MA 3972-MA-Book April 11, 2018 16:5
More Applications of Definite Integrals 343
Step 4: Verify result by
differentiating
y=2(2x+1)^1 /^2
dy
dx
= 2
(
1
2
)
( 2 x+ 1 )−^1 /^2 ( 2 )
=
2
√
2 x+ 1
.
Compare this with:
dy
dx
=
y
2 x+ 1
=
2 ( 2 x+ 1 )^1 /^2
2 x+ 1
=
√^2
2 x+ 1
.
Thus, the function is
y= f(x)=2(2x+1)^1 /^2.
(d) f(x)=2(2x+1)^1 /^2
f(0.1)=2(2(0.1)+1)^1 /^2 =2(1.2)^1 /^2
≈ 2. 191
- See Figure 14.10-1.
[–π, π] by [–4, 4]
Figure 14.10-1
(a) Intersection points: Using the
[Intersection] function on the
calculator, you havex=0 and
x= 1 .37131.
Area ofR=
∫ 1. 37131
0
[3 sinx−(ex−1)]dx.
Enter
∫
(3 sin(x))−(e∧(x)−1),x,0,
1.37131 and obtain 0.836303.
The area of regionRis approximately
0.836.
(b) Using the Washer Method, volume of
R=π
∫ 1. 37131
0
[
(3 sinx)^2 −(ex−1)^2
]
dx.
Enterπ
∫
((3 sin(x))∧ 2 −(e∧(x)−1)∧2,
x,0,1.37131) and obtain 2.54273π
or 7.98824.
Thus, the volume of the solid is 7.988.
(c) Volume of solid=π
∫ 1. 37131
0
(Area of
Cross Section)dx.
Area of Cross Section=
1
2
πr^2
=
1
2
π
(
1
2
(3 sinx−(ex−1))
) 2
.
Enter
(
π
2
)
1
4
∗
∫
((3 sin(x)−
(e∧(x)−1))∧2,x,0,1.37131)
and obtain 0.077184πor 0.24248.
Thus, the volume of the solid is
approximately 0.077184πor 0.242.
