AREA Between Curves - 1 PDF

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Kuta Software - Infinite Calculus

Name___________________________________

Area Between Curves

Date________________ Period____

For each problem, find the area of the region enclosed by the curves. 1) y = 2 x2 − 8 x + 10 x2 − 2x − 1 y= 2 x=1 x=3

2) x = 2y 2 + 12 y + 19 y2 − 4y − 10 x=− 2 y = −3 y = −2

y

−8

−6

−4

y

8

8

6

6

4

4

2

2

−2

2

4

6

8 x

−8

−6

−4

−2

−2

−2

−4

−4

−6

−6

−8

−8

x2 1 − 3x − 2 2 y=3

−4

8 x

2

4

6

8 x

y

8

8

6

6

4

4

2

2

−2

6

4) y = −

y

−6

4

x3 + 2x 2 2 y = −x 2 + 4 x

3) y =

−8

2

2

4

6

8 x

−8

−6

−4

−2

−2

−2

−4

−4

−6

−6

−8

−8

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 = −2x 2 − 1 y = −x + 3 x=0 x=1

6) y = 2 y= x

3

x2 y 8 6

y 8

4 6 2 4 −8

−6

−4

−2

2

4

6

2

8 x

−2 −8

−6

−4

−2

2

4

6

8 x

−4

−2 −6 −4 −8 −6 −8

7) y = − x3 + 6x y = −x2

8) y = −2 ⋅ sec 2 x y = 2cos x x=0 π x= 4

y 8 6 4

y 4

2

3 −8

−6

−4

−2

2

4

6

8 x 2

−2

1

−4 −6

−π −

−8

π 2

−1 −2 −3 −4

π 2

π

x

Kuta Software - Infinite Calculus

Name___________________________________

Area Between Curves

Date________________ Period____

For each problem, find the area of the region enclosed by the curves. 1) y = 2 x2 − 8 x + 10 x2 − 2x − 1 y= 2 x=1 x=3

2) x = 2y 2 + 12 y + 19 y2 − 4y − 10 x=− 2 y = −3 y = −2

y

−8

−6

−4

y

8

8

6

6

4

4

2

2

−2

2

4

6

8 x

−8

−6

−4

−2

2

−2

−2

−4

−4

−6

−6

−8

−8

4

−2

3

(

x2 − 2x − 1 2 x2 − 8x + 10 − 2 1 = 11

(

x2 1 − 3x − 2 2 y=3

))

(

(

2 y 2 + 12 y + 19 − −

dx

−3

29 =

−4

≈ 4.833

4) y = −

y

8

8

6

6

4

4

2

2

−2

2

4

6

8 x

−8

−6

−4

−2

( (

2

−2

−2

−4

−4

−6

−6

−8

7

y2 − 4y − 10 2

6

y

−6

8 x

x3 + 2x 2 2 y = −x 2 + 4 x

3) y =

−8

6

1 x2 3− − 3x − 2 2 −1 128 ≈ 42.667 = 3

4

6

8 x

−8

2

))

dx

( x  (− 2

(

− x 2 + 4x − −

0 4

2

=4

3

x3 + 2x 2 2

)) ))

dx +

+ 2x 2 − (−x 2 + 4x dx

))

dy

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 = −2x 2 − 1 y = −x + 3 x=0 x=1

6) y = 2 y= x

3

x2 y 8 6

y 8

4 6 2 4 −8

−6

−4

−2

2

4

6

2

8 x

−2 −8

−6

−4

−2

2

4

6

8 x

−4

−2 −6 −4 −8 −6



−8

1

 ( −x + 3 − (−2 x − 1)) dx

8

(2 3 x 2 − x) dx 0

2

=

0

=

32 = 6.4 5

25 ≈ 4.167 6

7) y = − x3 + 6x y = −x2

8) y = −2 ⋅ sec 2 x y = 2cos x x=0 π x= 4

y 8 6 4

y 4

2

3 −8

−6

−4

−2

2

4

6

8 x 2

−2

1

−4 −6

−π −

−8

π 2

−1

π 2

π

x

−2

0

 (−x − (−x + 6x)) dx +  ( −x + 6x + x ) dx 2

3

−3

−2 3

3

0

253 ≈ 21.083 = 12

−4

2



π 4

(2cos x + 2 ⋅ sec 2 x) dx 0

= 2+

2 ≈ 3.414

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