Download Problem Set 3 Partial Solutions

Math 142, Problem Set 3 Solutions
1
Math 142.009, Problem Set 3 Solutions (Partial)
Due Wednesday, Feb. 2, 2011
From Text Calculus: Concepts and Contexts, 3rd edition, by James Stewart
Section 5.3, pp. 374-376, # 3, 6, 8, 13, 17, 23, 25, 37, 38, 41
Section 5.4, pp. 383-385, # 3, 5, 7, 9, 11, 12
Section 5.5, pp. 392-393, # 2, 3, 4, 7, 8, 11, 21, 22, 41, 45, 47, 53, 63, 64
Additional Problems:
Z 1
2
1.) Evaluate
dx.
2
0 1+x
Solution: For this problem we need to recall the differentiation formula:
1
d
tan−1 x =
. Since integration is done by finding the antiderivative (by
dx
1 + x2
the “Evaluation Theorem” in the text, also known as the Fundamental Theorem of
Calculus, Part II),
Z 1
Z 1
1
1
π
π
2
dx = 2
dx = 2 tan−1 x 0 = 2 tan−1 1−2 tan−1 0 = 2· −0 = ,
2
2
4
2
0 1+x
0 1+x
using tan(π/4) = 1, hence tan−1 1 = π/4, and tan 0 = 0, hence tan−1 0 = 0.
2.) Verify by differentiation that the following formula is correct:
Z
1
1
xe2x dx = xe2x − e2x + C.
2
4
Solution: We can always check whether an indefinite integral is correct by taking
its derivative. Here, using the product rule (uv) 0 = u 0 v − uv 0 on the first term,
1
1
1
d 1 2x 1 2x
xe − e + C = · 1 · e2x + x · 2e2x − · 2e2x + 0
dx 2
4
2
2
4
1 2x
1
· e + xe2x − · e2x = xe2x .
2
2
Hence
since the indefinite integral is the antiderivative (by definition), the equation
Z
1
1
xe2x dx = xe2x − e2x + C is correct.
2
4
=
3.) Find the general indefinite integral
Z
(u + 2)(2u + 1) du.
Solution: The best way to do this integral is just to multiply out the integrand:
(u + 2)(2u + 1) = 2u2 + 5u + 2. Hence
Z
Z
2
5
(u + 2)(2u + 1) du = 2u2 + 5u + 2 du = u3 + u2 + 2u + C.
3
2
Math 142, Problem Set 3 Solutions
Z
4.) Let f (x) =
1
x
e3t +
2
1
dt.
3t
A.) Evaluate f (x) by computing the integral.
Solution: To integrate e3t , we guess the antiderivative e3t . However,
d 3t
e
dt Z
so to get the derivative to be e3t we need to multiply by 13 . In other words,
= 3e3t ,
e3t dt =
1
1
1 1
1 3t
e + C. For the integral of , write
= · and use the scalar linearity property
3
3t
3t
3 t
d
1
of the integral to take the 13 out of the integral. Recalling that
ln |t| = , we get
dt
t
Z
Z
1
1
1
1
dt =
dt = ln |t| + C.
3t
3
t
3
Hence by the additivity property of the integral,
x
Z x
1 3t 1
1
3t
e + ln |t| f (x) =
e + dt =
3t
3
3
1
1
1
1
1
1
1
1
1
= e3x + ln |x| − e3 − ln 1 = e3x + ln |x| − e3 ,
3
3
3
3
3
3
3
since ln 1 = 0.
B.) Evaluate f 0 (x) by taking the derivative of your answer to part A.
1
1
1
Solution: Since f (x) = e3x + ln |x| − e3 ,
3
3
3
1 1
1
1
f 0 (x) = · 3e3x + · − 0 = e3x + ,
3
3 x
3x
1
because e3 is a constant and hence has derivative 0.
3
C.) Compare your result in part B to what you get by applying the Fundamental
Theorem of Calculus to find f 0 (x).
d
dx
Solution: The Fundamental Theorem of Calculus (Part II) says that
Z
x
1
g(t) dt = g(x). Applying the Fundamental Theorem with g(t) = e3t + ,
3t
1
Z x
d
d
1
1
f (x) =
e3t + dt = e3x + ,
f 0 (x) =
dx
dx 1
3t
3x
which agrees with the answer in part b.
Z t s
e + ln(s2 + 10)
5.) Let f (t) =
ds. Find f 0 (t).
20 + cos s + 35
s
0
Z x
d
Solution: In the Fundamental Theorem of Calculus formula
g(t) dt =
dx 1
g(x),
variable x with t to get
Z we can replace the integration variable t with s and the
d t
es + ln(s2 + 10)
g(s) ds = g(t). Applying this formula with g(s) = 20
, we get
dt 1
s + cos s + 35
Z
d t es + ln(s2 + 10)
et + ln(t2 + 10)
d
0
f (t) = f (t) =
ds = 20
.
dt
dt 0 s20 + cos s + 35
t + cos t + 35
Math 142, Problem Set 3 Solutions
Z
x3 +x
6.) Let f (x) =
√
3
2 + sin t dt. Find f 0 (x).
0
√
Solution: Let u = x + x. Then f =
2 + sin t dt. By the Fundamental
0
Z u
√
√
df
d
Theorem of Calculus,
=
2 + sin t dt = 2 + sin u. Hence by the chain
du
du 0
rule,
Z
3
f 0 (x) =
u
p
df du √
df
=
= 2 + sin u · (3x2 + 1) = (3x2 + 1) 2 + sin(x3 + x).
dx
du dx
π/2
Z
7.) Evaluate
0
cos x
dx.
1 + sin2 x
Solution: We make the substitution u = sin x. Then du = cos x dx. When
x = 0, then u = sin 0 = 0. When x = π/2, then u = sin(π/2) = 1. Hence, changing
everything (integrand, differential, and limits) over in terms of u,
π/2
Z
0
cos x
dx =
1 + sin2 x
Z
0
1
1
1
π
π
du = tan−1 u0 = tan−1 1 = tan−1 0 = − 0 = ,
2
1+u
4
4
since tan(π/4) = 1 and tan 0 = 0.
Z e−1
x
8.) Evaluate
dx.
x+1
0
Solution: Method 1: substitution. Let u = x + 1. Then du = dx, and x = u − 1.
When x = 0, we have u = 1, and when x = e − 1, we have u = e − 1 + 1 = e. Hence
Z e
Z e
Z e−1
x
u−1
u 1
dx =
du =
− du
x+1
u
u
1
1 u
0
Z e
1
=
1 − du = (u − ln |u|)|e1 = e − ln e − (1 − ln 1) = e − 1 − 1 = e − 2,
u
1
since ln e = 1 and ln 1 = 0.
Method 2: algebra trick. Write
Hence
Z
0
e−1
x
dx =
x+1
Z
x
x+1−1
x+1
1
1
=
=
−
= 1−
.
x+1
x+1
x+1 x+1
x+1
e−1
1−
0
1
dx = (x − ln |x + 1|)|e−1
0
x+1
e − 1 − ln(e − 1 + 1) − (0 − ln 1) = e − 1 − ln e − 0 = e − 1 − 1 = e − 2.
x
Method 3: algebra method: division of polynomials. We can evaluate x+1
by
formally dividing x + 1 into x. We see that the leading term x of x + 1 goes into x
one time, and multiplying 1 times x + 1 and subtracting from x gives a remainder of
x
1
−1. That is, the division gives
= 1−
. From this point on, the rest is the
x+1
x+1
same as in Method 2.
Math 142, Problem Set 3 Solutions
4
Z
9.) Suppose f is continuous on R and
8
Z
f (x) dx = 12. Find
ln 2
e3x f (e3x ) dx.
0
1
a
Hint: Recall how to simplify ea ln b : either use ln(ba ) = a ln b or write ea ln b = eln b .
Z ln 2
e3x f (e3x ) dx, we make the substitution u = e3x . Then
Solution: To evaluate
0
du = 3e3x dx, so e3x dx = 31 du. When x = 0, then u = e3·0 = e0 = 1. When x = ln 2,
3
then u = e3 ln 2 = eln(2 ) = 23 = 8, using ln(ab ) = b ln a and the fact that eln t = t.
Hence by substitution,
Z 8
Z
Z ln 2
1
1 8
1
3x
3x
f (u) du =
f (u) du = · 12 = 4,
e f (e ) dx =
3 1
3
1 3
0
Z 8
f (x) dx = 12.
since we are given
1