Math 211
7–20–2007
Review Problems for Test 1
These problems are provided to help you study. The presence of a problem on this sheet does not imply
that a similar problem will appear on the test. And the absence of a problem from this sheet does not imply
that the test will not have a similar problem.
1. Compute the following derivatives:
(a)
d
(arctan x)5 + arctan(x5 ) .
dx
(b)
d
arcsin(ex + 1).
dx
(c)
d
(tan−1 (sec−1 x)).
dx
(d)
d cos x
.
dx cos−1 x
2. Compute the following limits:
2 − 2ex−1
.
x→1 sin(x − 1)
√
(b) lim+ x ln x.
(a) lim
x→0
cos 2x
.
x2 + 3x + 1
2x
3
(d) lim 1 +
.
x→∞
x
p
x2 + 4x − x .
(e) lim
(c) lim
x→0
x→+∞
(f) lim
x→0
sin 4x + tan 5x
.
x cos 3x + 12x
x − x cos x
.
x→0 x sin x + 2x
1/x
.
(h) lim+ e2x + x
(g) lim
x→0
3. Compute
Z
(cos x)3 1 + (sin x)1/2 dx.
4. Compute
Z
(cos 7x)4 dx.
5. Compute
Z
(sec 3x)5 tan 3x dx.
6. Compute
Z
(tan x)4 dx.
1
7. Compute
Z
(csc 2x)3 (cot 2x)3 dx.
Z p
8. Compute
x2 − 1 dx.
9. Compute
Z
ln(x2 + 5) dx.
10. Compute
Z
x2
11. Compute
Z
p
x 25 − x2 dx.
12. Compute
Z
x2
√
dx.
x2 + 1
13. Compute
Z
3 + 4x + 5x2 + 3x3
dx.
x2 (x + 3)
14. Compute
Z
x3 e4x dx.
15. Compute
Z
e4x cos 2x dx.
16. Compute
Z
cos 3x sin 2x dx.
17. Compute
Z
dx
dx.
x1/2 (x1/3 + x1/4 )
18. Compute
Z
x2
p
25 − x2 dx.
x−2
dx.
− 8x + 25
19. How would you try to decompose
2(x − 2)2
using partial fractions?
x3 (x2 + 4)2
20. Why do you try the partial fractions decomposition
5x
C
A
B
+
=
+
2
2
(x − 1) (x + 1)
x − 1 (x − 1)
x+1
rather than the decomposition
B
A
5x
+
=
?
2
2
(x − 1) (x + 1)
(x − 1)
x+1
21. What is wrong with the following “partial fractions decomposition”?
7
A B
C
= + +
.
x(x − 1)
x
x
x−1
22. Find the partial fractions decomposition of
−3x4 + x3 − 6x2 − 3
.
x(x2 + 1)2
2
23. (a) Compute
−x2 + 8x − 4
dx.
x(x2 + x − 2)
Z
(b) If you use the antiderivative in (a) to compute
Z
1/2
−1
−x2 + 8x − 4
dx,
x(x2 + x − 2)
you get
[2 ln |x| + ln |x − 1| − 4 ln |x + 2|]1/2
−1 =
1
1
2 ln + ln − 4 ln 32 − (2 ln 1 + ln 2 − 4 ln 1) ≈ −4.39445.
2
2
Does this computation make sense? Why or why not?
24. Find the area of the region under y =
Z
∞
25. Compute
11
26. Compute
Z
x
from x = 0 to ∞.
(x2 + 1)2
xe−3x dx.
0
√
3
2
27. Prove that
Z
1
dx.
x−3
∞
−x4
e
dx converges. [Hint: Compare the integral to
Z
∞
e−x dx.]
0
0
Solutions to the Review Problems for Test 1
1. Compute the following derivatives:
(a)
d
(arctan x)5 + arctan(x5 ) .
dx
1
5x4
d
(arctan x)5 + arctan(x5 ) = 5(arctan x)4 · 2
.
+
dx
x + 1 1 + x10
(b)
(c)
d
arcsin(ex + 1).
dx
d
(tan−1 (sec−1 x)).
dx
d
ex
.
arcsin(ex + 1) = p
dx
1 − (ex + 1)2
d
(tan−1 (sec−1 x)) =
dx
(d)
1
1 + (sec−1 x)2
d cos x
.
dx cos−1 x
(cos
−1
d cos x
=
dx cos−1 x
x)(− sin x) − (cos x)
(cos−1 x)2
3
1
√
|x| x2 − 1
−1
√
1 − x2
.
.
2. Compute the following limits:
2 − 2ex−1
.
x→1 sin(x − 1)
(a) lim
−2ex−1
−2
2 − 2ex−1
= lim
=
= −2.
x→1 cos(x − 1)
x→1 sin(x − 1)
1
lim
(b) lim+
x→0
√
x ln x.
lim
x→0+
(c) lim
x→0
x2
√
1
ln x
x
= lim+ −2x1/2 = 0.
x ln x = lim+
= lim+
1
1
x→0
x→0
x→0
√
− 3/2
x
2x
cos 2x
.
+ 3x + 1
lim
x→0
cos 2x
= 1.
x2 + 3x + 1
L’Hôpital’s rule doesn’t apply, because plugging in x = 0 gives 1, which is not an indeterminate form.
2x
3
(d) lim 1 +
.
x→∞
x
2x
3
Set y = 1 +
, so
x
2x
3
3
.
= 2x ln 1 +
ln y = ln 1 +
x
x
Then
3
ln 1 +
3
3
x
lim ln y = lim 2x ln 1 +
=
= 2 lim x ln 1 +
= 2 lim
1
x→∞
x→∞
x→∞
x→∞
x
x
x


3
 1 
− 2

3
x
1+
x
2 lim
= 6.
x→∞
1
− 2
x
So
lim
x→∞
(e) lim
x→+∞
3
1+
x
2x
= e6 .
p
x2 + 4x − x
√x2 + 4x + x
p
p
x2 + 4x − x2
2
2
lim
= lim √
=
x + 4x − x = lim
x + 4x − x · √
2
x→+∞
x→+∞
x + 4x + x x→+∞ x2 + 4x + x
1
4x
4
4x
x
= lim √
·
=
= lim
lim √
√
2
2
1
1
x→+∞
x→+∞
x→+∞
x + 4x + x
x + 4x + x
x2 + 4x + 1
x
x
4
lim r
x→+∞
(f) lim
x→0
4
4
= = 2.
= lim r
x→+∞
2
4
x + 4x
+
1
1
+
+
1
x
x2
4
2
sin 4x + tan 5x
.
x cos 3x + 12x
4 cos 4x + 5(sec 5x)2
9
sin 4x + tan 5x
= lim
=
.
x→0 −3x sin 3x + cos 3x + 12
x→0 x cos 3x + 12x
13
lim
(g) lim
x→0
x − x cos x
.
x sin x + 2x
x − x cos x
1 + x sin x − cos x
1+0−1
= lim
=
= 0.
x sin x + 2x x→0 x cos x + sin x + 2
0+0+2
lim
x→0
(h) lim+ e2x + x
x→0
1/x
Set y = e2x + x
.
1/x
. Then
2x
ln y = ln e
+x
1/x
ln e2x + x
1
2x
.
= ln e + x =
x
x
Therefore,
2e2x + 1
2x
ln e + x
lim+ ln y = lim+
= lim+ e + x = 3.
x
1
x→0
x→0
x→0
2x
Hence,
lim+ e2x + x
x→0
3. Compute
Z
= e3 .
(cos x)3 1 + (sin x)1/2 dx.
Z
Z
(cos x)3 1 + (sin x)1/2 dx = (cos x)2 1 + (sin x)1/2 cos x dx =
Z
Z
1/x
1 − (sin x)2
u = sin x,
(1 − u2 )(1 + u1/2 ) du =
Z
1 + (sin x)1/2 cos x dx =
du = cos x dx,
du
dx =
cos x
1
2
2
(1 + u1/2 − u2 − u5/2 ) du = u + u3/2 − u3 − u7/2 + C =
3
3
7
2
1
2
sin x + (sin x)3/2 − (sin x)3 − (sin x)7/2 + C.
3
3
7
4. Compute
Z
Z
(cos 7x)4 dx.
4
(cos 7x) dx =
Z
2
2
(cos 7x) (cos 7x) dx =
5
Z
1
1
(1 + cos 14x) · (1 + cos 14x) dx =
2
2
5. Compute
Z
Z 1
1
1 + 2 cos 14x + (1 + cos 28x) dx =
1 + 2 cos 14x + (cos 14x)2 dx =
4
2
1
1
1
1
x + sin 14x + (x +
sin 28x) + C.
4
7
2
28
1
4
Z
Z
(sec 3x)5 tan 3x dx.
5
(sec 3x) tan 3x dx =
6. Compute
Z
1
(sec 3x) (sec 3x tan 3x dx) =
3
4
u = sec 3x,
du = 3 sec 3x tan 3x dx,
(tan x)4 dx =
Z
(tan x)2 (tan x)2 dx =
(tan x)2 (sec x)2 dx −
Z
(tan x)2 dx =
Z
Z
Z
u2 du − tan x + x + C =
u4 du =
Z
(tan x)2 (sec x)2 − 1 dx =
(tan x)2 (sec x)2 dx −
Z
(sec x)2 dx +
du = (sec x)2 dx,
dx =
Z
Z
(sec x)2 − 1 dx =
dx =
du
(sec x)2
1
1 3
u − tan x + x + C = (tan x)3 − tan x + x + C.
3
3
(csc 2x)3 (cot 2x)3 dx.
Z
Z
Z
(tan x)2 (sec x)2 dx −
u = tan x,
7. Compute
1
1 5
u +C =
(sec 3x)5 + C.
15
15
du
dx =
3 sec 3x tan 3x
Z
(tan x)4 dx.
Z
Z
Z
3
3
(csc 2x) (cot 2x) dx =
Z
(csc 2x)2 (cot 2x)2 (csc 2x cot 2x) dx =
(csc 2x)2 (csc 2x)2 − 1 (csc 2x cot 2x) dx =
u = csc 2x,
Z
u2 (u2 − 1)(csc 2x cot 2x) ·
du
=
−2 csc 2x cot 2x
du
−2 csc 2x cot 2x
Z
Z
1
1 1
1
1
1 1 5 1 3
5
3
2 2
4
2
−
u − u +C = −
(csc 2x) − (csc 2x) + C.
u (u − 1) du = −
(u − u ) du = −
2
2
2 5
3
2 5
3
8. Compute
du = −2 csc 2x cot 2x dx,
dx =
Z p
x2 − 1 dx.
Z p
Z p
Z
Z p
2
2
2
x − 1 dx =
(sec θ) − 1 sec θ tan θ dθ =
(tan θ) sec θ tan θ dθ = sec θ(tan θ)2 dθ =
6
x
x = sec θ,
x 2- 1
dx = sec θ tan θ dθ
q
1
Z
sec θ (sec θ)2 − 1 dθ =
Z
1
1
(sec θ)3 − sec θ dθ = sec θ tan θ + ln | sec θ +tan θ|−ln | sec θ +tan θ|+C =
2
2
p
1
1
1
1 p
sec θ tan θ − ln | sec θ + tan θ| + C = x x2 − 1 − ln |x + x2 − 1| + C.
2
2
2
2
9. Compute
Z
ln(x2 + 5) dx.
Z
d
dx
+ ln(x2 + 5)
Z
ln(x2 + 5) dx = x ln(x2 + 5) −
1
ց
2x
+5
−
x
x2
Z
dx
2x2
dx = x ln(x2 + 5) −
2
x +5
Z 2−
10
x2 + 5
dx =
x
10
x ln(x2 + 5) − 2x + √ tan−1 √ + C.
5
5
The second equality comes from dividing 2x2 by x2 + 5 (long division). Alternatively, you can do this:
2x2
2x2 + 10 − 10
2x2 + 10
10
2(x2 + 5)
10
10
=
=
−
=
− 2
=2− 2
.
x2 + 5
x2 + 5
x2 + 5
x2 + 5
x2 + 5
x +5
x +5
10. Compute
Z
Z
x2
p
25 − x2 dx.
Z
Z
p
p
p
2
2
2
x 25 − x dx = 25(sin θ) 25 − 25(sin θ) (5 cos θ) dθ = 25(sin θ)2 25(cos θ)2 (5 cos θ) dθ =
2
5
x = 5 sin θ,
x
dx = 5 cos θ dθ
q
25 - x 2
Z
1
625
1
(1 − cos 2θ) · (1 + cos 2θ) dθ =
1 − (cos 2θ)2 dθ =
2
2
4
Z
Z
1
625
1
625
625
(1 − cos 4θ) dθ =
(θ − sin 4θ) + C =
(sin 2θ)2 dθ =
4
4
2
8
4
625
625
x 1 p
25 − x2 (25 − 2x2 ) + C.
θ − sin θ cos θ 2(cos θ)2 − 1 + C =
arcsin − x
8
8
5 8
625
Z
(sin θ)2 (cos θ)2 dθ = 625
Z
7
11. Compute
Z
Z
p
x 25 − x2 dx.
Z
Z
p
√
1
1
1
du
2
x 25 − x dx = x u ·
=−
u1/2 du = − u3/2 + C = − (25 − x2 )3/2 + C.
−2x
2
3
3
u = 25 − x2 ,
12. Compute
Z
Z
du = −2x dx,
du
dx =
−2x
x2
√
dx.
x2 + 1
x2
√
dx =
x2 + 1
Z
(tan θ)2
p
(sec θ)2 dθ =
(tan θ)2 + 1
x = tan θ,
Z
(tan θ)2
p
(sec θ)2 dθ =
(sec θ)2
x2 + 1
dx = (sec θ)2 dθ
Z
(tan θ)2
(sec θ)2 dθ =
sec θ
x
q
1
Z
sec θ(tan θ)2 dθ =
Z
sec θ (sec θ)2 − 1 dθ =
Z
(sec θ)3 − sec θ dθ =
p
1
1 p
1
1
sec θ tan θ − ln | sec θ + tan θ| + C = x x2 + 1 − ln | x2 + 1 + x| + C.
2
2
2
2
13. Compute
Z
3 + 4x + 5x2 + 3x3
dx.
x2 (x + 3)
The top and the bottom both have degree 3, so I must divide the top by the bottom:
3 + 4x + 5x2 + 3x3
3 + 4x − 4x2
=3+
.
2
x (x + 3)
x2 (x + 3)
I’ll put the 3 aside for now, and work on the fraction:
C
A
B
3 + 4x − 4x2
= + 2+
.
2
x (x + 3)
x
x
x+3
Clear denominators:
3 + 4x − 4x2 = Ax(x + 3) + B(x + 3) + Cx2 .
Set x = 0: I get 3 = 3B, so B = 1.
Set x = −3: I get −45 = 9C, so C = −5.
Plug B and C back in:
3 + 4x − 4x2 = Ax(x + 3) + (x + 3) − 5x2 .
Differentiate:
4 − 8x = A(x + 3) + Ax + 1 − 10x.
8
Set x = 0: I 4 = 3A + 1, so A = 1.
Therefore,
3 + 4x − 4x2
5
1
1
= + 2−
.
x2 (x + 3)
x x
x+3
Hence,
3 + 4x + 5x2 + 3x3
1
1
5
= 3+ + 2 −
.
x2 (x + 3)
x x
x+3
Finally,
Z
Z 3 + 4x + 5x2 + 3x3
1
1
5
1
3
+
dx = 3x + ln |x| − − 5 ln |x + 3| + C.
−
dx
=
+
x2 (x + 3)
x x2
x+3
x
14. Compute
Z
x3 e4x dx.
Z
d
dx
+
x3
− 3x
+
−
2
6x
6
dx
e4x
ց
ց
ց
ց
1 4x
e
4
1 4x
e
16
1 4x
e
64
1 4x
e
256
Z
1
3
6
6 4x
x3 e4x dx = x3 e4x − x2 e4x + xe4x −
e + C.
4
16
64
256
+
15. Compute
Z
0
e4x cos 2x dx.
Z
d
dx
+
−
e4x
cos 2x
−2 sin 2x
ց
ց
+ −4 cos 2x →
Z
dx
1 4x
e
4
1 4x
e
16
Z
1 4x
1 4x
1
e cos 2x dx = e cos 2x + e sin 2x −
e4x cos 2x dx,
4
8
4
Z
1
1
5
e4x cos 2x dx = e4x cos 2x + e4x sin 2x,
4
4
8
Z
1
1
e4x cos 2x dx = e4x cos 2x + e4x sin 2x + C.
5
10
4x
9
16. Compute
Z
cos 3x sin 2x dx.
Z
d
dx
+
−
cos 3x
−3 sin 3x
sin 2x
ց
1
− cos 2x
2
ց
1
− sin 2x
4
+ −9 cos 3x →
Z
17. Compute
Z
dx
Z
3
9
1
cos 3x sin 2x dx,
cos 3x sin 2x dx = − cos 3x cos 2x − sin 3x sin 2x +
2
4
4
Z
3
1
5
cos 3x sin 2x dx = − cos 3x cos 2x − sin 3x sin 2x,
−
4
2
4
Z
3
2
cos 3x sin 2x dx = cos 3x cos 2x + sin 3x sin 2x + C.
5
5
dx
dx.
x1/2 (x1/3 + x1/4 )
Since the least common multiple of 2, 3, and 4 is 12, I’ll let x = u12 :
Z
dx
dx =
x1/2 (x1/3 + x1/4 )
12u11 du
= 12
u6 (u4 + u3 )
Z
u2
du = 12
u+1
Z
Z u−1+
1
u+1
du =
x = u12 , dx = 12u11 du
1 1/6
1 2
u − u + ln |u + 1| + C = 12
x − x1/12 + ln |x1/12 + 1| + C.
12
2
2
18. Compute
Since
Z
x2
x−2
dx.
− 8x + 25
1
· (−8) = −4 and (−4)2 = 16, I have
2
x2 − 8x + 25 = x2 − 8x + 16 + 9 = (x − 4)2 + 9.
Therefore,
Z
x−2
dx =
x2 − 8x + 25
Z
x−2
dx =
(x − 4)2 + 9
[u = x − 4,
Z
u
du +
u2 + 9
Z
Z
du = dx;
(u + 4) − 2
du =
u2 + 9
Z
u+2
du =
u2 + 9
x = u + 4]
2
1
2
u
1
2
x−4
du = ln |u2 + 9| + tan−1 + C = ln |(x − 4)2 + 9| + tan−1
+ C.
u2 + 9
2
3
3
2
3
3
I did the first part of the u-integral using the substitution w = u2 + 9.
10
19. How would you try to decompose
2(x − 2)2
using partial fractions?
x3 (x2 + 4)2
2(x − 2)2
A
C
Dx + E
B
Fx + G
= + 2+ 3+ 2
.
+ 2
x3 (x2 + 4)2
x
x
x
x +4
(x + 4)2
20. Why do you try the partial fractions decomposition
C
A
B
5x
+
=
+
(x − 1)2 (x + 1)
x − 1 (x − 1)2
x+1
rather than the decomposition
5x
B
A
+
=
?
(x − 1)2 (x + 1)
(x − 1)2
x+1
Partial fractions is the opposite of combining fractions over a common denominator. In this case, the
5x
?” Both of the decompositions above could
question is: “What fractions would add up to
(x − 1)2 (x + 1)
occur, since both have (x − 1)2 (x + 1) as the common denominator.
However, since you don’t know beforehand what the fractions are, you must assume the “worst case”
A
— namely, that there might be an
term.
x−1
21. What is wrong with the following “partial fractions decomposition”?
A B
C
7
= + +
.
x(x − 1)
x
x
x−1
D
The first two terms could be combined into a single term , so they’re redundant. There is no reason
x
to list the same denominator twice.
22. Find the partial fractions decomposition of
−3x4 + x3 − 6x2 − 3
.
x(x2 + 1)2
Try the decomposition
A Bx + C
−3x4 + x3 − 6x2 − 3
Dx + E
= + 2
.
+ 2
x(x2 + 1)2
x
x +1
(x + 1)2
Clear denominators:
−3x4 + x3 − 6x2 − 3 = A(x2 + 1)2 + (Bx + C)(x)(x2 + 1) + (Dx + E)(x).
Set x = 0: I get A = −3. Plug it back in:
−3x4 + x3 − 6x2 − 3 = −3(x2 + 1)2 + (Bx + C)(x)(x2 + 1) + (Dx + E)(x).
11
Differentiate:
12x3 + 3x2 − 12x = −12x(x2 + 1) + B(x3 + x) + (Bx + C)(3x2 + 1) + 2Dx + E.
Set x = 0: I get C + E = 0.
Differentiate again:
−36x2 + 6x − 12 = −36x2 − 12 + B(3x2 + 1) + B(3x2 + 1) + (Bx + C)(6x) + 2D.
Set x = 0: I get B + D = 0.
Cancel the −36x2 and −12 terms in the previous equation, then differentiate:
6x = B(3x2 + 1) + B(3x2 + 1) + (Bx + C)(6x) + 2D,
6 = 6Bx + 6Bx + 6Bx + (Bx + C)(6).
Set x = 0: I get C = 1. Since C + E = 0, it follows that E = −1.
Plug C = 1 back in, then simplify the equation:
6 = 24Bx + 6,
or
0 = 24Bx.
Set x = 1: I get B = 0. But B + D = 0, so D = 0.
Hence,
−3x4 + x3 − 6x2 − 3
3
1
1
=− + 2
.
−
x(x2 + 1)2
x x + 1 (x2 + 1)2
23. (a) Compute
Z
−x2 + 8x − 4
dx.
x(x2 + x − 2)
x(x2 + x − 2) = x(x − 1)(x + 2), so I try
A
B
C
−x2 + 8x − 4
= +
+
.
2
x(x + x − 2)
x
x−1 x+2
Clear denominators:
−x2 + 8x − 4 = A(x − 1)(x + 2) + Bx(x + 2) + Cx(x − 1).
Set x = 0: I get −4 = −2A, or A = 2.
Set x = 1: I get 3 = 3B, or B = 1.
Set x = −2: I get −24 = 6C, or C = −4.
Therefore,
Z
Z −x2 + 8x − 4
2
1
4
dx = 2 ln |x| + ln |x − 1| − 4 ln |x + 2| + C.
dx =
+
−
x(x2 + x − 2)
x x−1 x+2
(b) If you use the antiderivative in (a) to compute
Z
1/2
−1
−x2 + 8x − 4
dx,
x(x2 + x − 2)
you get
1/2
[2 ln |x| + ln |x − 1| − 4 ln |x + 2|]−1 =
1
1
2 ln + ln − 4 ln 32 − (2 ln 1 + ln 2 − 4 ln 1) ≈ −4.39445.
2
2
12
Does this computation make sense? Why or why not?
The computation is incorrect, because the antiderivative is valid only within intervals which don’t
1
contain the singularities at x = −2, x = 0, and x = 1. The interval −1 ≤ x ≤ includes the singularity at
2
x = 0. It is not legal to simply “plug in the endpoints” — to do this definite integral correctly, you should
set it up as two improper integrals.
24. Find the area of the region under y =
(x2
x
from x = 0 to ∞.
+ 1)2
y=
x
(x 2+ 1)2
The area is
A=
Z
∞
0
x
dx = lim
2
c→∞
(x + 1)2
Z
c
0
c
x
1
1
1
1
dx
=
lim
−
lim
−
1
= .
=
−
2
2
2
2
c→∞
(x + 1)
2(x + 1) 0
2 c→∞ c + 1
2
Here’s the work for the antiderivative:
Z
Z
Z
x
x du
du
1
1
1
dx =
·
=−
=
+C =−
+ C.
2
2
2
2
2
(x + 1)
u 2x
2
u
2u
2(x + 1)
u = x2 + 1,
25. Compute
Z
du = 2x dx,
du
dx =
2x
∞
xe−3x dx.
0
Z
∞
xe−3x dx = lim
b→+∞
0
Z
b
1
1
− xe−3x − e−3x =
b→+∞
3
9
0
b
xe−3x dx = lim
0
1
1
1
1
1
= −0 − 0 + = .
− be−3b − e−3b +
b→+∞
3
9
9
9
9
lim
Here’s the work for the two limits:
lim e−3b = lim
b→+∞
b→+∞
lim be−3b = lim
b→+∞
b→+∞
1
= 0,
e3b
b
1
= lim
= 0.
b→+∞ 3e3b
e3b
I used L’Hôpital’s Rule to compute the second limit.
13
Here’s the work for the antiderivative:
d
dx
Z
x
e−3x
+
Z
26. Compute
Z
11
√
3
2
−
1
+
0
ց
ց
dx
1
− e−3x
3
1 −3x
e
9
1
1
xe−3x dx = − xe−3x − e−3x + C.
3
9
1
dx.
x−3
Z
11
√
3
2
1
dx =
x−3
Z
2
3
√
3
1
dx +
x−3
Z
11
3
√
3
1
dx.
x−3
The first integral is
3
Z
2
1
√
dx = lim
3
a→3−
x−3
a
Z
2
a
3
1
3
3
2/3
2/3
√
dx
=
lim
(x
−
3)
lim
(a
−
3)
−
1
=− .
=
3
a→3−
a→3−
2
2
2
x−3
2
The second integral is
Z
3
11
1
√
dx = lim
3
b→3+
x−3
Z
b
11
11
3
3
1
2/3
2/3
√
4
−
(b
−
3)
= 6.
(x
−
3)
lim
=
dx
=
lim
3
b→3+ 2
2 b→3+
x−3
b
Therefore,
Z
2
11
√
3
1
9
3
dx = − + 6 = .
2
2
x−3
y =3
1
x-3
1
has a vertical asymptote at x = 3, but the (signed) area on each side is finite.
x−3
The negative area to the left of x = 3 partially cancels the positive area to the right of x = 3. Thus, the
integral in this problem does not represent the actual area bounded by the graph, the x-axis, and the lines
x = 2, x = 3, and x = 11.
The graph of √
3
27. Prove that
Z
∞
4
e−x dx converges.
0
14
4
The interval of integration is x ≥ 0. On this interval, x ≤ x4 , so −x ≥ −x4 , and e−x ≥ e−x . Therefore,
Z
0
Now
Z
∞
−x
e
0
Since
Z
dx = lim
b→∞
Z
0
b
∞
e−x dx ≥
∞
4
e−x dx.
0
b
e−x dx = lim −e−x 0 = lim −e−b + 1 = −0 + 1 = 1.
b→∞
∞
e−x dx converges, (*) shows that
0
Z
Z
b→∞
∞
4
e−x dx converges as well.
0
Only a life lived for others is a life worthwhile. - Albert Einstein
c 2008 by Bruce Ikenaga
15
(∗)