CAS MA 124 Study Guide - Midterm Guide: Power Rule, Partial Fraction Decomposition

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10 Oct 2018
Professor
MA 124 - Spring 10 - Exam 1 - Prof. Meuser
1) (16 points) Find the derivative of
f(x) = Z0
ex
tan3t dt
State any rules and theorems you are applying and where you are applying them.
Answer:
Since
Za
b
f(x)dx =Zb
a
f(x)dx
we write
f(x) = Zex
0
tan3t dt
Now let u=ex. Then
f(x) = f(u) = Zu
0
tan3t dt
By the Fundamental Theorem of Calculus I we have
f0(u) = df
du =tan3u
Since u=ex, du/dx =ex. By the Chain Rule
df
dx =df
du ·du
dx
So
f0(x) = df
dx =tan3u·ex=extan3ex
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2) (16 points) Find the values of the following definite integrals. Simplify your answers
as much as possible.
a) (8 points)
Z1
0
(3x1)49 dx
Let u= 3x1. Then du = 3dx so (1/3)du =dx. When x= 0 , u =1 and when
x= 1 , u = 2. So we get
Z1
0
(3x1)49 dx =1
3Z2
1
u49 du
Integrating using the power rule we get
1
3Z2
1
u49 du =1
3·1
50u50
2
1=1
150u50
2
1
Substituting in the values we get
1
150(250 (1)50) = 1
150(250 1)
NOTE: 250 is too large to be evaluated by a calculator, hence we leave it in exponential
form.
b) (8 points)
Ze
1x+ 1
x2
dx
Simplify and expand to get:
x+ 1
x2
=1 + 1
x2
= 1 + 2
x+1
x2
So we get
Ze
1
1 + 2
x+1
x2dx =Ze
1
1dx + 2 Ze
1
1
xdx +Ze
1
x2dx
Z1dx =x;Z1
xdx = ln x;Zx2dx =x1=1
x
Evaluating get
e1 + 2(ln eln 1) (1
e1)
Sine ln e= 1 and ln 1 = 0 we get
e1+21
e+ 1 = e1
e+ 2
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