07 - Area Between Curves

8/12/2019 07 - Area Between Curves

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Kuta Software - Infinite Calculus Name__________________________________

Period___Date________________Area Between Curves

For each problem, find the area of the region enclosed by the curves.

1) y = 2 x2 − 8 x + 10

y = x2

2 − 2 x − 1

x =1

x = 3

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

2) x = 2 y2 + 12 y + 19

x = − y2

2 − 4 y − 10

y =−3

y = −2

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

3) y = x2

2 − 3 x −

1

2

y = 3

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

4) y = − x3

2 + 2 x

2

y = − x2 + 4 x

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

-1-



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For each problem, find the area of the region enclosed by the curves. You may use the provided graph

to sketch the curves and shade the enclosed region.

5) y = −2 x2 − 1

y = − x + 3

x = 0

x = 1

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

6) y = 23

x2

y = x

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

7) y = − x3 + 6 x

y = − x2

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

8) y = −2 ⋅ sec2 x

y = 2cos x

x = 0

x =π

4

x

y

−π−

π

2

π

2

π

−4

−3

−2

−1

1

2

3

4

-2-



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Kuta Software - Infinite Calculus Name__________________________________

Period___Date________________Area Between Curves

For each problem, find the area of the region enclosed by the curves.

1) y = 2 x2 − 8 x + 10

y = x2

2 − 2 x − 1

x =1

x = 3

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ 1

3

(2 x2 − 8 x + 10 − ( x

2

2 − 2 x − 1)) dx

= 11

2) x = 2 y2 + 12 y + 19

x = − y2

2 − 4 y − 10

y =−3

y = −2

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ −3

−2

(2 y2 + 12 y + 19 − (− y

2

2 − 4 y − 10)) dy

=29

6 ≈ 4.833

3) y = x2

2 − 3 x −

1

2

y = 3

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ −1

7

(3 − ( x2

2 − 3 x −

1

2 )) dx

=128

3 ≈ 42.667

4) y = − x3

2 + 2 x

2

y = − x2 + 4 x

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ 0

2

(− x2 + 4 x − (− x3

2 + 2 x2)) dx +

∫ 2

4

(− x3

2 + 2 x

2 − (− x2

+ 4 x)) dx

= 4

-1-



For each problem, find the area of the region enclosed by the curves. You may use the provided graph

to sketch the curves and shade the enclosed region.

5) y = −2 x2 − 1

y = − x + 3

x = 0

x = 1

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ 01

(− x + 3 − (−2 x2 − 1)) dx

=25

6 ≈ 4.167

6) y = 23

x2

y = x

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ 0

8

(23

x2

− x) dx

=32

5 = 6.4

7) y = − x3 + 6 x

y = − x2

x

y

−8 −6 −4 −2 2 4 6 8

−8

−6

−4

−2

2

4

6

8

∫ −2

0

(− x2 − (− x3

+ 6 x)) dx +

∫ 0

3

(− x3 + 6 x + x2) dx

=253

12 ≈ 21.083

8) y = −2 ⋅ sec2 x

y = 2cos x

x = 0

x =π

4

x

y

−π−

π

2

π

2

π

−4

−3

−2

−1

1

2

3

4

∫ 0

π

4

(2cos x + 2 ⋅ sec2 x) dx

= 2 + 2 ≈ 3.414

2

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07 - Area Between Curves