5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 16:1

Areas and Volumes 275

Solution:

(a) Sinces=

∫t

0

f(x)dx, thenv(t)=s′(t)=f(t).

Thus,v(4)=−8 ft/sec.

(b) s(3)=

∫ 3

0

f(x)dx=

∫ 2

0

f(x)dx+

∫ 3

2

f(x)dx=

1

2

(10)(2)−

1

2

(1)(5)=

15

2

ft.

(c) a(t)=v′(t). Sincev′(t)=f′(t),v′(t)=0att=4. Thus,a(4)=0 ft/sec^2.

(d) The particle is moving to the right whenv(t)>0. Thus, the particle is moving to the

right on intervals (0, 2) and (7, 8).

(e) The area offbelow thex-axis fromx=2tox=7 is larger than the area offabove the

x-axis fromx=0tox=2 andx=7tox=8. Thus,

∫ 8

0

f(x)dx<0 and the particle

is on the left side of the origin.

TIP • Do not forget that (fg)′= f′g+g′f andnot f′g′. However, lim(fg)=(limf)

(limg).

13.2 Approximating the Area Under a Curve

Main Concepts:Rectangular Approximations, Trapezoidal Approximations

Rectangular Approximations

Iff ≥0, the area under the curve offcan be approximated using three common types of

rectangles: left-endpoint rectangles, right-endpoint rectangles, or midpoint rectangles.

(See Figure 13.2-1.)

0 a

y f(x)

x

x 1 x 2 x 3 b

left-endpoint

0 a

y f(x)

x

x 1 x 2 x 3 b

right-endpoint

0 a

y f(x)

x

x 1 x 2 x 3 b

midpoint

Figure 13.2-1