5 Steps to a 5 AP Calculus BC 2019

Take a Diagnostic Exam 33

simplifies to

=x^2 sinx+ 2 xcosx− 2

∫

cosxdx,

and the final integration gives you

=x^2 sinx+ 2 xcosx−2 sinx+C.

Chapter 11

33.

∫ 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

34.

∫k

− 1

(2x−3)dx=x^2 − 3 x

]k

− 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.

KEY IDEA

You can check your answer by

evaluating

∫− 2

− 1

(2x−3)dxand

∫ 5

− 1

(2x−3)dx.

35. h(x)=

∫π

π/ 2

√

sintdt⇒h′(x)=

√

sinx

h′(π)=

√

sinπ=

√

0 = 0

36. 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).

37.

∫x

π

sin(2t)dt=

[

−cos(2t)

2

]x

π

=

−cos(2x)

2

−

(

−

cos(2π)

2

)

=−

1

2

cos(2x)+

1

2

38. 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.

39. 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

Figure DS-11.

+

• 
  

+

y= 2 sint

2 π

2

5 π

2 2

0 ππ^3 π

–2

2

y

t

Figure DS-11