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

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10 Oct 2018

School

Department

Course

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

2) (16 points) Find the values of the following deﬁnite integrals. Simplify your answers

as much as possible.

a) (8 points)

Z1

0

(3x−1)49 dx

Let u= 3x−1. Then du = 3dx so (1/3)du =dx. When x= 0 , u =−1 and when

x= 1 , u = 2. So we get

Z1

0

(3x−1)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

x−2dx

Z1dx =x;Z1

xdx = ln x;Zx−2dx =−x−1=−1

x

Evaluating get

e−1 + 2(ln e−ln 1) −(1

e−1)

Sine ln e= 1 and ln 1 = 0 we get

e−1+2−1

e+ 1 = e−1

e+ 2