by the x-axis. Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

92


Solution 1st method Integration with respect to y y2 = y+2 then y = 2 and y=-1 the point y=2 is the only point lie in region 2

∴ 𝐴𝐴 = � [𝑓𝑓1 (𝑦𝑦) − 𝑓𝑓2 (𝑦𝑦)]𝑑𝑑𝑑𝑑 4

0

= � [𝑦𝑦 + 2 − 𝑦𝑦 2 ]𝑑𝑑𝑑𝑑 0

𝐴𝐴 = nd

10 3

2 method Integration with respect to x x =y2 →y = ±√x = f 1 (x) y = x-2 → y = x-2 = f 2 (x) √x = x-2 Then x = 1 and x =4 The point x =1 does not satisfy the equation √x = x-2 For 0≤x≤2, [𝑓𝑓1 (𝑦𝑦) − 𝑓𝑓2 (𝑦𝑦)] = √𝑥𝑥 − 0 = √𝑥𝑥

For 2≤x≤4, [𝑓𝑓1 (𝑦𝑦) − 𝑓𝑓2 (𝑦𝑦)] = √𝑥𝑥 − 𝑥𝑥 + 2 2

4

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = � √𝑥𝑥𝑑𝑑𝑑𝑑 − � �√𝑥𝑥 − 𝑥𝑥 + 2�𝑑𝑑𝑑𝑑 𝐴𝐴 =

10 3

0

2

2- Lengths of Curves in the Plain If the function f has a continuous first derivative through the interval a≤x≤b, the length of the curve y = f(x) from a to b is:


93


𝑑𝑑𝑑𝑑 2 � . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑏𝑏

𝐿𝐿 = � �1 + � 𝑎𝑎

Example

1

𝑥𝑥

𝑥𝑥

Find the length of the curve 𝑦𝑦 = 𝑎𝑎(𝑒𝑒 𝑎𝑎 + 𝑒𝑒 −𝑎𝑎 ) from x =0 to x = a. 2

Solution


𝑏𝑏

𝐿𝐿 = � �1 + � 𝑦𝑦 =

𝑎𝑎

𝑥𝑥 𝑥𝑥 1 𝑎𝑎 �𝑒𝑒 𝑎𝑎 + 𝑒𝑒 −𝑎𝑎 � 2 𝑥𝑥 𝑎𝑎

1

𝑦𝑦� = �𝑒𝑒 + 𝑒𝑒 2

𝑥𝑥 𝑎𝑎

−

�

2

𝑏𝑏

𝑥𝑥 1 𝑥𝑥 𝐿𝐿 = � �1 + � �𝑒𝑒 𝑎𝑎 + 𝑒𝑒 −𝑎𝑎 �� . 𝑑𝑑𝑑𝑑 2 𝑎𝑎 1

1

𝐿𝐿 = 𝑎𝑎 �𝑒𝑒 + � 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑒𝑒

2

Example Find the length of the curve y = x2/3 between x=1 and x = 8 Solution 𝑏𝑏

𝐿𝐿 = � �1 + � 𝑎𝑎

𝑦𝑦 = 𝑥𝑥 2/3


2

𝑦𝑦� = 𝑥𝑥 −1/3 3

This give no answer at x =0 , so we will use x as f(y) 3

𝑥𝑥 = 𝑦𝑦 2

𝑑𝑑𝑑𝑑 3 = �𝑦𝑦 𝑑𝑑𝑑𝑑 2


94


At x =-1 then y = 1 At x = 8 then y =4 We have two intervals: 0≥y≥1 and 4≥y≥0 𝐿𝐿 = 𝐿𝐿1 + 𝐿𝐿2

4 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 2 � � = � 1 + � � . 𝑑𝑑𝑑𝑑 + � 1 + � � . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 0 0 1

1

4 9 9 = � �1 + 𝑦𝑦. 𝑑𝑑𝑑𝑑 + � �1 + 𝑦𝑦 . 𝑑𝑑𝑑𝑑 4 4 0 0

=105 unit

3- Area of Surface of Revolution If the function f has a continuous first derivative through the interval a≤x≤b, the area of surface generated by revolving the curve will be: 𝑏𝑏

𝑑𝑑𝑑𝑑

(revolution about x-axis)

𝑏𝑏

𝑑𝑑𝑑𝑑

(revolution about y-axis)

𝑆𝑆 = ∫𝑎𝑎 2𝜋𝜋𝜋𝜋�1 + ( )2 . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

𝑆𝑆 = ∫𝑎𝑎 2𝜋𝜋𝜋𝜋 �1 + ( )2 . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

Example Find the surface area of revolution generated by revolving about x-axis the area of the parabola y2 = 12x from x=0 to x=3. Solution 𝑑𝑑𝑑𝑑 2 𝑆𝑆 = � 2𝜋𝜋𝜋𝜋�1 + � � . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑏𝑏

y2 = 12x 𝑑𝑑𝑑𝑑 6 = 𝑑𝑑𝑑𝑑 𝑦𝑦


95

University of Technology____________ Mathematics_____________Chemical Engineering Department 3 1 6 2 � 𝑆𝑆 = � 2𝜋𝜋𝜋𝜋 1 + � � . 𝑑𝑑𝑑𝑑 = � (12𝑥𝑥 + 36)2 . 𝑑𝑑𝑑𝑑 𝑦𝑦 0 0 3

S = 43.9π square unit

Example Find the area of the surface of revolution generated by revolving about y-axis of x = y3 from y=0 to y =1. Solution 𝑑𝑑𝑑𝑑 2 𝑆𝑆 = � 2𝜋𝜋𝜋𝜋 �1 + � � . 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑎𝑎 𝑏𝑏

x =y3

𝑑𝑑𝑑𝑑 = 3𝑦𝑦 2 𝑑𝑑𝑑𝑑 𝑏𝑏

𝑆𝑆 = � 2𝜋𝜋𝑦𝑦 3 �1 + (3𝑦𝑦 2 )2 . 𝑑𝑑𝑑𝑑 𝑎𝑎

S = 1.13 π square unit

4- Volumes of Solids a- Disk Method The volume of the solid generated by revolving the region between the graph of a continuous function and axes are: 𝑏𝑏

if the revolution about x-axis

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = ∫𝑎𝑎 𝜋𝜋(𝑓𝑓(𝑦𝑦))2 𝑑𝑑𝑑𝑑

if the revolution about y-axis

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = ∫𝑎𝑎 𝜋𝜋(𝑓𝑓(𝑥𝑥))2 𝑑𝑑𝑑𝑑 𝑏𝑏


96


Example Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola y2=x and its latus rectum x=4 about the x-axis. Solution 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑓𝑓(𝑥𝑥))2 𝑑𝑑𝑑𝑑 𝑎𝑎

2

2

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋�√𝑥𝑥� 𝑑𝑑𝑑𝑑 0

= 8π cubic unit

Example The region enclosed by the semicircle 𝑦𝑦 = √𝑎𝑎2 + 𝑥𝑥 2 and the x-axis is revolved about the x-axis to generate a

sphere. Find the volume of the sphere. Solution 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑓𝑓(𝑥𝑥))2 𝑑𝑑𝑑𝑑 𝑎𝑎

𝑎𝑎

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(�𝑎𝑎2 + 𝑥𝑥 2 )2 𝑑𝑑𝑑𝑑 −𝑎𝑎

4

= 𝜋𝜋 𝑎𝑎2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 3

Example Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola 𝑦𝑦 = √𝑥𝑥 Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

97


and the lines y=1 and x=4 about the line y=1. Solution 4

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(√𝑥𝑥 − 1)2 𝑑𝑑𝑑𝑑 =

1

7 𝜋𝜋 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 6

Example The region between the curve 𝑥𝑥 =

1

√𝑦𝑦

1≤y≤4, is revolved about the y-axis to

generated a solid. Find the volume of the solid.

Solution 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑓𝑓(𝑦𝑦))2 𝑑𝑑𝑑𝑑 𝑎𝑎

4

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑒𝑒 = � 𝜋𝜋( 1

1

�𝑦𝑦

)2 𝑑𝑑𝑑𝑑

= 2𝜋𝜋𝜋𝜋𝜋𝜋2 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢

Example

Find the volume of the solid generated by revolving the first quadrant area bounded by the parabola 𝑦𝑦 2 = 8𝑥𝑥 and its latus rectum x=2 about the latus rectum.

Solution 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = ∫𝑎𝑎 𝜋𝜋(𝑓𝑓(𝑦𝑦))2 𝑑𝑑𝑑𝑑 4

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(2 − 𝑥𝑥)2 𝑑𝑑𝑑𝑑 −4 𝑏𝑏

𝑦𝑦 2 2 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(2 − ) 𝑑𝑑𝑑𝑑 8 𝑎𝑎 =



98


b- Washer Method Washer method for calculating volumes 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑅𝑅2 (𝑥𝑥) − 𝑟𝑟 2 (𝑥𝑥))𝑑𝑑𝑑𝑑 𝑎𝑎

R(x) = outer radius,

r(x) = inner radius

Example The region bounded by the curve y = x2 +1 and the line y = -x+3 is revolved about the x-axis to generated a solid. Find the volume of the solid. Solution x2 + 1 = -x+3 so x =-2 and x = 1 outer radius : R(x) = -x +3 inner radius : r(x) = x2 +1 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑅𝑅2 (𝑥𝑥) − 𝑟𝑟 2 (𝑥𝑥))𝑑𝑑𝑑𝑑 𝑎𝑎

1

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋((−𝑥𝑥 + 3)2 − (𝑥𝑥 2 + 1)2 )𝑑𝑑𝑑𝑑 =

−2


Example The region bounded by the curve y = x2 and the line y = 2x in the first quadrant is revolved about the y-axis to generated a solid. Find the volume of the solid. Solution x2 = 2x Chapter seven___________________ Integration ___________________Dr. Asawer A. Alwasiti

99


so x =0 → y=o and x = 2 → y = 4 outer radius : R(y) = √y inner radius : r(y) = y/2 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑅𝑅2 (𝑦𝑦) − 𝑟𝑟 2 (𝑦𝑦))𝑑𝑑𝑑𝑑 𝑎𝑎

4

𝑦𝑦 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋((�𝑦𝑦)2 − ( )2 )𝑑𝑑𝑑𝑑 2 0 =


Example The region bounded by the curve y = x2 and the line y = 2x in the first quadrant is revolved about the line x =2 parallel to the y-axis. Find the volume swept out. Solution x2 = 2x so x =0 → y=o and x = 1 → y = 4 outer radius : R(y) = 2 – y/2 inner radius : r(y) = 2 - √y 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋(𝑅𝑅2 (𝑦𝑦) − 𝑟𝑟 2 (𝑦𝑦))𝑑𝑑𝑑𝑑 𝑎𝑎

4

𝑦𝑦 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 𝜋𝜋((2 − )2 − (2 − �𝑦𝑦)2 )𝑑𝑑𝑑𝑑 2 0 =



100


c- Cylindrical Shells – Alternative to Washer The Shell Method (Axis the y-axis) Suppose y = f(x) is continuous throughout an interval a ≤ x ≤ b that does not cross the y-axis. Then, the volume of the solid generated by revolving the region between the graph of f and the interval a ≤ x ≤ b about the y-axis is found by integrating 2πxf(x) with respect to x from a to b. 𝑏𝑏

𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ℎ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡). 𝑑𝑑𝑑𝑑 = � 2𝜋𝜋𝜋𝜋𝜋𝜋 (𝑥𝑥). 𝑑𝑑𝑑𝑑 Example

𝑎𝑎

𝑎𝑎

The region bounded by the curve y =√x, the x-axis and the line x = 4 is revolved about the y-axis to generated a solid. Find the volume of the solid. Solution 𝑏𝑏

𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = ∫𝑎𝑎 2𝜋𝜋(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ℎ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡). 𝑑𝑑𝑑𝑑 = ∫𝑎𝑎 2𝜋𝜋𝜋𝜋𝜋𝜋 (𝑥𝑥). 𝑑𝑑𝑑𝑑

4

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋𝜋𝜋 √𝑥𝑥. 𝑑𝑑𝑑𝑑 =

0



101


Example The region bounded by the curve y =√x, the x-axis and the line x = 4 is revolved about the x-axis to generated a solid. Find the volume of the solid. Solution y = 0 to y =2 cylinder radius = y cylinder height = 4 – y2

𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ℎ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡). 𝑑𝑑𝑑𝑑 = 𝑎𝑎

2

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋𝜋𝜋(4 − 𝑦𝑦 2 ). 𝑑𝑑𝑑𝑑 0

= 8𝜋𝜋 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢

Example The region bounded by the curve y =x2, the y-axis and the line y = 1 is revolved about the line x =2 to generated a solid. Find the volume of the solid. Solution 𝑏𝑏

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)(𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒 ℎ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡). 𝑑𝑑𝑑𝑑 = 𝑎𝑎

1

𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 = � 2𝜋𝜋(2 − 𝑥𝑥)(1 − 𝑥𝑥 2 ). 𝑑𝑑𝑑𝑑 =

0



102


Chapter Eight Determinates A rectangular array of numbers like: 2 1

𝐴𝐴 = �

1 0

3 � −2

Is called a matrix of order 2 by 3 because it has 2 rows and 3 columns. For matrix A: a 12 = 1 a 11 =2 , a 21 =1 , a 22 = 0

, ,

a 13 = 3 a 23 = -2

Note: The first subscript denotes the numbers of the rows and the second subscript the number of columns. A matrix with the same number of rows and columns is a square matrix. A determinate: is a scalar-valued function whose domain is a set of square matrix. For matrix of 2 order 𝑎𝑎11 𝐴𝐴 = �𝑎𝑎

21

For matrix of 2 order 𝑎𝑎11 𝑎𝑎12 𝐴𝐴 = �𝑎𝑎 𝑎𝑎22 � 21 𝑎𝑎11 det(A)=|A|=�𝑎𝑎 21

𝑎𝑎12 𝑎𝑎22 � = a 11 a 22 - a 21 a 12

𝑎𝑎11 det(A)=𝑑𝑑𝑑𝑑𝑑𝑑 �𝑎𝑎21 𝑎𝑎31

𝑎𝑎12 𝑎𝑎22 𝑎𝑎32

For matrix of 3 order 𝑎𝑎11 𝑎𝑎12 𝑎𝑎13 𝐴𝐴 = �𝑎𝑎21 𝑎𝑎22 𝑎𝑎22 � 𝑎𝑎31 𝑎𝑎32 𝑎𝑎33

𝑎𝑎22 |A|=𝑎𝑎11 �𝑎𝑎 32

𝑎𝑎13 𝑎𝑎22 �= 𝑎𝑎33

𝑎𝑎23 𝑎𝑎21 � − 𝑎𝑎 � 12 𝑎𝑎33 𝑎𝑎31

𝑎𝑎12 𝑎𝑎22 �

𝑎𝑎23 𝑎𝑎21 � + 𝑎𝑎 � 13 𝑎𝑎33 𝑎𝑎31

𝑎𝑎22 𝑎𝑎32 �

Chapter eight___________________ Determinates _________________Dr. Asawer A. Alwasiti

103


= a 11 A 11 – a 12 A 12 + a 13 A 13 in which A 11 , A 12 , A 13 are the minors of a 11 , a 12 , a 13 respectively. |A|=a 11 (a 22 a 33 – a 23 a 3 ) – a 12 (a 21 a 33 - a 23 a 31 ) + a 13 (a 21 a 32 – a 22 a 33 ) Example Evaluate the determinate of the matrix 2 𝐴𝐴 = �2 4

3 4 −1 3� −5 6

det(A) = |A|=2 � =6

−1 −5

3 2 � − 3� 6 4

3 2 � + 4� 6 4

−1 � −5

H.W 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 Evaluate � 0 −𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

Properties of Determinates

𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1 � 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

ans. = -1

1- The value of the determinate is unchanged if all the corresponding rows and columns are interchanged. 𝑎𝑎11 𝑎𝑎12 𝑎𝑎11 𝑎𝑎21 i.e. �𝑎𝑎 � = � 𝑎𝑎22 𝑎𝑎12 𝑎𝑎22 � 21 Example 1 𝐴𝐴 = �2 3

2 4 1

1 3� 2

Transport matrix 1 2 3 𝐴𝐴 = �2 4 1� 1 3 2 1 2 1 4 𝑑𝑑𝑑𝑑𝑑𝑑 �2 4 3� = 1 � 1 3 1 2

2 3 � − 2� 3 2

3 2 � + 1� 2 3

4 �=5 1


104


1 𝑑𝑑𝑑𝑑𝑑𝑑 �2 1

2 4 3

3 4 1� = 1 � 3 2

1 2 � − 2� 2 1

2 1 � + 3� 1 2

4 �=5 3

2- If all the elements of a row or a column of a determinate are multiplied by the same number k, then the value of the determinate is multiplied by k.

Example 3 �4 −1

9 5 3 6 0� = 3 � 4 −3 0 −1

3 2 −1

5 3 0� = 3 ∗ 2 � 2 0 −1

3 1 −1

5 0� 0

3- If any two rows or any two columns of a determinate are interchanged then the sign of the value of the determinate is changed. Example 2 𝐴𝐴 = �3 2

1 3 −1 −2� 𝑖𝑖𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑖𝑖𝑖𝑖 36 3 1 1 2 3 Prove that det(B) = -36 if 𝐵𝐵 = �−1 3 −2� 3 2 1 −1 −2 −1 3 3 −2 det(B)=1 � � − 2� � + 3� � = −36 3 1 3 2 2 1

4- The determinate is zero if

a- Two rows or columns of a matrix is identical Example Find the determinate of the matrix 2 𝐴𝐴 = �3 1

2 3 1

1 4� 3 3 4 3 det(A)=|A|=2� � − 2� 1 3 1

4 3 � + 1� 3 1

3 �=0 1


105


b- Every element in any row or any column is zero Example 1 𝐴𝐴 = �0 5

2 0 7

3 0� 2

0 0 0

3 2� 5

det(A)= |A| =1 � Example 1 𝐴𝐴 = �6 3

det(A)= |A| =1 �

Cramers Rule

0 7

0 0 � − 2� 2 5

0 0 � + 3� 2 5

0 �=0 7

0 0

2 6 � − 0� 5 3

2 6 � + 3� 5 3

0 �=0 0

a 11 x + a 12 y =b 1 a 21 x + a 22 y =b 2 If D ≠ 0 So D x =x =

𝑏𝑏1 𝑏𝑏2

𝑎𝑎 12 � 𝑎𝑎 22 𝐷𝐷 𝑎𝑎 11 𝑏𝑏1 � � 𝑎𝑎 21 𝑏𝑏2 𝐷𝐷 �

D y =y = Example

Solve the system 3x – y =9 x + 2y = -4 solution D=�

3 1

x=

𝐷𝐷𝑥𝑥

y=

𝐷𝐷𝑦𝑦

𝐷𝐷

𝐷𝐷

=

=

−1 �=7 2 �

9 −4

�

3 1

7

−1 � 2

9 � −4 7

=2

= −3


106


Systems of three equations in three unknowns work, it can be solved in the same way: a 11 x + a 12 y + a 13 z = b 1 a 21 x + a 22 y + a 23 z = b 2 a 31 x + a 32 y + a 33 z = b 3 𝑏𝑏 𝐷𝐷𝑥𝑥 1 1 ∴ = 𝑥𝑥 = �𝑏𝑏2 𝐷𝐷 𝐷𝐷 𝑏𝑏3

𝑎𝑎 𝐷𝐷𝑦𝑦 1 11 ∴ = 𝑦𝑦 = �𝑎𝑎21 𝐷𝐷 𝐷𝐷 𝑎𝑎31

Example

𝑎𝑎 𝐷𝐷𝑥𝑥 1 11 ∴ = 𝑥𝑥 = �𝑎𝑎21 𝐷𝐷 𝐷𝐷 𝑎𝑎31

Solve the equations by Cramers rule


𝑏𝑏1 𝑏𝑏2 𝑏𝑏3


𝑎𝑎13 𝑎𝑎23 � 𝑎𝑎33

𝑎𝑎13 𝑎𝑎23 � 𝑎𝑎33 𝑏𝑏1 𝑏𝑏2 � 𝑏𝑏3

2x 1 + 4x 2 +2x 3 = 16 2x 1 - x 2 -2x 3 = -6 4x 1 + x 2 -2x 3 = 0 Solution 2 4 2 𝐷𝐷 = �2 −1 −2� = 4 4 1 −2 16 4 2 1 x 1 = �−6 −1 −2� = 1 4 0 1 −2 2 16 2 1 x 2 = �2 −6 −2� = 2 4 4 0 −2 2 4 16 1 x 3 = �2 −1 −6� = 3 4 4 1 0 Chapter eight___________________ Determinates _________________Dr. Asawer A. Alwasiti

107


H.W 1 – Solve 5x 1 + 4x 3 +2x 4 = 3 x 1 - x 2 +2x 3 +x 4 = 1 4x 1 + x 2 +2x 3 = 1 x 1 + x 2 +x 3 + x 4 = 0 2- Solve 3y +2x = z +1 3x +2z = 8 – 5y 3z - 1 = x – 2y

ans. x 1 =1, x 2 =-1, x 3 =-1, x 4 =1

ans. x=3, y=-1, z=2


108


Chapter Nine Polar Coordinate Definition: p(r , θ)

r: the direct distance from 0 to p. θ: the direct angle from the initial ray to the segment op.

Chapter nine___________________ Polar Coordinate_________________Dr. Asawer A. Alwasiti

109


Cartesian Versus Polar Coordinates:

Equations relating polar and Cartesian coordinates: x = r cos θ

,

y = r sin θ

so x2 + y2 = r2 and

𝑦𝑦 𝑥𝑥

=

Example

𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 θ

𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 θ

= 𝑡𝑡𝑡𝑡𝑡𝑡θ

find a Cartesian equation for the curve r2cos θ.sin θ=4 Solution (rcos θ)(rsin θ) =4 xy =4 Example Find the Cartesian equation for r = 1+2r cos θ Solution �𝑥𝑥 2 + 𝑦𝑦 2 = 1 + 2𝑥𝑥

𝑥𝑥 2 + 𝑦𝑦 2 = 1 + 4𝑥𝑥 + 4𝑥𝑥 2

𝑦𝑦 2 − 3𝑥𝑥 2 − 4𝑥𝑥 − 1 = 0 Example

Find a Cartesian equation for r =1-cos θ Solution r =1-cos θ r2 = r - rcos θ 𝑥𝑥 2 + 𝑦𝑦 2 = �𝑥𝑥 2 + 𝑦𝑦 2 − 𝑥𝑥

[(𝑥𝑥 2 + 𝑦𝑦 2 ) + 𝑥𝑥]2 = 𝑥𝑥 2 + 𝑦𝑦 2 = 0 Example

Find the Cartesian equation for r cos( θ-π/3)=3 Solution 𝑟𝑟 �𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

𝜋𝜋 𝜋𝜋 + 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � = 3 3 3


110


1 √3 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟. + 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟. =3 2 2 𝑥𝑥 + √3𝑦𝑦 = 6

Example

Replace the polar equations by equivalent Cartesian equations and identify their graphs. a- r2 = 4r cosθ b- 𝑟𝑟 =

Solution

4

2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 −𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

a- r2 = 4r cosθ x2 + y2=4x x2 - 4x + y2 =0 (x – 2)2 + y2 =4 The graph is circle with radius 2 and center (2, 0) b- 𝑟𝑟 ==

4

2𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 −𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

2𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 − 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 4 y = 2x – 4

The graph is line with slope 2

Graphing in Polar Coordinates The graph of the equation f(r ,θ) consists of the points whose polar coordinates in some form satisfy the equation Symmetry tests for Graph: 1- Symmetry about the x-axis : if the point (r,θ) lies on the graph , the point (r,θ) lies on the graph (fig.1) 2- Symmetry about the y-axis : if the point (r,θ) lies on the graph the point (r,πθ) or (-r, -θ)lies on the graph (fig.2) 3- Symmetry about the origin : if the point (r,θ) lies on the graph, the point (r,θ) or (r, θ+π) lies on the graph (fig.3) Chapter nine___________________ Polar Coordinate_________________Dr. Asawer A. Alwasiti

111


Note: If a curve has any two of the symmetry listed above, it also has the third.

fig.(1)

fig.(2)

fig.(3)

Example Graph the curve r = 1-cosθ Solution The curve is symmetric about the x-axis because : (r,θ) on graph

r = 1-cosθ r = 1- cos (-θ) = 1-cosθ (r, -θ) on the graph

The cosine has period 2π The curve is called a cardioids because of its heart shape. θ r =1-cosθ

0 0

π/3 1/2

π/2 1

2π/3 3/2

π 2

4π/3 3/2

3π/2 1

5π/3 1/2

2π 0


112


Example Graph the curve r=1-sinθ Solution The curve is symmetric about y-axis θ r =1sinθ

0 1

π/6 0.5

π/4 0.3

π/3 π/2 0.13 0

2π/3 0.13

5π/6 0.5

π 1

4π/3 1.86

3π/2 5π/3 2π 2 1.86 1

Example Graph the curve r=1+ cosθ/2 Solution Since the graph has period 2π, we must let θ run from 0 to 4π to produce the entire graph. The graph is symmetric about x-axis. θ 0 π/3 π/2 2π/3 π 4π/3 3π/2 5π/4 2π r 2 1.86 1.7 1.5 1 0.5 0.3 0.13 1 =1+cosθ/2

5π/2 3π 0.3 1

7π/2 4π/2 1.7 2

H.W Graph the functions a- r = -1+cosθ b- r = 3/2 - sinθ


113


Chapter Ten Complex Number Complex Number has a form : (a + ib) Where a is a real number ib is an imaginary unit Example Write the expression √-9-5 in the form of a + ib Solution real part is -5 imaginary part is √-9 =-5+ 3i

Properties of Complex Number 1- Addition and Substraction Example (2+3i) + (4 -5i) = 6 -2i Example -3i – (2+3i) = -2 -6i 2- Multiplication Example (5 -2i) ( 3 +4i) = 23 + 14i Example (2 +3i) (1+i) = -1 + 5i 3- Equality a + ib = c + id

if a = c and b = d

Example 2 +3i = x +yi Chapter ten___________________ Complex Number_________________Dr. Asawer A. Alwasiti

114


Is equal if x =2 and y=3 Example Solve x +(x+y)i = 2+5i x = 2 and y = 3 4- Division To reduce any rational combination of a complex number to a single complex number we usually multiply by conjugate.

Example Solve

Example Simplify

Hint

H.W Find x and y if

Graphing Representation of Complex Number U

The complex number x+iy can be represented graphically by using the x-axis as the real axis and y-axis as imaginary axis. The origin then is (0+0i) and the number (x+0i) is on the real axis and the number (0+yi) as the imaginary part. Chapter ten___________________ Complex Number_________________Dr. Asawer A. Alwasiti

115


There are two geometric representation of the complex number z = x+iy 1-

As point p(x,y) in x-y plane As the vector op→ from

2the origin to p.

The two above representations are called (Arg and Diagram) The distance from the origin to point p is called (absolute value or modulus) of the complex number and denoted by (r). r2 = x 2 + y 2 If z = x +iy Then |z| = The angle θ of initial side of positive x-axis and terminal side of op→ is called Argument of the complex number. θ = argz = arg (x+yi) = Example find the absolute value and the argument of z = 1Solution |z| =

=2

θ = argz =

= tan-1(-√3) = 300

Chapter ten___________________ Complex Number_________________Dr. Asawer A. Alwasiti

116


Polar Form of Complex Number

y =r cosθ

,

x = r sinθ

z= r (cosθ + isinθ) For exponential form or Euler form

Example Express 1 – i in polar form z = x + yi

x =1

and y = -1

|z| =

= √2

θ = argz =

= tan-1(-1) = 315

z = √2(cos315 + isin315) H.W Express (

in polar and Euler form

Multiplication and Division If z 1 = r 1 (cosθ 1 +isinθ 1 ) and z 2 = r 2 (cosθ 2 +isinθ 2 ) then: |z 1 .z 2 | = r 1 .r 2

,

arg(z 1 .z 2 ) = (θ 1 +θ 2 )

z 1. .z 2 = r 1 .r 2 [ cos(θ 1 +θ 2 ) + isin(θ 1 +θ 2 )] z 1 .z 2 = r 1 .r 2 ei(θ1+θ2)

polar form

Euler form , polar form Euler form


117


Example Let z 1 = 1 +i , z 2 = √3 – i find z 1 .z 2 , arg , Modulus Solution |z 1 | = r 1 = √2 θ 1 = tan-1y/x = 45 z 1 = √2eπ/4i as the same z 2 = 2e-π/6θ arg (z 1 .z 2 ) = π/12 modulus = |z 1 .z 2 | = r 1 .r 2 = 2√2 z 1 .z 2 = 2√2eπ/12i

Euler form

z 1 .z 2 = 2√2(cosπ/12 + isinπ/12)

Polar form

z 1 .z 2 = 2.73 + 0.73i H.W Find z 1 .z 2 , z 1 /z 2 in Euler and polar form if z 1 = 2+ 3i , z 2 = 1- 2i

Power of Complex Number (zn) z2 = z.z z3 = z.z.z zn = z.z.z……z

to nth of z when is positive integer

z = reiθ zn = rn einθ einθ = ( cosnθ +isinnθ) = ( cosθ +isinθ)n

DeMoivers theorem


118


Example Evaluate (-1 –i)6 Solution |z| = r =√2 θ = π/4 (-1 - i)6 = [√2 (cos π/4 + isinπ/4)]6 = -8i Example If n =3 by using DeMoivers theorem find the solution of cos3θ , sin3θ Solution ( cosθ +isinθ)3 = cos3θ + isin3θ Also ( cosθ +isinθ)3 = cos3θ + 3icos2θ.isinθ -3sin2θ.cosθ – isin3θ cos3θ + isin3θ = (cos3θ-3sin2θ.cosθ) + i(3cos2θ.sinθ – sin3θ) so cos3θ = cos3θ-3sin2θ.cosθ and sin3θ = 3cos2θ.sinθ – sin3θ

Roots of Complex Number Wn =z W = z1/n The polar form of z = r(cosθ +isinθ) can be write as z = r[cos(θ + 2πk) +isin(θ + 2πk)] in which k = 1, 2, 3,…….. z1/n = [r[cos(θ + 2πk) +isin(θ + 2πk)]]1/n by DeMoivers theorem Chapter ten___________________ Complex Number_________________Dr. Asawer A. Alwasiti

119


Also

Example Find the solution of x3 +1 =0 Solution x3 +1 =0

gives x = (-1)1/3 = (-1 +0i)1/3

|z| = r =1 θ =180 (-1 +0i) = 1 (cos180 + isin180)

The three roots of (-1) are: For k=0

= 1/2 +√3/2i

For k =1

= -1

For k = 2

= 1/2 - √3/2i

H.W Find the forth roots of (-16) Ans. k=0 √2(1+i) k=0 √2(-1+i) k=0 √2(-1-i) k=0 √2(1-i)


120