Calculus II – Math 121
Test #2
Spring 2007
For the last 10 points on this test, please hand in correc-
tion problem for each problem on the test that you lost any
points.
Problem 1 – Section 5.7 #18
Problem 2 – Section 5.7 #60
Problem 3 – Section 5.7 #50
Problem 4 – Section 5.9 #22
Problem 5 – Section 6.2 #24
Problem 6 – Section 6.3 #22
1. [10 points each] Compute the following indefinite inte-
grals:
a. Zx7+ (1 + x)7+ (1 −x)7dx
You could do this by inspection. That is, you could guess
that the answer should look something like 1
8x8+1
8(1+ x)8+
1
8(1 −x)8+C, check by taking the derivative and see that
the third term is off by a factor of −1 and negate it to fix it.
Alternatively, you could break it into three integrals.
Zx7dx +Z(1 + x)7dx +Z(1 −x)7dx
The first integral is clear, the second you could do with a
substitution u= 1 + x, and the third with a substitution
u= 1 −x.
Either way, you should get the answer
1
8x8+1
8(1 + x)8
−
1
8(1 −x)8+C.
b. Z5xp1−3x2dx
Again, you could find the integral by inspection, but here’s
a formal way using substitution. Since there is a composition
where the inner function is 1 −3x2, you should try the sub-
stitution u= 1−3x2. For that substitution, du =−6x dx, so
the integral becomes R−
5
6√u du which is clearly −
5
9u3/2+C,
which, upon substituting back gives the answer
−
5
9(1 −3x2)3/2+C.
c. Z2 sin 3xcos 3x dx
You might do this with two substitutions, the first being
u= 3x,du = 3 dx. That simplifies the integral a little bit
to Z2
3sin ucos u du.
At this point you might see note that the derivative of sine is
cosine, suggesting the substitution v= sin u,dv = cos u du,
which gives the integral
Z2
3v dv.
Alternatively, you might do just one substitution v= sin 3x,
dv = 3 cos 3x dx, which does the same thing as the two sub-
stitutions mentioned above. In any case, that last integral
equals 1
3v2+C=1
3sin2u+C=1
3sin23x+C.
2. [10 points each] Compute the following definite integrals:
a. Z1
−1
t
(1 + t2)3dt
The easiest way to find the integral is to note that the
integrand is an odd function and it’s evaluated from −ato
a, so its integral is 0. But if you didn’t notice that, there
are other ways to evaluate it.
Since there’s a composition where the inner function is
1 + t2, the substitution u= 1 + t2,du = 2t dt ought to help.
Note that when t=−1, u= 2, and when t= 1, u= 2 also,
so the integral becomes
Z2
2
1
2u3du.
Now since the limits of integration are the same, therefore
the value of the integral is 0.
b. Z4
1
(1 + √x)3
√xdx
Either of the substitutions u=√xor u= 1 + √xwork,
but the second one happens to work a little better, and it
probably is the one you chose since there’s a composition
where the inner function is 1 + √x.
If u= 1 + √x, then du =1
2√xdx, then when x= 1,
u= 2, and when x= 4, u= 3, so the integral becomes
Z3
2
2u3du =1
2u4
3
2=1
2(34
−24) = 65
2.
3. [10] Use the substitution u= 1 + x3to transform the
integral Z2
1
x2sin(1 + x3)
sin(1 −x3)dx into an integral with respect
to u. (Do not evaluate the integral.)
If u= 1 + x3, then du = 3x2dx. For x, the limits of
integration are 1 and 2, so for uthe limits of integration are
1 + 13= 2 and 1 + 23= 9. The factor sin(1 + x3) in the
numerator will b ecome sin u. The factor x2dx will become
1
3du. That leaves the denominator sin(1 −x3) left to be
converted. But u= 1 +x3, so x3=u−1, and 1 −x3= 2−u,
so sin(1 −x3) = sin(2 −u). Thus, the original integral is
converted to
Z9
2
sin u
3 sin(2 −u)du.
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