5 Steps to a 5 AP Calculus BC 2019

Definite Integrals 251

11.8 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1.

∫ 0

− 1

(

1 +x−x^3

)

dx

=x+

x^2

2

−

x^4

4

] 0

− 1

= 0 −

[

(−1)+

(−1)^2

2

−

(−1)^4

4

]

=

3

4

2. Letu=x−2;du=dx.
   ∫
   (x−2)^1 /^2 dx=

∫

u^1 /^2 du

=

2 u^1 /^2

3

+C

=

2

3

(x−2)^3 /^2 +C

Thus,

∫ 11

6

(x−2)^1 /^2 dx=

2

3

(x−2)^3 /^2

] 11

6

=

2

3

[

(11−2)^3 /^2

−(6−2)^3 /^2

]

=

2

3

(27−8)=

38

3

.

3. Letu=t+1;du=dtandt=u−1.

Rewrite:

∫

t

t+ 1

dt=

∫

u− 1

u

du

=

∫ (

1 −

1

u

)

du

=u−ln|u|+C

=t+ 1 −ln

∣

∣t+ 1

∣

∣+C

∫ 3

1

t

t+ 1

dt=

[

t+ 1 −ln

∣

∣t+ 1

∣

∣]^31

=

[

(3)+ 1 −ln

∣∣

3 + 1

∣∣]

−

(

(1)+ 1 −ln

∣∣

1 + 1

∣∣)

= 4 −ln 4− 2 +ln 2

= 2 −ln 4+ln 2

= 2 −ln(2)^2 +ln 2

= 2 −2ln2+ln 2

= 2 −ln 2.

4.Setx− 3 =0;x=3.

∣∣

x− 3

∣∣

=

{

(x−3) ifx≥ 3

−(x−3) ifx< 3

∫ 6

0

∣∣

x− 3

∣∣

dx=

∫ 3

0

−(x−3)dx

+

∫ 6

3

(x−3)dx

=

[

−x^2

2

+ 3 x

] 3

0

+

[

x^2

2

− 3 x

] 6

3

=

(

−

(3)^2

2

+3(3)

)

− 0

+

(

62

2

−3(6)

)

−

(

32

2

−3(3)

)

=

9

2

+

9

2

= 9

5.

∫k

0

(6x−1)dx= 3 x^2 −x

]k

0 =^3 k

(^2) −k
Set 3k^2 −k= 4 ⇒ 3 k^2 −k− 4 = 0
⇒(3k−4)(k+1)= 0
⇒k=

4

3

ork=−1.

Verify your results by evaluating

∫ 4 / 3

0

(6x−1)dxand

∫− 1

0

(6x−1)dx.