Download Test 1 Solutions

Solutions to MATH 271 Test #1H
This is the 2:40 class’s version of the test. See pages 4–6 for the 4:40 class’s version.
(1) (20 points) Evaluate
Z
x2 sin(4x) dx.
Solution: This is a mixture of a polynomial and a trig function, so Integration by
Parts needs to be used. The substitution y = 4x (u will be used in the Integration by Parts later, so it’s an iffy letter to use) could be done first; see below for
information about this.
If we do Integration by Parts right away, then we want u = x2 and
v 0 = sin(4x), so that the degree of the polynomial will go down. Then u0 = 2x
− cos(4x)
, and
and v =
4
Z
Z
− cos(4x)
− cos(4x)
2
2
x sin(4x) dx = (x )
− 2x ·
dx
4
4
Z
−x2 cos(4x) 1
=
+
x cos(4x) dx.
4
2
We do this new integral with Integration by Parts as well. Now u = x,
sin(4x)
v 0 = cos(4x), and so u0 = 1 and v =
. Then
4
Z
Z
−x2 cos(4x) 1
2
+
x cos(4x) dx (from above)
x sin(4x) dx =
4
2
Z
−x2 cos(4x) 1
sin(4x)
sin(4x)
=
+
x·
− 1·
dx
4
2
4
4
Z
−x2 cos(4x) x sin(4x) 1
=
+
−
cos(4x) dx.
4
8
8
The last integral (and the one above for sin(4x)) can be done with the substitution
du
du
= 4, so dx =
. This gives us the final answer
u = 4x, which means that
dx
4
Z
Z
cos(u)
−x2 cos(4x) x sin(4x) 1
2
x sin(4x) dx =
+
−
du
4
8
8
4
−x cos(4x) x sin(4x) 1 1
=
+
− · sin(u) + C
4
8
8 4
−x cos(4x) x sin(4x) cos(4x)
=
+
−
+ C.
4
8
32
1
dy
If you do the substitution y = 4x first, then you end up with
= 4, so
dx
dy
dx =
, and we get the integral
4
Z
Z
Z 2
Z
dy
y
dy
1
2
2
x sin(4x) dx = x sin(y) ·
=
sin y ·
=
y 2 sin y dy.
4
4
4
64
(Note that we had to solve for x in terms of y to get rid of it after doing the
substitution; this is what I have sometimes called “substitution plus.”) Now the
problem can be done by using Integration by Parts twice; then the y’s have to be
replaced by 4x’s in the final answer.
Grading: +8 points for each Integration by Parts, +4 points for the substitution y = 4x. Grading for common mistakes usually resulted in a few points off
for algebra mistakes, or forgetting the “ + C” at the end.
(NOTE: Yes, the explanation took more than a page. You didn’t have to
include the text parts on the test, though, and you didn’t have to say how you
would have graded the problem, either.
(2) (15 points) Evaluate
Z
sin5 θ cos20 θ dθ.
Solution: This is trigonometric integral of the form
Z
sinm θ cosn θ dθ, so we check
to see whether this is an “easy case”, i.e., whether one of the exponents is odd.
Since the exponent of sin5 θ is odd, we keep one sin θ factor and convert the rest
to cos θ’s. Then we can do the substitution u = cos θ to turn the integral into a
polynomial, integrate it, then replace the u’s with cos θ.
So we do this:
Z
Z
5
20
sin θ cos θ dθ = sin4 θ cos20 θ · sin θ dθ
Z
Z
2
2
20
= (1 − cos θ) cos θ · sin θ dθ = −(1 − u2 )2 u20 du
Z
Z
2
4 2
= −(1 − 2u + u )u 0 du = u20 − 2u22 + u24 du
=
2u23
u25
− cos21 θ 2 cos23 θ cos25 θ
−u21
+
−
+C =
+
−
+ C.
21
23
25
21
23
25
Grading: +5 points for determining that the exponent of sin θ was odd,
+5 points for setting up the substitution u = cos θ, +5 points for the final answer.
Common mistakes were usually algebraic mistakes, with a few points taken off.
2
(3) (20 points) Evaluate
Z
tan−1 (4t) dt.
Solution:
This is the integral of a single function, but we can write it as
Z
1 · tan−1 (4t) dt to turn it into an Integration by Parts problem. The substitution y = 4t (dy/dt = 4) makes life a little easier, so we actually do that first.
1
Then we do integration by parts with u = tan−1 y and v 0 = 1, so that u0 =
1 + y2
and v = y:
Z
−1
tan
Z
Z
dy
1
(4t) dt = tan (y)
=
1 · tan−1 y dy
4
4
Z
Z
1
1
1
y
−1
−1
(tan y)(y) −
· y dy =
y tan y −
dy .
=
4
1 + y2
4
1 + y2
−1
Now, we can do this integral by substitution; let u = 1 + y 2 . Then du/dy = 2y,
so dy = du/2y. Then
Z
y
dy =
1 + y2
Z
y du
1
·
=
u 2y
2
Z
1
1
du = ln u.
u
2
Now we put this into the equation above it, and convert everything back into t’s,
getting:
Z
tan−1 (4t) dt =
2
y tan−1 y 1
(4t) tan−1 (4t) 1
− ln u + C =
− ln 1 + (4t)2 + C.
4
8
4
8
Grading: This was a hard one to grade. I had intended +7 points for the
(eventual) substitution
for 4t, +7 points for Integration by Parts, and +6 points to
Z
y
do the integral
dy. Some people didn’t know the derivative of tan−1 (x);
1 + y2
if they gave the correct procedure, then they got +12 points (total). Some thought
1
that tan−1 x meant
and generally got +10 points (total) if they included
tan x
this as part of the problem.
3
(4) (15 points) Give the form of the Partial Fraction Decomposition of the following
rational function. You do not have to find the constants.
(3x − 4)2 (x − 5)2 (x2 − x + 1)
(x − 1)2 (x + 2)(2x2 + 4x + 4)2
Solution: Since 2x2 + 4x + 4 has no real roots, the form of the Partial Fraction
Decomposition is:
A
C
Fx + G
B
Dx + E
+
+
+ 2
+
.
2
2
x − 1 (x − 1)
x + 2 2x + 4x + 4 (2x + 4x + 4)2
You did not have to find A, B, C, etc.; this is the form.
Grading: Points were taken off based on the number of types of mistakes
made, not the actual number. (So if you forgot the Dx and F x in the decomposition above, you were only penalized once.) If you made one type of mistake, you
got +12 points (total); if you made two types of mistakes, you got +10 points
(total); et cetera. (For this problem, the only possible scores were 0, 3, 5, 7, 10,
12, 14 (for an arithmetic or algebra mistake), and 15.)
(5) (15 points) Evaluate
Z
3x −
4
x+2
+ 2
dx.
x − 3 x + 4x + 6
Solution: A lot of people made this harder than it actually was. First of all,
Z
Z
Z
Z
4
x+2
4
x+2
+ 2
dx = 3x dx −
dx +
dx
3x −
2
x − 3 x + 4x + 6
x−3
x + 4x + 6
Z
3x2
x+2
=
− 4 ln(x − 3) +
dx.
2
2
x + 4x + 6
The last integral can be done with the substitution u = x2 + 4x + 6, since
du/dx = 2x + 4 = 2(x + 2). Then
Z
x+2
dx =
2
x + 4x + 6
Z
x+2
du
1
·
=
u
2(x + 2)
2
Z
1
1
1
du = ln u = ln(x2 + 4x + 6),
u
2
2
3x2
1
− 4 ln(x − 3) + ln(x2 + 4x + 6) + C.
2
2
Grading: +4 points for the integrals of 3x and 4/(x − 3), +3 points for the
substitution u = x2 + 4x + 6, +4 points for putting everything together. Partial
credit: +8 points (total) if no work was done on the last term; +12 points if some
work was done on the last term, but there was no final answer.
so the final answer is
4
(6) (15 points) Do a trigonometric substitution with the following integral. You do not
need to find an antiderivative. Your final answer should involve only trigonometric
functions and not have any x’s in it.
Z
√
3x 4 − x2
√
dx
2x + 4 − x2 + 1
p
p
Solution: Since this integral has 4 − x2 = 22 − x2 in it, the substitution to
try is x = 2 sin θ. (Using x = −2 sin θ or x = 2 cos θ as a trigonometric substitution works, but is not “standard.” You will get a slightly
different answer for
p
each of these.) Then we also have dx = 2 cos θ dθ and 4 − x2 = 3 cos θ. We use
these equations to replace x’s with θ’s everywhere:
Z
√
Z
3x 4 − x2
3(2 sin θ)(2 cos θ)
√
· (2 cos θ) dθ
dx =
2
2(2 sin θ) + (2 cos θ) + 1
2x + 4 − x + 1
Z
sin θ cos2 θ
dθ.
= 24
4 sin θ + 2 cos θ + 1
(You did not have to do the simplifying after the second = sign.)
p Grading: +5 points for choosing x = 3 sin θ, +5 points for stating dx and
4 − x2 , +5 points for doing the actual substitution. I graded the integral you
got right after substitution; some people continued from here, but always incorrectly. The function has a nice antiderivative but requires a lot of work (or
Maple).*
* In case you’re interested, after you enter
int (24 * sin(theta)*cos(theta)^2/(4*sin(theta)+2*cos(theta)+1), theta);
into Maple 9.5, you get
!
2
tan 21θ
207
1
12 4 tan 21θ − 2
72
ln
tan
+1 +
2 +
2
5 tan 1 2 + 1
125
2θ
5
1
tan 2 θ
+1
2θ
!
2
6 −88 tan 21θ + 14 207
1
1
+
−
ln −8 tan
+ tan
−3
25 tan 1 2 + 1
125
2θ
2θ
2θ
√
468 √
1
1
174
−1
−
19 tanh
−8 + 2 tan
19 +
θ,
2375
38
2θ
125
in terms of θ. The antiderivative in terms of x is an even bigger mess.
5
Solutions to MATH 271 Test #1J
(1) (20 points) Evaluate
Z
t2 e6t dt.
Solution: This is a mixture of a polynomial and a trig function, so Integration by
Parts needs to be used. The substitution y = 6t (u will be used in the Integration by Parts later, so it’s an iffy letter to use) could be done first; see below for
information about this.
If we do Integration by Parts right away, then we want u = t2 and v 0 = e6t ,
e6t
so that the degree of the polynomial will go down. Then u0 = 2t and v =
, and
6
6t Z
Z
Z
e
e6t
te6t
1
2 6t
2
t e dt = (t )
− 2t ·
dt =
−
te6t dt.
6
6
6
3
We do this new integral with Integration by Parts as well. Now u = t, v 0 = e6t ,
e6t
and so u0 = 1 and v =
. Then
6
Z
Z
t2 e6t
1
2 6t
t e dt =
−
te6t dt (from above)
6
3
Z
Z
2 6t
t e
1
e6t
e6t
t2 e6t
te6t
1
=
−
t·
− 1·
dx =
−
+
e6t dt.
6
3
6
6
6
18
18
The last integral (and the one above for sin(4x)) can be done with the substitution
du
du
u = 6t, which means that
= 6, so dt =
. This gives us the final answer
dt
6
Z
Z u
te6t
1
e
t2 e6t
2 6t
t e dt =
−
+
du
6
18
18
6
t2 e6t
te6t
1 1 u
t2 e6t
te6t
e6t
=
−
+
· e +C =
−
+
+ C.
6
18
18 6
6
18
108
dy
If you do the substitution y = 6t first, then you end up with
= 6, so
dt
dy
dt =
, and we get the integral
6
Z 2
Z
Z
Z
y
1
2 6t
2 y dy
y dy
t e dt = t e ·
=
e ·
=
y 2 ey dy.
6
6
6
216
(Note that we had to solve for t in terms of y to get rid of it after doing the
substitution; this is what I have sometimes called “substitution plus.”) Now the
1
problem can be done by using Integration by Parts twice; then the y’s have to be
replaced by 6t’s in the final answer.
Grading: +8 points for each Integration by Parts, +4 points for the substitution y = 6t (where needed). Grading for common mistakes usually resulted in
a few points off for algebra mistakes, or forgetting the “ + C” at the end.
(NOTE: Yes, the explanation took more than a page. You didn’t have to
include the text parts on the test, though, and you didn’t have to say how you
would have graded the problem, either.
Z
(2) (20 points) Evaluate sin−1 (3t) dt.
Solution:
This is the integral of a single function, but we can write it as
Z
1 · sin−1 (3t) dt to turn it into an Integration by Parts problem. The substitution y = 3t (dy/dt = 3) makes life a little easier, so we actually do that first. Then
1
we do integration by parts with u = sin−1 y and v 0 = 1, so that u0 = p
1 − y2
and v = y:
Z
Z
Z
dy
1
−1
−1
=
1 · sin−1 y dy
sin (3t) dt = sin (y)
3
3
!
Z
1
1
p
=
(sin−1 y)(y) −
· y dy
3
1 − y2
!
Z
1
y
−1
p
y sin y −
dy .
=
3
1 − y2
Now, we can do this integral by substitution; let u = 1 − y 2 . Then du/dy = −2y,
so dy = du/(−2y). Then
Z
Z
Z
√
y
y
du
−1
−1 u1/2
−1/2
p
dy =
·
=
u
du
=
·
=
−
u.
2
2
1/2
u1/2 −2y
1 − y2
Now we put this into the equation above it, and convert everything back into t’s,
getting:
Z
y sin−1 y 1 √
(3t) sin−1 (4t) 1 p
sin−1 (3t) dt =
+
u+C =
+
1 − (3t)2 + C.
3
3
3
3
Grading: This was a hard one to grade. I had intended +7 points for the
(eventual) substitution
Z for 3t, +7 points for Integration by Parts, and +6 points
y
p
to do the integral
dy. Some people didn’t know the derivative of
1 − y2
sin−1 (x); if they gave the correct procedure, then they got +12 points (total).
1
Some thought that sin−1 x meant
and generally got +10 points (total) if
sin x
they included this as part of the problem.
2
(3) (15 points) Evaluate
Z
sin10 θ cos5 θ dθ.
Solution: This is trigonometric integral of the form
Z
sinm θ cosn θ dθ, so we check
to see whether this is an “easy case”, i.e., whether one of the exponents is odd.
Since the exponent of cos5 θ is odd, we keep one cos θ factor and convert the rest
to sin θ’s. Then we can do the substitution u = sin θ to turn the integral into a
polynomial, integrate it, then replace the u’s with sin θ.
So we do this:
Z
Z
10
5
sin θ cos θ dθ = sin10 θ cos4 θ · cos θ dθ
Z
Z
10
2
2
= sin θ(1 − sin θ) · cos θ dθ = u10 (1 − u2 )2 du
Z
Z
10
2
4
= u (1 − 2u + u ) du = u10 − 2u12 + u14 du
u11
2u13
u15
sin11 θ 2 sin13 θ sin15 θ
=
−
+
+C =
−
+
+ C.
11
13
15
11
13
15
Grading: +5 points for determining that the exponent of cos θ was odd,
+5 points for setting up the substitution u = sin θ, +5 points for the final answer.
Common mistakes were usually algebraic mistakes, with a few points taken off.
3
(4) (15 points) Do a trigonometric substitution with the following integral. You do not
need to find an antiderivative. Your final answer should involve only trigonometric
functions and not have any x’s in it.
Z
√
2x x2 − 9
√
dx
3x + x2 − 9 − 4
p
p
Solution: Since this integral has x2 − 9 = x2 − 32 in it, the substitution to
try is x = 3 sec θ. (It is also possible to start off with x = −3 sec θ, and you
get
p a slightly different integral.) Then we also have dx = 3 sec θ tan θ dθ and
x2 − 9 = 3 tan θ. We use these equations to replace x’s with θ’s everywhere:
Z
√
Z
2x x2 − 9
2(3 sec θ)(3 tan θ)
√
dx =
(3 sec θ tan θ) dθ
2
3(3 sec θ) + (3 tan θ) − 4
3x + x − 9 − 4
Z
sec2 θ tan2 θ
= 54
dθ.
9 sec θ + 3 tan θ − 4
(You did not have to do the simplifying after the second = sign.)
p Grading: +5 points for choosing x = 3 sec θ, +5 points for stating dx and
x2 − 9, +5 points for doing the actual substitution. I graded the integral you
got right after substitution; some people continued from here, but always incorrectly. The function has a nice antiderivative but requires a lot of work (or
Maple).*
* In case you’re interested, after you enter
int (54 * sec(theta)^2 * tan(theta)^2 / (9 * sec(theta)+3*tan(theta)-4), theta);
into Maple 9.5, you get
−1
−2
−2
3
1
9
1
9
1
tan
+1
−
tan
+1
+
tan
−1
−
2
2θ
2
2θ
4
2θ
!
2
√
1
1
225 √
1
25
1
−1
+
ln 13 tan
+ 5 + 6 tan
−
14 tan
26 tan
+6
14
32
2θ
2θ
112
56
2θ
−1
3
1
11
1
19
1
+
tan
−1
−
ln tan
+1 +
ln tan
−1 ,
4
2θ
4
2θ
16
2θ
in terms of θ. The antiderivative in terms of x is an even bigger mess.
4
(5) (15 points) Evaluate
Z
x2 +
7
4x + 6
− 2
dx.
x + 4 x + 3x + 10
Solution: A lot of people made this harder than it actually was. First of all,
Z
Z
Z
Z
7
4x + 6
7
4x + 6
2
2
x +
− 2
dx = x dx +
dx −
dx
2
x + 4 x + 3x + 10
x+4
x + 3x + 10
Z
x3
4x + 6
=
+ 7 ln(x + 4) −
dx.
2
3
x + 3x + 10
The last integral can be done with the substitution u = x2 + 3x + 10, since
du/dx = 2x + 3. Then
Z
4x + 6
dx =
2
x + 3x + 10
Z
2(2x + 3)
du
·
=2
u
2x + 3
Z
1
du = 2 ln u = 2 ln(x2 + 3x + 10),
u
x3
+ 7 ln(x + 4) − 2 ln(x2 + 3x + 10) + C.
3
Grading: +4 points for the integrals of 3x and 7/(x + 4), +3 points for the
substitution u = x2 + 3x + 10, +4 points for putting everything together. Partial
credit: +8 points (total) if no work was done on the last term; +12 points if some
work was done on the last term, but there was no final answer.
so the final answer is
(6) (15 points) Give the form of the Partial Fraction Decomposition of the following
rational function. You do not have to find the constants.
(2x + 3)(x − 4)2 (x2 + 1)63 1
(x + 1)(x − 2)2 (3x2 + 6x + 9)2
Solution: Since 3x2 + 6x + 9 has no real roots, the form of the Partial Fraction
Decomposition is:
A
B
C
Dx + E
Fx + G
+
+
+ 2
+
.
2
2
x + 1 x − 2 (x − 2)
3x + 6x + 9 (3x + 6x + 9)2
You did not have to find A, B, C, etc.; this is the form.
Grading: Points were taken off based on the number of types of mistakes
made, not the actual number. (So if you forgot the Dx and F x in the decomposition above, you were only penalized once.) If you made one type of mistake, you
got +12 points (total); if you made two types of mistakes, you got +10 points
(total); et cetera. (For this problem, the only possible scores were 0, 3, 5, 7, 10,
12, 14 (for an arithmetic or algebra mistake), and 15.)
5