5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 12:11

30 STEP 2. Determine Your Test Readiness

28. f

(

π

2

)

= 3 ⇒

(

π

2

,3

)

is on the graph.

f′

(

π

2

)

=− 1 ⇒slope of the tangent atx=

π

2

is− 1.

Equation of tangent line:y− 3 =

− 1

(

x−

π

2

)

ory=−x+

π

2

+ 3.

Thus, f

(

π

2

+

π

180

)

≈−

(

π

2

+

π

180

)

+

π

2

+ 3

≈ 3 −

π

180

≈ 2. 98255

≈ 2. 983.

Chapter 11

29.

∫

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.

30. 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.

31. See the figure below.
    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.

Chapter 12

32.

∫ 4

1

1

√

x

dx=

∫ 4

1

x−^1 /^2 dx=

x^1 /^2

1 / 2

] 4

1

= 2 x^1 /^2

] 4

1

=2(4)^1 /^2 −2(1)^1 /^2 = 4 − 2 = 2

33.

∫k

− 1

(2x−3)dx=x^2 − 3 x

]k

− 1

=(k^2 − 3 k)−((−1)^2

−3(−1))

=k^2 − 3 k−(1+3)

=k^2 − 3 k− 4

Setk^2 − 3 k− 4 = 6 ⇒k^2 − 3 k− 10 = 0

⇒(k−5)(k+2)= 0 ⇒k=5ork=− 2.