Math 461

Introduction to Probability

A.J. Hildebrand

Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f (x, R ∗ y) R ∗over a giving region Rin∗ the R ∗ xy-plane. This means writing the integral as an iterated integral of the form ∗ ∗ f (x, y)dxdy and/or ∗ ∗ f (x, y)dydx, with specific limits in place of the asterisks. To do this, follow the steps above (most importantly, sketch the given region). The remaining questions are evaluations of integrals over concrete functions. 1. Set up a double integral of f (x, y) over the region given by 0 < x < 1, x < y < x + 1. Solution: Z 1 Z x+1 f (x, y)dydx x=0

y=x

2. Set up a double integral of f (x, y) over the part of the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, on which y ≤ x/2. Solution: Z 1 Z x/2 Z 1/2 Z 1 f (x, y)dydx or f (x, y)dxdy x=0

y=0

y=0

x=2y

3. Set up a double integral of f (x, y) over the part of the unit square on which x + y > 0.5. Solution: Z 1/2 Z 1 Z 1 Z 1 f (x, y)dydx + f (x, y)dydx x=0

y=1/2−x

x=1/2

y=0

4. Set up a double integral of f (x, y) over the part of the unit square on which both x and y are greater than 0.5. Solution: Z 1 Z 1 f (x, y)dydx x=1/2

y=1/2

5. Set up a double integral of f (x, y) over the part of the unit square on which at least one of x and y is greater than 0.5. Solution: Z 1/2 Z 1 Z 1 Z 1 f (x, y)dydx + f (x, y)dydx x=0

y=1/2

x=1/2

y=0

6. Set up a double integral of f (x, y) over the part of the region given by 0 < x < 50 − y < 50 on which both x and y are greater than 20. Solution: Z 30 Z 50−x f (x, y)dydx x=20

y=20

7. Set up a double integral of f (x, y) over the set of all points (x, y) in the first quadrant with |x − y| ≤ 1. Solution: Z ∞ Z x+1 Z 1 Z x+1 f (x, y)dydx + f (x, y)dydx x=0

y=0

x=1

1

y=x−1

Math 461

Introduction to Probability

A.J. Hildebrand

RR 8. Evaluate R e−x−y dxdy, where R is the region in the first quadrant in which x + y ≤ 1. Solution: Z 1 Z 1−x Z 1 −x −y e e dydx = e−x (1 − e−(1−x) )dx x=0

y=0

0

Z =

1

(e−x − e−1 )dx = 1 − 2e−1 .

0

RR 9. Evaluate R e−x−2y dxdy, where R is the region in the first quadrant in which x ≤ y Solution: Z ∞ Z ∞ Z ∞ 1 1 −3x e dx = e−x−2y dydx = 2 6 x=0 y=x 0 RR 10. Evaluate R (x2 + y 2 )dxdy, where R is the region 0 ≤ x ≤ y ≤ L Solution: Z L Z y Z L 1 3 y 2 2 2 (x + y )dydx = x + y x dydx x=0 y=0 x=0 y=0 3 Z L 4 L 4 3 = y dy = . 3 0 3 RR 11. Evaluate R (x − y + 1)dxdy, where R is the region inside the unit square in which x + y ≥ 0.5. Solution: Z 0.5 Z 1 Z 1 Z 1 (x − y + 1)dydx + (x − y + 1)dydx x=0

y=0.5−x 0.5

x=0.5

y=0

Z 1 1 2 1 2 1 1 = xy − y + y dx + xy − y + y dx 2 2 y=0.5−x y=0 x=0 x=0.5 Z 0.5 1 1 1 1 = x(1 − + x) − (1 − ( − x)2 ) + (1 − + x) dx 2 2 2 2 0 Z 1 1 + x+ dx 2 0.5 Z 0.5 1 3 1 2 1 1 = + x + x2 dx + x + x 8 2 2 2 0.5 0 1 1 3 3 1 7 1 1 = · + + · + + = 2 3 2 8 2·2 3·2 2 8 4 8 Z

R1R1 12. Evaluate 0 0 x max(x, y)dydx. Solution: Z 1 Z x Z x2 dydx + x=0

y=0

1

x=0

Z

1

y=x

2

1

1 − x2 x3 + x 2 0 1 1 1 1 = + − = 4 2 2 4 Z

xydydx =

dx 3 . 8

Introduction to Probability

A.J. Hildebrand

Practice problems on double integrals The problems below illustrate the kind of double integrals that frequently arise in probability applications. The first group of questions asks to set up a double integral of a general function f (x, R ∗ y) R ∗over a giving region Rin∗ the R ∗ xy-plane. This means writing the integral as an iterated integral of the form ∗ ∗ f (x, y)dxdy and/or ∗ ∗ f (x, y)dydx, with specific limits in place of the asterisks. To do this, follow the steps above (most importantly, sketch the given region). The remaining questions are evaluations of integrals over concrete functions. 1. Set up a double integral of f (x, y) over the region given by 0 < x < 1, x < y < x + 1. Solution: Z 1 Z x+1 f (x, y)dydx x=0

y=x

2. Set up a double integral of f (x, y) over the part of the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, on which y ≤ x/2. Solution: Z 1 Z x/2 Z 1/2 Z 1 f (x, y)dydx or f (x, y)dxdy x=0

y=0

y=0

x=2y

3. Set up a double integral of f (x, y) over the part of the unit square on which x + y > 0.5. Solution: Z 1/2 Z 1 Z 1 Z 1 f (x, y)dydx + f (x, y)dydx x=0

y=1/2−x

x=1/2

y=0

4. Set up a double integral of f (x, y) over the part of the unit square on which both x and y are greater than 0.5. Solution: Z 1 Z 1 f (x, y)dydx x=1/2

y=1/2

5. Set up a double integral of f (x, y) over the part of the unit square on which at least one of x and y is greater than 0.5. Solution: Z 1/2 Z 1 Z 1 Z 1 f (x, y)dydx + f (x, y)dydx x=0

y=1/2

x=1/2

y=0

6. Set up a double integral of f (x, y) over the part of the region given by 0 < x < 50 − y < 50 on which both x and y are greater than 20. Solution: Z 30 Z 50−x f (x, y)dydx x=20

y=20

7. Set up a double integral of f (x, y) over the set of all points (x, y) in the first quadrant with |x − y| ≤ 1. Solution: Z ∞ Z x+1 Z 1 Z x+1 f (x, y)dydx + f (x, y)dydx x=0

y=0

x=1

1

y=x−1

Math 461

Introduction to Probability

A.J. Hildebrand

RR 8. Evaluate R e−x−y dxdy, where R is the region in the first quadrant in which x + y ≤ 1. Solution: Z 1 Z 1−x Z 1 −x −y e e dydx = e−x (1 − e−(1−x) )dx x=0

y=0

0

Z =

1

(e−x − e−1 )dx = 1 − 2e−1 .

0

RR 9. Evaluate R e−x−2y dxdy, where R is the region in the first quadrant in which x ≤ y Solution: Z ∞ Z ∞ Z ∞ 1 1 −3x e dx = e−x−2y dydx = 2 6 x=0 y=x 0 RR 10. Evaluate R (x2 + y 2 )dxdy, where R is the region 0 ≤ x ≤ y ≤ L Solution: Z L Z y Z L 1 3 y 2 2 2 (x + y )dydx = x + y x dydx x=0 y=0 x=0 y=0 3 Z L 4 L 4 3 = y dy = . 3 0 3 RR 11. Evaluate R (x − y + 1)dxdy, where R is the region inside the unit square in which x + y ≥ 0.5. Solution: Z 0.5 Z 1 Z 1 Z 1 (x − y + 1)dydx + (x − y + 1)dydx x=0

y=0.5−x 0.5

x=0.5

y=0

Z 1 1 2 1 2 1 1 = xy − y + y dx + xy − y + y dx 2 2 y=0.5−x y=0 x=0 x=0.5 Z 0.5 1 1 1 1 = x(1 − + x) − (1 − ( − x)2 ) + (1 − + x) dx 2 2 2 2 0 Z 1 1 + x+ dx 2 0.5 Z 0.5 1 3 1 2 1 1 = + x + x2 dx + x + x 8 2 2 2 0.5 0 1 1 3 3 1 7 1 1 = · + + · + + = 2 3 2 8 2·2 3·2 2 8 4 8 Z

R1R1 12. Evaluate 0 0 x max(x, y)dydx. Solution: Z 1 Z x Z x2 dydx + x=0

y=0

1

x=0

Z

1

y=x

2

1

1 − x2 x3 + x 2 0 1 1 1 1 = + − = 4 2 2 4 Z

xydydx =

dx 3 . 8