Math 148 Exam I Practice Problems
This review should not be used as your sole source for preparation for the exam. You
should also re-work all examples given in lecture, all homework problems, all lab assignment
problems, and all quiz problems. These problems were provided courtesy of Dr. Peter Howard.
Solutions were written by Dr. Glenn Lahodny Jr.
Z
1. Evaluate (ln x)2 dx.
Z
2. Evaluate
0
2
2x
dx.
(x2 − 1)1/3
Z
4x2 + 13x − 9
dx.
x3 + 2x2 − 3x
Z
5x2 + 4x + 3
dx.
x(x2 + x + 1)
3. Evaluate
4. Evaluate
Z
5. Evaluate
√1
3
tan−1 x dx.
0
Z
6. Evaluate
2
x5 ex dx.
7. (a) Find the third order Taylor polynomial for f (x) = tan−1 x at x = 0.
π
(b) Use your result in part (a) to approximate the values of and π.
4
8. Suppose that f (x) = ln x is approximated by a second order Taylor polynomial near
x = 1. Find the maximum possible error of this approximation for x ∈ [1, 2].
9. Determine the smallest order n that will ensure that the nth order Taylor polynomial
for f (x) = x ln x − x near x = 1 has a maximum error less than 0.01 for x ∈ [1, 2].
10. Solve the differential equation
dy
= y cos x with initial condition y(0) = 1.
dx
1
Solutions
Z
1. Evaluate
(ln x)2 dx.
Method 1: Let w = ln x, so that dw = x1 dx. Then
Z
Z
2
(ln x) dx =
w2 x dw
Z
=
w2 ew dw.
Let u = w2 and dv = ew dw, from which du = 2w dw and v = ew . Then
Z
Z
2 w
2 w
w e dw = w e − 2wew dw.
Let u = 2w and dv = ew dw, from which du = 2 dw and v = ew . Then
Z
Z
2 w
2 w
w
w
w e dw = w e − 2we − 2e dw
= w2 ew − 2wew + 2ew + C
= (ln x)2 eln x − 2(ln x)eln x + 2eln x + C
= x(ln x)2 − 2x ln x + 2x + C.
Method 2: Let u = (ln x)2 and dv = dx, from which du = x2 ln x dx and v = x. Then
Z
Z
2
2
(ln x) dx = x(ln x) − 2 ln x dx.
2
x
dx and v = x. Then
Z
Z
2
2
(ln x) dx = x(ln x) − 2x ln x − 2 dx
Let u = 2 ln x and dv = dx, from which du =
= x(ln x)2 − 2x ln x + 2x + C.
2
Z
2
2. Evaluate
(x2
0
2x
dx.
− 1)1/3
This is an improper integral, with a discontinuity at x = 1. Therefore,
Z 2
Z 1
Z 2
2x
2x
2x
dx =
dx +
dx
2
1/3
2
1/3
2
1/3
0 (x − 1)
0 (x − 1)
1 (x − 1)
Z N
Z 2
2x
2x
= lim−
dx + lim+
dx
2
1/3
2
1/3
N →1
N →1
(x − 1)
N (x − 1)
#
"
"0
N
2 #
3 2
3 2
(x − 1)2/3 + lim+
(x − 1)2/3 = lim−
N →1
N →1
2
2
0
N
3
3 2/3 3 2
3 2
2/3
2/3
+ lim+ 3 − (N − 1)
= lim− (N − 1) −
N →1
N →1
2
2
2
2
5/3
3 3
= − +
.
2
2
Z
3. Evaluate
4x2 + 13x − 9
dx.
x3 + 2x2 − 3x
Since the integrand is a rational function, we use partial fractions. The integrand can
be decomposed as
4x2 + 13x − 9
4x2 + 13x − 9
A
B
C
=
=
+
+
.
x3 + 2x2 − 3x
x(x + 3)(x − 1)
x x+3 x−1
Multiplying by the denominator, we have
4x2 + 13x − 9 = A(x + 3)(x − 1) + Bx(x − 1) + Cx(x + 3).
This is a case in which it’s easiest to proceed by choosing particular values of x.
If x = 0 : −9 = −3A
A=3
If x = 1 :
8 = 4C
C=2
If x = −3 : −12 = 12B B = −1.
Finally,
Z
Z 4x2 + 13x − 9
3
1
2
dx =
−
+
dx
x3 + 2x2 − 3x
x x+3 x−1
= 3 ln |x| − ln |x + 3| + 2 ln |x − 1| + C.
3
Z
4. Evaluate
5x2 + 4x + 3
dx.
x(x2 + x + 1)
Since the integrand is a rational function, we use partial fractions. The integrand can
be decomposed as
A
Bx + C
5x2 + 4x + 3
=
+
.
x(x2 + x + 1)
x x2 + x + 1
Multiplying by the denominator, we have
5x2 + 4x + 3 = A(x2 + x + 1) + (Bx + C)x
= (A + B)x2 + (A + C)x + A.
Equating the coefficients, we have
A+B =5
A+C =4
A = 3.
It follows that A = 3, B = 2, and C = 1. Therefore,
Z
Z 5x2 + 4x + 3
2x + 1
3
dx =
+
dx
x(x2 + x + 1)
x x2 + x + 1
= 3 ln |x| + ln |x2 + x + 1| + C.
(Note: The second integral was evaluated using the substitution u = x2 + x + 1.)
Z
5. Evaluate
√1
3
tan−1 x dx.
0
Let u = tan−1 x and dv = dx, from which du =
Z
1
√
3
−1
tan
x dx = x tan
=
=
=
=
Z
1
dx and v = x. Then
+1
√1
3
x
dx
+1
0
√1
3
1
−1
2
x tan x − ln |x + 1|
2
0
1
1
1
4
−1
√ tan
√
−0−
ln − ln 1
2
3
3
3
1 π 1 4
√
− ln
2 3
3 6
π
1 4
√ − ln .
6 3 2 3
−1
0
x2
√1
x 0 3 −
x2
(Note: The integral obtained from the integration by parts formula was evaluated using
the substitution w = x2 + 1.)
4
Z
6. Evaluate
2
x5 ex dx.
Let w = x2 , so that dw = 2x dx. Then
Z
Z
1
5 x2
5 w
dw
x e dx =
xe
2x
Z
1
=
x4 ew dw
2
Z
1
(x2 )2 ew dw
=
2
Z
1
=
w2 ew dw.
2
Let u = w2 and dv = ew dw, from which du = 2w dw and v = ew . Then
Z
Z
1
1
2 2
2 w
w
w e dw =
w e − 2we dw
2
2
Z
1 2 w
=
w e − wew dw.
2
Let u = w and dv = ew dw, from which du = dw and v = ew . Then
Z
Z
1 2 w
1
2 2
w
w
w e dw =
w e − we − e dw
2
2
1 2 w
=
w e − wew + ew + C
2
1 4 x2
2
2
=
x e − x2 ex + ex + C.
2
5
7. (a) Find the third order Taylor polynomial for f (x) = tan−1 x at x = 0.
For the third order Taylor polynomial, we have
T3 (x) = f (0) + f 0 (0)x +
f 00 (0) 2 f 000 (0) 3
x +
x,
2!
3!
for which we compute
f (x) = tan−1 x
1
f 0 (x) = 2
x +1
2x
f 00 (x) = − 2
(x + 1)2
2(x2 + 1)2 − 8x2 (x2 + 1)
f 000 (x) = −
(x2 + 1)4
It follows that
f (0)
f 0 (0)
f 00 (0)
f 000 (0)
=
=
=
=
0
1
0
−2
Therefore,
1
T3 (x) = x − x3 .
3
(b) Use your result in part (a) to approximate the values of
π
and π.
4
By part (a),
1
tan−1 x ≈ x − x3 .
3
Since tan
π
π
= 1, we see that = tan−1 1, and so
4
4
π
1
2
≈ 1 − (1)3 =
4
3
3 2
8
π≈4
= .
3
3
(The approximation can be improved by taking higher order polynomials.)
6
8. Suppose that f (x) = ln x is approximated by a second order Taylor polynomial near
x = 1. Find the maximum possible error of this approximation for x ∈ [1, 2].
By Taylor’s Inequality,
Error = |Rn (x)| ≤
M
|x − a|n+1 ,
(n + 1)!
where |f (n+1) (x)| ≤ M .
Using a second order (n = 2) Taylor polynomial at x = 1,
Error = |R2 (x)| ≤
M
|x − 1|3 ,
3!
where |f 000 (x)| ≤ M for 1 ≤ x ≤ 2.
Here, f 000 (x) =
2
and so for 1 ≤ x ≤ 2,
x3
2
2
2
000
|f (x)| = 3 = 3 ≤ 3 = 2.
x
|x|
1
Therefore,
Error ≤
2
1
1
1
|x − 1|3 = |x − 1|3 ≤ |2 − 1|3 = .
3!
3
3
3
The maximum error is 13 .
7
9. Determine the smallest order n that will ensure that the nth order Taylor polynomial
for f (x) = x ln x − x near x = 1 has a maximum error less than 0.01 for x ∈ [1, 2].
By Taylor’s Inequality,
Error = |Rn (x)| ≤
M
|x − a|n+1 ,
(n + 1)!
where |f (n+1) (x)| ≤ M .
We must find a general form for f (n+1) (x). We start by finding a general form for the
nth derivative:
f 0 (x) = ln x
1
f 00 (x) =
x
1
000
f (x) = − 2
x
2
(4)
f (x) = 3
x
3!
(5)
f (x) = − 4
x
f (n) (x) = (−1)n
(n − 2)!
.
xn−1
Therefore,
f (n+1) (x) = (−1)n+1
(n − 1)!
.
xn
For 1 ≤ x ≤ 2,
|f
(n+1)
(n − 1)!
(n − 1)!
(n
−
1)!
n+1
=
≤
= (n − 1)! = M.
(x)| = (−1)
n
n
x
|x|
1n
Thus,
(n − 1)!
|x − 1|n+1
(n + 1)!
1
=
|x − 1|n+1
(n + 1)n
1
≤
|2 − 1|n+1
(n + 1)n
1
=
.
(n + 1)n
Error ≤
We want Error < 0.01 =
1
,
100
so we are trying to find n such that
1
1
<
.
(n + 1)n
100
We see that n = 10 is the minimum value that works.
8
10. Solve the differential equation
dy
= y cos x with initial condition y(0) = 1.
dx
Using separation of variables,
dy
= y cos x
dx
1
dy
y
Z
1
dy
y
ln |y|
y
= cos x dx
Z
=
cos x dx
= sin x + C
= Cesin x .
Applying the initial condition,
y(0) = C = 1.
Therefore,
y = esin x .
9