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

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

Take a Diagnostic Exam 31

KEY IDEA

You can check your answer by

evaluating

∫− 2

− 1

(2x−3)dxand

∫ 5

− 1

(2x−3)dx.

34. h(x)=

∫π

π/ 2

√

sintdt⇒h′(x)=

√

sinx

h′(π)=

√

sinπ=

√

0 = 0

35. Letu= 3 x;du= 3 dxor
    du
    3

=dx.

∫

g(3x)dx=

∫

g(u)

du

3

=

1

3

∫

g(u)du

=

1

3

f(u)+c=

1

3

f(3x)+c

∫ 2

0

g(3x)dx=

1

3

[f(3x)]^20

=

1

3

f(6)−

1

3

f(0)

Thus, the correct choice is (A).

36.

∫x

π

sin(2t)dt=

[

−cos(2t)

2

]x

π

=

−cos(2x)

2

−

(

−

cos(2π)

2

)

=−

1

2

cos(2x)+

1

2

37. I.

∫c

a

f(x)dx=

∫b

a

f(x)dx+

∫c

b

f(x)dx

The statement is true, since the upper and

lower limits of the integrals are in sequence,

i.e.,a→c=a→b→c.

II.

∫b

a

f(x)dx=

∫c

a

f(x)dx−

∫b

c

f(x)dx

=

∫c

a

f(x)dx+

∫c

b

f(x)dx

The statement is not always true.

III.

∫c

b

f(x)dx=

∫a

b

f(x)dx−

∫a

c

f(x)dx

=

∫a

b

f(x)dx+

∫c

a

f(x)dx

The statement is true.

Thus, only statements I and III are true.

38. Sinceg(x)=

∫ x

π/ 2

2 sintdt, then

g′(x)=2 sinx.

Setg′(x)= 0 ⇒2 sinx= 0 ⇒x=πor 2π

g′′(x)=2 cosxandg′′(π)=2 cosπ=

−2 andg′′(2π)= 1.

Thusg has a local minimum atx= 2 π. You

can also approach the problem geometrically

by looking at the area under the curve. See the

figure below.

+

−

+

y = 2sint

3 π 2 π

2

5 π

2

π π

(^02)
− 2
2
y
t
Chapter 13

39. Total distance=

∫ 4

0

v(t)dt+

∣∣

∣

∣

∫ 6

4

v(t)dt

∣∣

∣

∣

=

1

2

(4)(20)+

∣∣

∣∣^1

2

(2)(−10)

∣∣

∣∣

= 40 + 10 =50 feet