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(MATH014)[2011](s)midterm~4274^_79766.pdf
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Math014 Calculus II, Spring 2011 Midterm Exam Solution
1. ([18 points]) Consider the shaded region bounded between the curves given by y = 2sin2 x and y = sin2x over the interval 0 x .
(i) Find the area of the shaded region. [8 points]
Solution
The intersection of the curves:
2sin2 x .sin2x =2sin2 x .2sinxcosx =0
2sinx(sinx .cosx)=0


i.e. the graphs intersect when x =0, x = 4 and x = .
y
Using cos2A =1.2sin2 A to integrate: 2
. .

area = 4 [sin2x .2sin2 x]dx+ [2sin2 x .sin2x]dx 1

0
4
x
. .

4

= [sin2x + cos2x .1]dx+ [1.cos2x .sin2x]dx

0
.. . 4 .1
11 4 11 .
= . cos2x+ sin2x.x + x. sin2x+ cos2x

22 0 22 4
1 11 1

=( . )+ +(+).( . )= +2
242 2422
1
Using reduction formula sin2 xdx = .1 sinxcosx + sinxdx to integrate:
22
. .

area = 4 [sin2x .2sin2 x]dx+ [2sin2 x .sin2x]dx

0
4
. .
.. .
1 44 1 .
= . cos2x + sinxcosx . dx+ .sinxcosx + cos2x + dx

2 0 2
0 4
4

= +2
2
(ii) Rotate the shaded region about the y-axis to generate a solid of revolution. Express the volume of the solid region by a de.nite integral. Do not evaluate the integral. [4 points]

Solution volume = 2x .2sin2 x .sin2x.dx
0
(iii) UseSimpsonsRulewith n =6subintervalsof equallength to estimatethevolume of thesolid regioninpart(ii). (Recallthattheareaunderthegraph of a quadraticpolynomial p(x) on the interval[a,b]is given by b.a[p(a)+4p(a+b)+p(b)].) [6 points]
62
Solution Applying Simpsons Rule with n =6 subintervals of length /6 to the function
or
f(x)=2x|2sin2 x .sin2x =4xsinx|sinx .cosx|
2 5
we have x 0
6323 6

2(3.1) 2 (3. 3) 22 (3+ 3) 52 (1+ 3)
f(x)0 22 0
63 36
0.0000 1.2042 4.1714 19.7392 31.1356 22.4702 0.0000
..
volume f(0)+4f(/6)+2f(/3)+4f(/2)+2f(2/3)+4f(5/6)+f()
18
25+7 3 or
= 3 42.6329
27
2. ([15 points]) Evaluate the followingintegrals. (Write on the backside of this page if necessary.)
.
2
(i) sin4xcos2xdx [5 points]
0
Solution
1
Using product to sum formula sinAcosB = 2 [sin(A+ B)+sin(A.B)]to integrate:
. .
22 1
sin4xcos2xdx = [sin(4x +2x)+sin(4x .2x)]dx
2
00
..
11 2 11 112
= . cos6x . cos2x =( +).(.. )=
12 4 0 12 412 43 Using double angle formula sin2A =2sinAcos A to integrate:
. . ..
22 2
1 2
sin4xcos2xdx = 2sin2xcos 2 2xdx = . cos 3 2x =
3 0 3
00
4
1
(ii) dx [5 points]
22 .1
2 xx
Solution Let x = secu such that dx = secutanudu.

.. .11 . .11
4 coscos
1 4 secutanu 4 tanu
dx = du = du
2 1 2 u 1
2 xx2 .1 cos.1 secsec2 u .1 cos.1 secutanu
22
. .11
cos..11
4 cos4 15 3
= cosudu =[sinu = .
1
cos.11 cos.12 42
2
11
Or, let u = such that du = . x2 dx. Then
x
41/41/4
1 dx = . .1 du = .u(1.u 2).1/2du
22 .1
2 xx1/21 1/2
.1
u2
.
1/4
. 15 3
2)1/2 .
=(1.u = .
44
1/2