5 Steps to a 5 AP Calculus BC 2019

400 STEP 5. Build Your Test-Taking Confidence

Section II Part B

3. (A) Midpoints of 5 subintervals of equal
   length aret=3, 9, 15, 21, and 27.
   The length of each subinterval is
   30 − 0
   5

= 6.

Thus,

∫ 30

0

v(t)dt≈6[v(3)+v(9)+v(15)

+v(21)+v(27)]

=6[7. 5 + 12 + 13. 5

+ 14 +13]

=6[60]= 360.

(B) Average velocity=

1

30 − 0

∫ 30

0

v(t)dt

≈

1

30

(360)

≈12 m/sec.

(C) Average acceleration=

12 − 0

30 − 0

m/sec^2

= 0 .4m/sec^2.

(D) Approximate acceleration att= 6

=

v(9)−v(3)

9 − 3

=

12 − 7. 5

6

= 0 .75 m/sec^2.

(E) Looking at the velocity in the table, you

see that the velocity decreases fromt= 18

tot=30. Thus, the acceleration is

negative for 18<t<30.

4.

02

R

y

x

y = x^3

x = 2

(A) Area of R=

∫ 2

0

x^3 dx=

x^4

4

] 2

0

=

24

4

− 0 = 4

(B) Volume of solid=π

∫ 2

0

(

x^3

) 2

dx

=π

[

x^7

7

] 2

0

=

27 (π)

7

=

128 π

7

.

(C) π

∫a

0

(

x^3

) 2

dx=

1

2

(

128 π

7

)

π

[

x^7

7

]a

0

=

64 π

7

;

πa^7

7

=

64 π

7

;

a^7 = 64 = 26 ; a= 26 /^7

(D) Area of cross section=(x^3 )^2 =x^6.

Volume of solid=

∫ 2

0

x^6 dx=

x^7

7

] 2

0

=

128

7

.