# MTH 151 Study Guide - Midterm Guide: Tasmanian Government Railways Y Class, Scilab, Hyperbolic Function

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Exam 3 - Math 151

Dec 14/15, 2009

R. Ketchersid

Instruction. I would like you to perform your scratch work on scratch paper, I want to see neat work on

the exams, however, all relevant work must be shown unless I believe the problem is so simple that it does

not require any work. When you do a substitution you should write out what you are doing carefully so that

I can follow your line of reasoning.

Put a box around your ﬁnal answer so it is clear to me what you want me to grade. If your answer is hard

to ﬁnd or ambiguous, then you will loose points.

You may skip the ﬁnal step in the evaluation of a deﬁnite integral. So if your problem is to evaluate

R3

2t√t2−1dt, then I expect something like the following for a deﬁnite integral:

Let u=t2−1so du/2 = t dt and

Z3

2

tpt2−1dt =1

2Z8

3

√u du = (1/2)(2/3)u3/2

8

3

and something like the following for an indeﬁnite integral:

Let u=t2−1so du/2 = t dt and

Ztpt2−1dt =1

2Z√u du = (1/2)(2/3)u3/2+C= 1/3(t2−1)3/2+C

If you have questions on the form of your answer please ask during class. As a rule of thumb, there is no

question on this exam that requires a calculator; in fact I expect exact answers to questions not numeric

approximations.

d

dx cosh−1(u) = 1

√u2−1

du

dx (u > 1) d

dx sinh−1(u) = 1

√u2+ 1

du

dx

d

dx coth−1(u) = 1

1−u2

du

dx (|u|>1) d

dx tanh−1(u) = 1

1−u2

du

dx (|u|<1)

d

dx csch−1(u) = −1

|u|√1 + u2

du

dx (u6= 0) d

dx sech−1(u) = −1

u√1−u2

du

dx (0 < u < 1)

-2-

1)[42pts, 3pts each] Evaluate the following deﬁnite and indeﬁnite integrals.

a) Zt2tan(3t3+ 2) dt

Let u= 3t3+ 2 so du/dt = 9t2hence 1/9du =t2dt and

Zt2tan(3t3+ 2) dt =1

9Ztan(u)du =1

9Zsin(u)

cos(u)du

Let v= cos(u)so dv =−sin(u)du and

1

9Zsin(u)

cos(u)du =−1

9Zdv

v=−1

9ln |v|+C=−1

9ln |cos(u)|+C=−1

9ln |cos(3t3+ 2)|+C

b) Zy dy

4y2−9

Let u= 4y2−9so du/dy = 8yhence 1/8du =y dy and

Zy dy

4y2−9=1

8Zdu

u=1

8ln |u|+C=1

8ln |4y2−9|+C

c) Zdu

4u2−9with 4u2−9>0

Zdu

4u2−9=1

9Zdu

(4/9)u2−1=1

9Zdu

(2/3u)2−1

Let v= 2/3uso 3/2dv =du and

1

9Zdu

(2/3u)2−1=1

93

2Zdv

v2−1=−1

6Zdv

1−v2

Notice 4u2−9>0is equivalent to |v|>1so we have

−1

6Zdv

1−v2=−1

6coth−1(v) + C=−1

6coth−1(2/3u) + C

d) Z3

√u(2u−1/u)du

Z3

√u(2u−1/u)du =Z2u4/3−u−2/3du = 6/7u7/3−3u1/3+C

e) Z2−3t

tdt

Z2−3t

tdt =Z2

t−3dt = 2 ln |t| − 3t+C