5 Steps to a 5 AP Calculus BC 2019

34 STEP 2. Determine Your Test Readiness

40.

∫∞

0

e−xdx=klim→∞

∫k

0

e−xdx=klim→∞[−e−x]k 0

=klim→∞

[

−e−k+e^0

]

=klim→∞

[

−e−k+ 1

]

= 1

Chapter 12

41. Total distance=

∫ 4

0

v(t)dt+

∣∣

∣

∣

∫ 6

4

v(t)dt

∣∣

∣

∣

=

1

2

(4)(20)+

∣∣

∣∣^1

2

(2)(−10)

∣∣

∣∣

= 40 + 10 =50 feet

42.

∫ 5

− 1

f(x)dx=

∫ 1

− 1

f(x)dx+

∫ 5

1

f(x)dx

=−

1

2

(2)(1)+

1

2

(2+4)(1)

=− 1 + 3 = 2

43. To find points of intersection, set
    y=x^2 −x= 0
    ⇒x(x−1)= 0 ⇒x=0orx=1.
    See Figure DS-12.

01

y

x

y=x^2 – x

Figure DS-12

Area=

∣

∣∣

∣

∫ 1

0

(

x^2 −x

)

dx

∣

∣∣

∣=

∣∣

∣∣

∣

x^3

3

−

x^2

2

] 1

0

∣∣

∣∣

∣

=

∣∣

∣∣

(

1

3

−

1

2

)

− 0

∣∣

∣∣=

∣∣

∣∣−^1

6

∣∣

∣∣

=

1

6

44.

∫k

−k

f(x)dx= 0 ⇒ f(x) is an odd function,

i.e., f(x)=−f(−x). Thus the graph in choice

(D) is the only odd function.

45. Area=

∫k

1

√

xdx=

∫k

1

x^1 /^2 dx

=

[

x^3 /^2

3 / 2

]k

1

=

[

2

3

x^3 /^2

]k

1

=

2

3

k^3 /^2 −

2

3

(1)^3 /^2

=

2

3

k^3 /^2 −

2

3

=

2

3

(

k^3 /^2 − 1

)

.

Since A=8, set

2

3

(

k^3 /^2 − 1

)

= 8 ⇒k^3 /^2 − 1

= 12 ⇒k^3 /^2 =13 ork= 132 /^3.

46. See Figure DS-13.

Figure DS-13

Using the [Intersection] function of the

calculator, you obtain the intersection

points atx= 0 .785398, 3.92699, and

7.06858.

Area=

∫ 3. 92699

0. 785398

(sinx−cosx)dx

+

∫ 7. 06858

3. 92699

(cosx−sinx)dx

= 2. 82843 + 2. 82843 ≈ 5. 65685 ≈ 5. 657

You can also find the area by:

Area=

∫ 7. 06858

. 785398

∣∣

sinx−cosx

∣∣

dx

≈ 5. 65685 ≈ 5. 657.