Engineering Math II – Spring 2015
Solutions for Class Activity #1
Problem 1 (p. 399, #85). Find the derivative of g(x) =
R0
R 3x
2x f (u)du + 0 f (u)du.
R 3x
u−1
2x u+1 du.
Hint:
R 3x
2x
f (u)du =
Solution. Following the hint, we first split the integral into two parts:
Z 0
Z 3x
Z 3x
u−1
u−1
u−1
du =
du +
du.
g(x) =
u+1
2x u + 1
0
2x u + 1
Since
Rb
a
f (u)du = −
Ra
b
f (u)du, we can rewrite the above equation as:
Z
g(x) = −
0
2x
u−1
du +
u+1
Z
0
3x
u−1
du.
u+1
d
u−1
Now let A(u) be the antiderivative of u−1
u+1 ; that is, du A(u) = u+1 (note that we do not explicitly
find A(u) since, as we will soon see, this is unnecessary). Then, by the Fundamental Theorem
of Calculus (FTC), we have:
g(x) = − A(2x) − A(0) + A(3x) − A(0).
Taking the derivative of both sides, we see that
d
g(x) =
dx
d
− A(2x) − A(0) + A(3x) − A(0)
dx
d
d
d
d
A(0) +
A(3x) +
A(0).
= − A(2x) +
dx
dx
dx
dx
d
Since A(0) is a constant, dx
A(0) = 0. Moreover, by the chain rule and the fact that A(u) is the
u−1
antiderivative of u+1 , we have:
d
2x − 1
3x − 1
4x − 2 9x − 3
g(x) = −2 ·
+3·
=−
+
.
dx
2x + 1
3x + 1
2x + 1 3x + 1
We are done, since we have now found an explicit formula for the derivative of g(x).
Class Activity #1 Solutions
2
Problem 2 (p. 399, #89). If F (x) =
Rx
1
f (t)dt, where f (t) =
R t2
1
√
1+u4
du,
u
find F 00 (2).
Solution. First we find F 0 (x). Let A(t) be the antiderivative of f (t). Then by the FTC, we
have:
F (x) = A(x) − A(1).
Taking the derivative of both sides, and observing that A0 (x) = f (x) (since A(t) is the antiderivative of f (t)) and A0 (1) = 0, we see that:
F 0 (x) = f (x).
It
now follows that F 00 (x) = f 0 (x). In order to find f 0 (x), let B(u) be the antiderivative of
√
1+u4
. Then by the FTC, we see that:
u
f (x) = B(x2 ) − B(1).
Taking the derivative of both sides, and using the chain rule, we have:
p
1 + (x2 )4
00
0
0 2
.
F (x) = f (x) = 2x · B (x ) = 2x ·
x2
Finally, substituting x = 2 into the above equation gives
√
1 + 28 √
00
F (2) = 4 ·
= 257.
4
Class Activity #1 Solutions
3
Problem 3 (p. 399, #80). The acceleration function (in m/s2 ) and the initial velocity for a
particle moving along a line are given below. Find (a) the velocity at time t and (b) the distance
traveled during the interval 0 ≤ t ≤ 3.
a(t) = 2t + 3;
v(0) = −4
Solution. (a) Let v(t) be the velocity of the particle at time t. Recall that acceleration is the
derivative of velocity, i.e. v 0 (t) = a(t). Thus in order to find v(t), we integrate both sides of the
equation v 0 (t) = 2t + 3 to get
Z
v(t) = (2t + 3)dt = t2 + 3t + C.
where C is the constant of integration. In order to determine C, we use the given initial
condition that v(0) = −4. That is,
v(0) = 02 + 3 · 0 + C = C = −4.
Thus v(t) = t2 + 3t − 4.
(b) There are two notions of distance: the “net” distance traveled by the particle, and the
total distance. The problem does not state which distance we are to compute, so we do both
here. If such an ambiguity arises on a quiz or test, you should ask a question to clarify what
exactly is expected of you.
The net distance traveled by the particle on the interval 0 ≤ t ≤ 3 is just the distance
between the position of the particle at t = 0 and the position of the particle when t = 3. That
is, if s(t) is the position of the particle at time t, then the net distance is just |s(3) − s(0)|.
Recall that velocity is the derivative of position, so that s0 (t) = v(t). Thus in order to find s(t),
we integrate both sides of the equation s0 (t) = t2 + 3t − 4 so get
Z
1
3
s(t) = (t2 + 3t − 4)dt = t3 + t2 − 4t + C,
3
2
where C is the constant of integration. At first glance, we now seem to be stuck, since we have
no initial conditions that could be used to determine C. In fact we do not need to determine
C, since we are interested in the difference s(3) − s(0), in which case the constant is canceled
during the subtraction. That is,
1
1
3
3
|s(3) − s(0)| = · 33 + · 32 − 4 · 3 + C −
· 03 + · 02 − 4 · 0 + C 3
2
3
2
1
3
= · 27 + · 9 − 12
3
2
21
=
.
2
Thus the net distance traveled by the particle is 21
2 meters.
In order to find the total distance traveled, we first note that since the particle is moving
along a line, it can only move in two directions: a positive direction and a negative direction.
Hence the total distance traveled is the sum of the distance traveled in the positive direction
and the distance traveled in the negative direction. Our first task, then, is to determine the
intervals for which the particle is moving in the positive and negative directions. The intervals
can be found by determining when v(t) < 0 and v(t) > 0.
Class Activity #1 Solutions
4
We determined part (a) that v(t) = t2 + 3t − 4. This function has two critical points
determined by
t2 + 3t − 4 = (t + 4)(t − 1) = 0.
So the critical points occur when t = −4 and t = 1. The only critical point that occurs in the
interval 0 ≤ t ≤ 3 is when t = 1, and we split this interval into the two intervals 0 ≤ t < 1 and
1 < t ≤ 3. If 0 ≤ t < 1, then t2 < 1 and 3t < 3 so that t2 + 3t − 4 < 1 + 3 − 4 = 0. That is,
v(t) < 0, so that the particle is moving in the negative direction for 0 ≤ t < 1. On the other
hand, if 1 < t ≤ 3, the t2 > 1 and 3t > 3 so that t2 + 3t − 4 > 1 + 3 − 4 = 0. This means that
the particle is moving in the positive direction for 1 < t ≤ 3.
Now that we know the intervals when the particle is moving in the positive and negative
directions, we see that the total distance traveled is given by |s(1)−s(0)|+|s(3)−s(1)|. Working
these out, we have:
1
1
3
3
|s(1) − s(0)| = · 13 + · 12 − 4 · 1 + C −
· 03 + · 02 − 4 · 0 + C 3
2
3
2
1 3
= + − 4
3 2
13
,
=
6
and
1
1
3
3
|s(3) − s(1)| = · 33 + · 32 − 4 · 3 + C −
· 13 + · 12 − 4 · 1 + C 3
2
3
2
1 3
27
− 12 − − + 4
= 9 +
2
3 2
76
=
.
6
Thus the total distance traveled by the particle is
13
6
+
76
6
=
89
6
meters.
Class Activity #1 Solutions
5
Problem 4 (p. 400, #104). Suppose h is a function such that h(1) = −2, hR0 (1) = 2, h00 (1) =
2
3, h(2) = 6, h0 (2) = 5, h00 (2) = 13, and h00 is continuous everywhere. Evaluate 1 h00 (u)du.
0
Solution. We know that h00 (u) = h0 (u) , which means that h0 (u) is an antiderivative of h00 (u).
Since h00 (u) is continuous everywhere, the FTC applies, and we see that
Z
2
h00 (u)du = h0 (2) − h0 (1) = 5 − 2 = 3.
1
Observe that much of the information given in the statement of the problem is superfluous.
Class Activity #1 Solutions
6
Problem 5 (p. 407, #46). Evaluate the indefinite integral
R
sin(x)
dx.
1+cos2 (x)
Solution. We make
substitution u = cos(x). Then we have du = − sin(x)dx so that the
R the
1
d
1
integral becomes 1+u
du.
Recall that du
arctan(u) = 1+u
2
2 ; i.e., arctan(u) is an antiderivative
1
of 1+u2 . Hence, by the FTC, we have
Z
1
du = arctan(u) + C,
1 + u2
where C is the constant of integration. Substituting u = cos x into the above equation gives
Z
sin(x)
dx = arctan(cos(x)) + C.
1 + cos2 (x)
1
The key to solving this problem was recognizing that the derivative of arctan(u) is 1+u
2 . Make
sure that you re-familiarize yourself with the derivatives of trigonometric functions and their
inverses.
Class Activity #1 Solutions
7
Problem 6 (p. 407, #66). Evaluate the definite integral, if it exists:
R π/3
5
−π/3 sin (θ)dθ.
We present two solutions to this problem. The first solution uses the method of u substitution.
Solution 1. First, can we tell right off the bat whether or not the integral exists? What does it
mean for an integral to exist? One way to guarantee that an integral exists is for the integrand
to be continuous on whatever interval is determined by the limits of integration. Is sin5 (θ)
continuous on [−π/3, π/3]? We know that sin(θ) is continuous everywhere, and so is x5 . Since
sin5 (θ) is the composition of x5 with sin(θ), it is also continuous everywhere, and thus certainly
on the interval [−π/3, π/3]. So we are guaranteed that the integral does, in fact, exist.
2
Now observe that sin5 (θ) = sin2 (θ) · sin(θ). We make use of the Pythagorean trigonometric identity sin2 (θ) + cos2 (θ) = 1 and rewrite the integral as
Z π/3
Z π/3
2
5
sin (θ)dθ =
1 − cos2 (θ) · sin(θ)dθ.
−π/3
−π/3
Now we let u = cos(θ). Then du = − sin(θ)dθ. The lower limit of integration becomes
cos(−π/3) = 1/2 and the upper limit becomes cos(π/3) = 1/2. Hence the integral can be
re-written as
Z π/3
Z 1/2
5
sin (θ)dθ = −
(1 − u2 )2 du = 0.
−π/3
1/2
The reason that this last integral evaluates to zero is that the limits of integration coincide (this
means that the ∆x term in the Reimann sum for this integral is 0, forcing the entire sum to be
0).
The second solution uses a more general fact concerning the integral of an odd function on
the interval [−a, a], where a is any positive real number. Recall that a function f is odd if
f (−x) = −f (x) for all x in the domain of f . Geometrically, this means that the graph f is
symmetric
R a about the line y = −x. We will show that if f (x) is odd (and continuous on [−a, a]),
then −a f (x)dx = 0. First we split the integral into two pieces:
Z
a
Z
0
f (x)dx =
Z
f (x)dx +
−a
a
f (x)dx
−a
(1)
0
R0
Consider the integral −a f (x)dx from above. We make the substitution u = −x. Then du =
−dx, and the lower limit of integration becomes −(−a) = a, while the upper limit remains 0.
That is,
Z
Z
Z
Z
0
0
f (x)dx = −
−a
a
a
a
f (−u)du = −
f (−u)du =
0
f (u)du,
0
where the last equality in the above chain follows from the assumption that f is an odd function.
Replacing the first integral on the right hand side of equation (1) with the last integral in the
far right had side of the above chain of inequalities yields
Z a
Z a
Z a
f (x)dx = −
f (u)du +
f (x)dx = 0.
−a
0
0
Take a minute now to think about what we have just done. Without knowing anything about
the function f other than it being odd, we have shown that for any positive real number a, if f
Class Activity #1 Solutions
8
Ra
is continuous on [−a, a], then −a f (x)dx = 0. The power in what we have just proven is that
if we are confronted with an integral whose upper and lower limits are negatives of each other,
and we can recognize the integrand as an odd function, then we automatically know that the
integral evaluates to 0.
Solution 2. Consider sin5 (−θ). Recall that the sine function is an odd function. That is,
sin(−θ) = − sin(θ). Thus
sin5 (−θ) = (−1)5 sin5 (θ) = − sin5 (θ).
This means that sin5 (θ) is an odd function, and therefore
Z
π/3
sin5 (θ)dθ = 0.
−π/3
If some functions are called odd functions, it would make sense that some functions be called
even. You may recall that a function f is even if f (−x) = f (x) for all x in the domain of f .
Geometrically, this means that the graph of f is symmetric about the y-axis. Examples of even
functions include y = |x| and h = x2 . In this case, one can show that if f is an even function
(and continuous on the interval [−a, a]), then
Z a
Z a
f (x)dx = 2
f (x)dx.
−a
0
Can you prove this identity yourself using what you have learned about integration so far?
Although not as nice as the identity for odd functions, it is sometimes useful in simplifying
hand computations.
Class Activity #1 Solutions
9
Problem 7 (p. 408, #82). If f is continuous and
R9
0
f (x)dx = 4, find
R3
0
xf (x2 )dx.
Solution. This problem may look menacing at first, but the solution is actually quite simple.
The key to solving
this problem is figuring out how to use the informationRthat is provided. We
R3
9
need to find 0 xf (x2 )dx. We don’t know what f is, but we are told that 0 f (x)dx = 4. If you
look carefully, you may notice that 9 = 32 , and in one of our integrals we have the term x2 . It
may seem like a long shot at this point, but we could try making the substitution u = x2 in the
integral we are asked to find, and see if we get something we can use.
If u = x2 , then du = 2xdx which means that du/2 = xdx. We have an xdx in the integral
we need to find – so far, so good. Now we need to adjust our limits of integration. When x = 0
then u = 0, and when x = 3 then u = 9. Thus, after making the substitution, we get
Z 3
Z
1 9
xf (x2 )dx =
f (u)du.
2 0
0
R9
Now we make use of integral that was given to us. Namely, we know that 0 f (u)du = 4, which
means that
Z
3
xf (x2 )dx = 2.
0
Class Activity #1 Solutions
10
Problem 8 (p. 408, #86). If f is continuous on [0, π], use the substitution u = π − x to show
that
Z π
Z
π π
xf (sin(x))dx =
f (sin(x))dx.
2 0
0
Solution. This problem really tests your understanding of the method of u substitution. First,
whenever you are told to show that two things are equal (remember proving trig identities in
pre-cal or trig class?), you usually manipulate one side by known rules to make it look like the
other side. Often there is one side of the identity that is easier to work with. This will be
out strategy in this problem, so we first need to figure out which side to work with and then
determine how to manipulate it.
We are told which substitution to make, but it may not be clear at first how the substitution
is useful, since there is no π − x term appearing in either of the integrals. However, if u = π − x,
then that means that x = π − u, so we can replace x in our integrals with π − u. Since there
is only one x in the integral on the right hand side, it is tempting to try an manipulate this
side. This would not be the best strategy, however, since we would need to somehow introduce
a multiple of x. This is not easy. Instead, let’s manipulate start with the left hand side of the
identity.
If u = π − x, then du = −dx. When we make the substitution, we need to change our limits
of integration. When x = 0 then u = π, and when x = π then u = 0, so we get (using some of
the basic properties of integrals)
Z
0
π
Z
0
xf sin(x))dx = −
(π − u)f (sin(π − u))du
Z ππ
=
(π − u)f (sin(π − u))du
0
Z π
Z π
f (sin(π − u))du −
uf (sin(π − u))du.
= π
0
0
It might look like we’ve just made things unnecessarily complicated, but we are starting
to see things that look sort of like what we ultimately want. We can make things simpler for
ourselves if we observe that sin(π − u) = − sin(−u) since adding π to an angle sends it to the
opposite quadrant. But the sine function is an odd function – that is, sin(−θ) = − sin(θ). This
means that − sin(−u) = sin(u) and allows us to rewrite our integral as
Z π
Z π
Z π
xf (sin(x))dx = π
f (sin(u))du −
uf (sin(u))du.
0
If we add
Rπ
0
0
0
uf (sin(u))du to both sides, we equation becomes
Z π
2
uf (sin(u))du = πf (sin(u))du.
0
Finally, dividing both sides by 2 gives
Z π
Z
π π
uf (sin(u))du =
f (sin(u))du.
2 0
0