Example (WW 4 # 1):
Use cylindrical shells to find the volume of the solid formed by rotating the
region bounded by y = e 4x + 1 , y = 0, x = 0, and x = 0.8 about the
y -axis.
Using shells,
Z
V
0.8
2πx(e 4x + 1) dx
=
0
Z
= 2π
0.8
xe 4x + x dx
0
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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Reminder:
Integration by parts:
Z
I
If you have an integral of a product, think of it as
u dv . Let one
factor be u and the other factor, along with dx, be dv .
I
Then
Z
Z
u dv = uv −
v du
I
Must choose dv to be something you can integrate. Most likely to
work if v is not more complex than dv was.
I
Integration by parts is most likely to work if you can also choose u so
that du is less complex than u is.
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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In Class Work
Evaluate the following integrals using integration by parts, and check your
answers!!
Z
xe 4x dx
1.
Z
2.
x ln(x) dx
Z
3.
1
x sin(πx) dx
Z0
4.
Z
5.
x 2 cos(2x) dx
x sec2 (4x) dx
Z
6.
ln(x) dx
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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Solutions
Z
1.
xe 4x dx
dv = e 4x dx
1
v = e 4x
4
u=x
du = dx
Z
xe
4x
Z
dx
= uv −
=
=
1 4x
xe
4
1 4x
xe
4
v du
Z
1
−
e 4x dx
4
1
− e 4x + C
16
Verify:
1
d 1 4x
1
1
( xe − e 4x + C ) = (x)(4e 4x + e 4x −
· 4e 4x = xe 4x
dx 4
16
4
16
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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Solutions
Z
2.
x ln(x) dx
u = ln(x)
dv = x dx
x2
1
v=
du = dx
x
2
Z
Z
x ln(x) dx = uv − v du
Z 2
Z
x2
x 1
x 2 ln(x) 1
=
ln(x) −
· dx =
−
x dx
2
2 x
2
2
x 2 ln(x) x 2
=
−
+C
2
4
Verify:
1
1 1
1
d x 2 ln(x) x 2
(
−
+ C ) = · x 2 · + ln(x) · 2x − · 2x
dx
2
4
2
x
2
4
x
x
= + x ln(x) −
2
2
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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Solutions
Z
1
3.
x sin(πx) dx
0
u=x
du = dx
Z
dv = sin(πx) dx
v = − π1 cos(πx)
1
Z
x sin(πx) dx
= uv −
v du
0
1
Z
1
x
− cos(πx) +
cos(πx) dx π
π
0
1
1
x
=
− cos(πx) + 2 sin(πx) π
π
0
1
1
1
= − cos(π) + 2 sin(π) − 0 + 2 sin(0)
π
π
π
1
1
= − · −1 + 0 =
π
π
=
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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Solutions
Z
4.
x 2 cos(2x) dx
u = x2
dv = cos(2x) dx
1
du = 2x dx
v = sin(2x)
2
Z
Z
Z
x2
2
x cos(2x) dx = uv − v du =
sin(2x) −
x sin(2x) dx
2
u=x
dv = sin(2x) dx
du = dx
v = − 21 cos(2x)
Z
Z
x2
sin(2x) − [uv − v du]
x 2 cos(2x) dx =
2
x2
x
1
=
sin(2x) − [− cos(2x) +
2
2
2
=
Math 104-Calculus 2 (Sklensky)
Z
cos(2x) dx]
x2
x
1
sin(2x) + cos(2x) − sin(2x) + C
2
4
In-Class 2
Work
February 24, 2012
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Solutions
Z
x sec2 (4x) dx
5.
u=x
du = dx
Z
x sec2 (4x) dx
Z
x sec2 (4x) dx
Math 104-Calculus 2 (Sklensky)
dv = sec2 (4x) dx
v = 14 tan(4x)
Z
= uv − v du
Z
1
=
x tan(4x) − tan(4x) dx
4
Z
sin(4x)
1
x tan(4x) −
dx
=
4
cos(4x)
1
Let u = cos(4x).Then − du = sin(4x) dx
4
Z
1
1
1
=
x tan(4x) +
du
4
4
u
x
1
=
tan(4x) +
ln | cos(4x)| + C
4
16
In-Class Work
February 24, 2012
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Solutions
Z
6.
ln(x) dx
u = ln(x)
du = x1 dx
dv = dx
v =x
Z
Z
ln(x) dx
= uv −
v du
Z
1
= x ln(x) − x · dx
x
Z
= x ln(x) − 1 dx
= x ln(x) − x + C
Verify:
d
1
(x ln(x) − x + C ) = x · + ln(x) − 1
dx
x
= ln(x)
Math 104-Calculus 2 (Sklensky)
In-Class Work
February 24, 2012
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