5 Steps to a 5 AP Calculus BC 2019

More Applications of Definite Integrals 337

13.11 Solutions to Practice Problems

Part A The use of a calculator is not

allowed.

1.

∫ 4

2

x^3 dx= f(c)(4−2)

∫ 4

2

x^3 dx=

x^4

4

] 4

2

=

(

44

4

)

−

(

24

4

)

= 60

2 f(c)= 60 ⇒ f(c)= 30

c^3 = 30 ⇒C= 30 (1/3).

2. Average Value=

1

8 − 0

∫ 1

0

f(x)dx

=

1

8

(

1

2

(8)(4)

)

= 2.

3. Displacement=s(4)−s(1)=− 15 −(−12)=− 3.
   Distance Traveled=

∫ 4

1

∣

∣v(t)

∣

∣dt.

v(t)=s′(t)= 2 t− 6

Set 2t− 6 = 0 ⇒t= 3

∣∣

2 t− 6

∣∣

=

{

−(2t−6) if 0≤t< 3

2 t−6if3≤t≤ 10

∫ 4

1

|v(t)|dt=

∫ 3

1

−(2t−6)dt+

∫ 4

3

(2t−6)dt

=

[

−t^2 + 6 t

] 3

1 +

[

t^2 − 6 t

] 4

3

= 4 + 1 = 5.

4. Position Functions(t)=

∫

v(t)dt

=

∫

(2t+1)dt

=t^2 +t+C

s(1)=− 4 ⇒(1)^2

+ 1 +C

=−4orC=− 6

s(t)=t^2 +t− 6

s(5)= 52 + 5 − 6 = 24.

5.Total Loss=

∫ 5

0

p(t)dt

=

∫ 5

0

(50t−600)dt

= 25 t^2 − 600 t

] 5

0 =−$2375.

6.v(t)=

∫

a(t)dt=

∫

− 2 dt=− 2 t+C

v 0 = 10 ⇒−2(0)+C=10 orC= 10

v(t)=− 2 t+ 10

Distance Traveled=

∫ 4

0

∣∣

v(t)

∣∣

dt.

Set∣ v(t)= 0 ⇒− 2 t+ 10 =0ort=5.

∣− 2 t+ 10 ∣∣=− 2 t+10 if 0≤t< 5

∫ 4

0

∣∣

v(t)

∣∣

dt=

∫ 4

0

(− 2 t+10)dt

=−t^2 + 10 t

] 4

0 =^24

7.Total Distance Traveled

=

∫ 8

0

∣∣

v(t)

∣∣

dt+

∣∣

∣∣

∫ 12

8

v(t)

∣∣

∣∣

=

1

2

(8) (10)+

1

2

(4) (10)

=60 meters.

8.Total Leakage=

∫ 3

0

10 e^0.^2 t= 50 e^0.^2 t

] 3

0

= 91. 1059 − 50

= 41. 1059 =41 gallons.

9.Total change in temperature

=

∫ 5

0

−cos

(

t

4

)

dt

=−4 sin

(

t

4

)] 5

0

=− 3. 79594 − 0

=− 3. 79594 ◦F.

Thus, the temperature of coffee after 5

minutes is (92− 3 .79594)≈ 88. 204 ◦F.