sol6

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CHAPTER 6

Applications of the Definite Integral in Geometry, Science, and Engineering EXERCISE SET 6.1

2

1. A = −1

4

2. A =

2 (x2 + 1 − x)dx = (x3 /3 + x − x2 /2) = 9/2 −1

4 √ 3/2 2 ( x + x/4)dx = (2x /3 + x /8) = 22/3

0

0 2

2 (y − 1/y )dy = (y /2 + 1/y) = 1 2

3. A =

2

1

1 2

2 (2 − y + y)dy = (2y − y /3 + y /2) = 10/3 2

4. A = 0

3

2

0

4

(4x − x2 )dx = 32/3

5. (a) A =

16

(b) A =

0

√ ( y − y/4)dy = 32/3

0

y (4, 16) y = 4x y = x2 5 x 1

6. Eliminate x to get y 2 = 4(y + 4)/2, y 2 − 2y − 8 = 0, (y − 4)(y + 2) = 0; y = −2, 4 with corresponding values of x = 1, 4. 4 1 √ √ √ [2 x − (−2 x)]dx + [2 x − (2x − 4)]dx (a) A = 1 0 1 4 √ √ = 4 xdx + (2 x − 2x + 4)dx = 8/3 + 19/3 = 9 0 1 4 (b) A = [(y/2 + 2) − y 2 /4]dy = 9

y

y = 2x – 4 x

(1, -2)

−2

1

7. A =

√ ( x − x2 )dx = 49/192

y

1/4

y = √x

(1, 1) y = x2 x

1 4

231

(4, 4)

y2 = 4x


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