5 Steps to a 5 AP Calculus BC 2019

32 STEP 2. Determine Your Test Readiness

Chapter 10

28.

∫

1 −x^2

x^2

dx=

∫ (

1

x^2

−

x^2

x^2

)

dx

=

∫ (

1

x^2

− 1

)

dx

=

∫

(x−^2 −1)dx=

x−^1

− 1

−x+C

=−

1

x

−x+C

KEY IDEA

You can check the answer by

differentiating your result.

29. Letu=ex+1;du=exdx.

f(x)=

∫

ex

ex+ 1

dx=

∫

1

u

du

=ln|u|+C=ln|ex+ 1 |+C

f(0)=ln|e^0 + 1 |+C=ln (2)+C

Sincef(0)=ln 2⇒ln (2)+C

=ln 2⇒C= 0.

Thus, f(x)=ln (ex+1) andf(ln 2)

=ln (eln 2+1)=ln (2+1)

=ln 3.

30. See Figure DS-10.
    To find the points of intersection, set

sin 2x=

1

2

⇒ 2 x=sin−^1

(

1

2

)

⇒ 2 x=

π

6

or 2x=

5 π

6

⇒x=

π

12

orx=

5 π

12

.

Volume of solid

=π

∫ 5 π/ 12

π/ 12

[

(sin 2x)^2 −

(

1

2

) 2 ]

dx.

Using your calculator, you obtain:

Volume of solid≈(0.478306)π

≈ 1. 50264 ≈ 1. 503.

Figure DS-10

31.

∫ 5

2

1

x^2 + 2 x− 3

dx=

∫ 5

2

1

(x+3)(x−1)

dx

Use a partial fraction decomposition with

A

(x+3)

+

B

(x−1)

=

1

(x+3)(x−1)

, which

givesA=

− 1

4

andB=

1

4

. Then the integral
becomes

=

∫ 5

2

− 1 / 4

(x+3)

dx+

∫ 5

2

1 / 4

(x−1)

dx

=

− 1

4

∫ 5

2

1

(x+3)

dx+

1

4

∫ 5

2

1

(x−1)

dx

=

− 1

4

ln

∣∣

x+ 3

∣∣

+

1

4

ln

∣∣

x− 1

∣∣]^5

2

=

1

4

(

ln

∣∣

∣∣x−^1

x+ 3

∣∣

∣∣

)] 5

2

=

1

4

[(

ln

∣∣

∣∣^4

8

∣∣

∣∣

)

−

(

ln

∣∣

∣∣^1

5

∣∣

∣∣

)]

=

1

4

ln

(

5

2

)

32. Integrate

∫

x^2 cosxdxby parts withu=x^2 ,

du= 2 xdx,dv=cosxdx, andv=sinx.

The integral becomes

=x^2 sinx−

∫

sinx(2x)dx

=x^2 sinx− 2

∫

xsinxdx.

Use parts again for the remaining integral,

lettingu=x,du=dx,dv=sinxdx, and

v=−cosx. The integral

=x^2 sinx− 2

[

−xcosx−

∫

(−cosx)dx

]