Areas, Volumes, and Arc Lengths 293
Answer:Graphs intersect atx=−1 andx=1. (See Figure 12.6-2.)
Area =
∫ 0
− 1
(
x^3 −x
)
dx+
∫ 1
0
(
x−x^3
)
dx.
Or, using symmetry, Area= 2
∫ 1
0
(
x−x^3
)
dx.
(–1,–1)
0
(1,1)
y = x
x
y
y = x^3
Figure 12.6-2
- The base of a solid is the region bounded by the linesy=x,x=1, and thex-axis. The
cross sections are squares perpendicular to thex-axis. Set up an integral to find the
volume of the solid. Do not evaluate the integral.
Answer:Area of cross section=x^2.
Volume of solid=
∫ 1
0
x^2 dx.
- Set up an integral to find the volume of a solid generated by revolving the region
bounded by the graph ofy=sinx, where 0≤x≤πand thex-axis, about thex-axis.
Do not evaluate the integral.
Answer:Volume =π
∫π
0
(sinx)^2 dx.
- The area under the curve ofy=
1
x
fromx=atox=5 is approximately 0.916, where
1 ≤a<5. Using your calculator, finda.
Answer:
∫ 5
a
1
x
dx=lnx
∣∣ 5
a=ln 5−lna=^0.^916
lna=ln 5− 0. 916 ≈. 693
a≈e^0.^693 ≈ 2
