MTH 151 Study Guide - Midterm Guide: Asymptote, Indeterminate Form

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Published on 12 Oct 2018
Department
Course
Professor
Name:
Exam 4 - Math 151
R. Ketchersid
1)[18pts, 2pts each] Evaluate the following indefinite integrals.
a) Zt·sin(t2)dt
u=t2;du/2 = t dt
Zt·sin(t2)dt =1
2Zsin(u)du =cos(u)
2+C=cos(t2)
2+C
b) Zy dy
4+9y2
u= 4 9y2;du/18 = y dy
Zy dy
4+9y2=1
18 Zdu
u=ln |u|
18 +C=ln(4 + 9y2)
18 +C
c) Zdy
4+9y2
Zdy
4+9y2=1
4Zdy
1+9y2/4=1
4Zdy
1 + (3/2y)2
u= 3/2y;2/3du =dy
1
4Zdy
1 + (3/2y)2=1
6Zdu
1 + u2=arctan(u)
6+C=arctan(3/2y)
6+C
d) Z(u2)2
udu
Z(u24u+ 4)
udu =Z(u4+4/u)du =u2
24u+ 4 ln |u|+C
e) Z1 + x
1xdx
u= 1 x;du =dx;x= 1 u
Z1 + x
1xdx =Z2u
udu =Z(u1/22u1/2)du = 2/3u3/24u1/2+C= 2/3(1 x)3/24(1 x)1/2+C
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-2-
f) Zπ/4
0
tan(x) sec2(x)dx
u= tan(x);du = sec2(x)dx
Zπ/4
0
tan(x) sec2(x)dx =Z1
0
u du =u2
2
1
0
=1
2
g) Z1
0
x+x3
1x4dx
Correct: The function x+x3
1x4has a vertical asymptote at x= 1 so the correct answer is DNE.
However, if you went through the formalism, I also accepted the following:
Z1
0
x+x3
1x4dx =Z1
0
x dx
1x4+Z1
0
x3dx
1x4
For the first integral, u=x2and du/2 = x dx
Z1
0
x dx
1x4=1
2Z1
0
du
1u2=1
2arcsin(u)
1
0=π
4
For the second integral, v= 1 x4and du/4 = x3dx
Z1
0
x3dx
1x4=1
4Z0
1
dv
v=1
4Z1
0
v1/2dv =1
42v1/2
1
0=1
2
So the answer is
1
2arcsin(u)
1
0+1
42v1/2
1
0=π
4+1
2
h)1Zπ/6
π/3
cos(x)
sin(x)dx
u= sin(x);du = cos(x)dx
Zπ/6
π/3
cos(x)
sin(x)dx =Z1/2
32
du
u= ln |u|
1/2
3/2= ln(1/2) ln(3/2) = ln(3)
2
i) Z2
1
d
dxp4x2dx
Z2
1
d
dxp4x2dx =p4x2
2
1=3
1A number of students made precisely the mistakes I warned you about. For example in (h) many wrote:
Zπ/6
π/3
cos(x)
sin(x)dx =Zπ/6
π/3
du
u= ln |u|
π/6
π/3
and many wrote
Zπ/6
π/3
cos(x)
sin(x)dx =Z1/2
32
du
u= ln |u|
1/2
3/2
= ln |sin(x)|
1/2
3/2
.
These are both incorrect.
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Document Summary

1)[18pts, 2pts each] evaluate the following inde nite integrals. (cid:90) 4 + 9y2 u = 3/2y; 2/3 du = dy (cid:90) (u 2)2 u d) (cid:90) (cid:90) 1 x dx e) u = 1 x; du = dx; x = 1 u (cid:90) 2 u u. 1 x dx = (cid:90) du u. 4 du(cid:90) (u2 4u + 4) u (cid:90) du = (u 4 + 4/u) du = u2. 4u + 4 ln|u| + c (cid:90) /4. 2 u = tan(x); du = sec2(x) dx(cid:90) /4 (cid:90) 1. 1 x4 has a vertical asymptote at x = 1 so the correct answer is dne. However, if you went through the formalism, i also accepted the following: For the rst integral, u = x2 and du/2 = x dx du . 1 i) d dx cos(x) sin(x) dx (cid:90) /6 (cid:112)4 x2 dx. /3 u = sin(x); du = cos(x) dx (cid:90) 1/2.

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