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