5 Steps to a 5 AP Calculus BC 2019

314 STEP 4. Review the Knowledge You Need to Score High

(a) Displacement=

∫t 2

t 1

v(t)dt

=

∫ 6

0

(t^2 + 3 t−10)dt=

t^3

3

+

3 t^2

2

− 10 t

] 6

0

= 66.

(b) Total Distance Traveled=

∫t 2

t 1

∣∣

v(t)

∣∣

dt

=

∫ 6

0

|t^2 + 3 t− 10 |dt.

Lett^2 + 3 t− 10 = 0 ⇒(t+5) (t−2)= 0 ⇒t=−5ort= 2

|t^2 + 3 t− 10 |=

{

−

(

t^2 + 3 t− 10

)

if 0≤t≤ 2

t^2 + 3 t− 10 ift> 2

∫ 6

0

|t^2 + 3 t− 10 |dt=

∫ 2

0

−(t^2 + 3 t−10)dt+

∫ 6

2

(t^2 + 3 t−10)dt

=

[

−t^3

3

−

3 t^2

2

+ 10 t

] 2

0

+

[

t^3

3

+

3 t^2

2

− 10 t

] 6

2

=

34

3

+

232

3

=

266

3

≈ 88. 667.

The total distance traveled by the particle is

266

3

or approximately 88.667.

Example 3

The velocity function of a moving particle on a coordinate line isv(t)=t^3 − 6 t^2 + 11 t−6.

Using a calculator, find (a) the displacement by the particle during 1≤t≤4, and (b) the

total distance traveled by the particle during 1≤t≤4.

(a) Displacement=

∫t 2

t 1

v(t)dt

=

∫ 4

1

(t^3 − 6 t^2 + 11 t−6)dt.

Enter

∫ (

x^3 − 6 x^2 + 11 x−6,x,1,4

)

and obtain

9

4

.

(b) Total Distance Traveled=

∫t 2

t 1

∣

∣v(t)

∣

∣dt.

Entery 1 =x∧ 3 − 6 x∧ 2 + 11 x−6 and use the [Zero] function to obtainx-intercepts at

x=1, 2, 3.

|v(t)|=

{

v(t)if1≤t≤2 and 3≤t≤ 4

−v(t)if2<t < 3