5 Steps to a 5 AP Calculus BC 2019

Areas, Volumes, and Arc Lengths 267

Area under the curve:

≈

π

4

·

1

2

[

cos(0)+2 cos

(

π/ 4

2

)

+2 cos

(

π/ 2

2

)

+2 cos

(

3 π/ 4

2

)

+cos

(

π

2

)]

≈

π

8

[

cos(0)+2 cos

(

π

8

)

+2 cos

(

π

4

)

+2 cos

(

3 π

8

)

+cos

(

π

2

)]

≈

π

8

[

1 +2(.9239)+ 2

(√

2

2

)

+2(.3827)+ 0

]

≈ 1. 9743.

TIP • When using a graphing calculator in solving a problem, you are required to write the

setup that leads to the answer. For example, if you are finding the volume of a solid, you

must write the definite integral and then use the calculator to compute the numerical

value, e.g., Volume=π

∫ 3

0 (5x)

(^2) dx= 225 π. Simply indicating the answer without
writing the integral would get you only one point for the answer. And you will not get
full credit for the problem.
12.3 Area and Definite Integrals
Main Concepts:Area Under a Curve, Area Between Two Curves
Area Under a Curve
Ify=f(x) is continuous and non-negative on [a,b], then the area under the curve of f
fromatobis:
Area=
∫b
a
f(x)dx.
Iff is continuous andf <0on[a,b], then the area under the curve fromatobis:
Area=−
∫b
a
f(x)dx.See Figure 12.3-1.
y f
x
0 ab
(+)
y
x
0
ab
(–)
f
Figure 12.3-1