Solutions to Math 41 Final Exam — December 10, 2012
1. (10 points) Find each of the following limits, with justification. If there is an infinite limit, then explain
whether it is ∞ or −∞.
Rx
ln(t + 1) dt
(a) lim 0
x→0
x2
(5 points) We are dealing with a 00 indeterminate, since the numerator and denominator are
R0
continuous and 02 = 0 and 0 ln(t + 1)dt = 0. We can apply L’Hospital:
Rx
ln(t + 1)dt
ln(x + 1)
= lim
(now this is again indeterminate 00 )
lim 0
2
x→0
x→0
x
2x
= lim
1
x+1
x→0
(using l’Hospital again)
2
1
2
=
Note that in the first equality we used FTC:
(b) lim (1 + sin x)
d
dx
Rx
0
ln(t + 1)dt = ln(x + 1).
1/x
x→0+
1
(5 points) Note that this is an indeterminate limit of form 1∞ . So, let L = lim (1 + sin x) x . We
x→0+
will compute ln L:
1
ln L = ln lim (1 + sin x) x
x→0+
1
= lim ln((1 + sin x) x )
x→0+
= lim
x→0+
= lim
ln(1 + sin x)
x
cos x
1+sin x
1
cos x
= lim
x→0+ 1 + sin x
1
= =1
1
x→0+
Therefore L = eln L = e .
(now this is indeterminate 00 )
(by L’Hospital)
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 2 of 16
2. (10 points) In each part below, use the method of your choice, but show the steps in your computations.
(a) The curve with equation xy = 2y − 1 passes through (x, y) = (1, 1). Find the equation of the line
tangent to the curve at this point.
(5 points) If we view y implicitly as a function of x with the goal of finding
need to take the natural log of both sides in order to differentiate xy :
dy
dx ,
then we first
ln(xy ) = ln(2y − 1) ⇐⇒ y ln x = ln(2y − 1)
We now differentiate both sides with respect to x and obtain:
y 0 ln x + y
2y 0
2
y
1
=
⇐⇒ y 0 (ln x −
)=−
x
2y − 1
2y − 1
x
y
(2y − 1)y
⇐⇒ y 0 = −
=
2
2 − x(2y − 1) ln x
x(ln x − 2y−1
)
At the point (x, y) = (1, 1) we obtain y 0 = 21 . This is the slope of the tangent line to that point.
The equation is then:
1
y − 1 = (x − 1)
2
(b) Find
h0 (x)
Z
if h(x) =
2
x2
et
dt
t2 + 3
x
(5 points) Let F (x) be an antiderivative of x2e+3 (which exists by the Fundamental Theorem of
x
Calculus, since x2e+3 is continuous). We know, by the fundamental theorem of calculus that:
Z
h(x) =
2
x2
et
dt = F (x2 ) − F (2)
t2 + 3
This implies the following formula for h0 (x):
2
d
d
d
2xex
h (x) =
F (x2 ) −
F (2) = F 0 (x2 ) x2 − 0 = 2xF 0 (x2 ) = 4
dx
dx
dx
x +3
0
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 3 of 16
3. (10 points) For this problem, we try to solve the equation x4 − 4x + 2 = 0.
(a) Show that this equation has at least one solution between x = 0 and x = 1. (State clearly what
results you use and why they apply.)
(5 points) Notice that f (0) = 0 − 0 + 2 = 2 and f (1) = 1 − 4 + 2 = −1. We will use the
Intermediate Value Theorem. The function f (x) = x4 − 4x + 2 is a polynomial, therefore it is
continuous. Because −1 < 0 < 2, by the IVT we obtain that there is a c ∈ (0, 1) such that
f (c) = 0. This implies that the polynomial has at least one solution between x = 0 and x = 1.
(b) It is a fact that this equation has a unique solution x = c between x = 0 and x = 1 (but you do not
need to prove this). Use Newton’s method, with initial guess x1 = 0, to compute the successive
approximations x2 and x3 for c.
(5 points) The formula for Newton’s method is the following:
xn+1 = xn −
f (xn )
f 0 (xn )
In our case f 0 (x) = 4x3 − 4. We compute x2 and x3 now taking x1 = 0:
x2 = 0 −
x3 =
1
2
=
−4
2
1
1
1
1
29
− 167 = +
=
2 −2
2 56
56
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 4 of 16
4. (10 points)
(a) Melted ice cream drips into a large waffle-style cone of height 16 cm and radius 3 cm. (This waffle
cone is shaped like a standard right circular cone.) At the moment when the depth of ice cream
is 8 cm, the rate of inflow of ice cream is 0.5 cm3 /sec. How fast is the level of liquid ice cream
rising in the cone at this moment?
(6 points) Let V be the volume of the ice-cream in the cone. Let h be the level in the cone and
r the radius of the surface of ice-cream in the cone. From the assumption, we are given that
dV
dh
3
dt = −0.5 cm /sec. We need to find out dt .
Next, the volume of ice-cream inside the cone is also given by
V =
3h
16 ,
Using similar triangles, we see that r =
V =
πr2 h
.
3
and plugging this in the formula above we get
3πh3
π 3 2
( h) h =
.
3 16
256
After taking the derivatives of this volume with respect to t, we find
dV
9π 2 dh
=
h
.
dt
256 dt
When h = 8, we get
dV
9π 2 dh
9π dh
=
(8 )
=
.
dt
256
dt
4 dt
Since
dV
dt
= 0.5 cm3 /sec, we finally get
dh
dt
=
2
9π
cm/sec.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 5 of 16
(b) (Note: ice cream imagery aside, this part is independent of any fact from part (a).)
The source of the melted ice cream from part (a) is a large spherical scoop; the scoop is melting in
such a way that it always maintains its spherical shape as it shrinks in size, but at any instant its
rate of change of volume, dV
dt , is equal to the product of its surface area, A, times a fixed constant:
specifically,
dVsphere
1
=−
Asphere
dt
250
(Here A and V are measured in square centimeters and cubic centimeters, respectively, and time
is measured in seconds.) Show that the radius of the scoop is shrinking at a constant rate, and
find this rate.
(4 points) Let r be the radius of the sphere scoop. First we have
Vsphere =
4πr3
3
from the volume formula of a ball. Taking derivatives of the volume with respect to t, we get
dVsphere
dr
= 4πr2 .
dt
dt
Using the assumption
dVsphere
dt
1
= − 250
Asphere and area formula Asphere = 4πr2 , we get
4πr2
dr
1
=−
(4πr2 ).
dt
250
After dividing 4πr2 at both sides, one can obtain
1
dr
=−
.
dt
250
This implies that the radius of the scoop is shrinking at a constant rate of
1
250
cm/sec.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 6 of 16
Z
x
cos(πt2 ) dt.
5. (19 points) Let g be the function defined on the interval {−1 ≤ x ≤ 1} by: g(x) =
−1
(a) Find g(−1) and g 0 (−1).
Z
−1
cos(πt2 )dt = 0, by the properties of definite integrals.
(2 points) We have g(−1) =
−1
By the Fundamental Theorem of Calculus, g 0 (x) = cos(πx2 ). Plugging in x = −1, we get
g 0 (−1) = cos(π(−1)2 ) = cos(π) = −1.
(b) It is a fact that g(1) = 2g(0); explain why this is true.
(3 points) For the sake of notational convenience we set f (t) = cos(πt2 ), so that g(x) =
cos(π(−t)2 )
Rx
−1 f (t)dt.
cos(πt2 )
Now notice that f is an even function: f (−t) =
=
= f (t). This means that
the graph of y = cos(πt2 ) is symmetric around the y-axis. We then conclude that
The signed area
f (t)dt = underneath f
−1
on [−1, 0]
Z
The signed area
= underneath f
on [0, 1]
0
Z
1
f (t)dt.
=
This fact can also be proved by substitution: u = −t. du = −dt.
Z −1
Z
Z 1
Z −1
Z −1
f (t)dt =
f (−u)(−du) = −
f (−u)du = −
f (u)du =
0
Now g(0) =
have g(1) =
0
0
R0
R1
f (t)dt and g(1) = −1 f (t)dt = −1 f (t)dt
−1
R0
R0
R0
−1 f (t)dt + −1 f (t)dt = 2 −1 f (t)dt = 2g(0).
+
R1
0
0
f (u)du.
−1
0
R0
(1)
0
f (t)dt. Using equation (1) we
(c) On what parts of the domain {−1 ≤ x ≤ 1} is g increasing? decreasing? Explain completely.
(4 points) Again, by FTC, g 0 (x) = cos(πx2 ). Recall that g is increasing when g 0 is positive,
decreasing when g 0 is negative. Thus we need to find out the subinterval of [−1, 1] on which
cos(πx2 ) stays positive or negative. We know cos θ is positive when 0 ≤ θ ≤ π2 . Therefore,
cos(πx2 ) is positive when
0 ≤ πx2 <
π
2
x2 <
⇔
Similarly, we know cos θ is negative when
π
3π
< πx2 <
2
2
⇔
π
2
1
2
<θ<
1
3
< x2 <
2
2
⇔
1
1
−√ < x < √ .
2
2
3π
2 .
Therefore, cos(πx2 ) is negative when
q
3
− √12 , or
2 < x <q
⇔
3
√1 < x <
2,
2
p
p
However, notice that 3/2 greater than 1 and − 3/2 is less than −1, and we do not care about
the range of x beyond [−1, 1]. Hence let us put: cos(πx2 ) is negative when
1
1
−1 < x < − √ , or √ < x < 1.
2
2
x −1
2
cos(πx ) −
−
− √12
0
√1
2
+
0
−
1
−
Therefore, g is increasing on (− √12 , √12 ), decreasing on [−1, − √12 ), ( √12 , 1]. Including or excluding
the endpoints ±1, ± √12 is a matter of definition. We accept answers of all kind regarding
in/excluding the endpoints.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 7 of 16
(d) On what parts of this domain is the graph of g concave upward? downward? Explain completely.
(4 points) By chain rule g 00 (x) = (cos(πx2 ))0 = − sin(πx2 )(x2 )0 = − sin(πx2 )2πx. g is concave
up if g 00 when positive, concave down when g 00 is negative, so we need to find out the subinterval
of [−1, 1] on which − sin(πx2 )2πx stays positive or negative. We know sin θ stays positive when
0 < θ < π. Therefore, sin(πx2 ) is positive when
0 < πx2 < π
⇔
0 < x2 < 1
⇔
−1 < x < 0, or 0 < x < 1.
We then analysis the sign of g 00 (x) = − sin(πx2 )2πx (−1, 0) and (0, 1). On (−1, 0), from the above
we know − sin(πx2 ) is negative, and 2πx is negative. Hence altogether g 00 (x) = − sin(πx2 )2πx
is positive on (−1, 0). Similarly, on (0, 1), − sin(πx2 ) is negative, and 2πx is positive. Hence
g 00 (x) = − sin(πx2 )2πx is negative on (0, 1).
x
−1
− sin(πx2 )
0
2πx
−
g 00 (x) = − sin(πx2 )2πx 0
−
−
+
0
0
0
0
−
+
−
1
0
+
0
Therefore, g is concave up on (−1, 0), concave down on (0, 1). We accept answers of all kind
regarding in/excluding the endpoints ±1 as (c).
(e) Using the information you’ve found in parts (a)-(d), sketch the graph of g on the domain [−1, 1].
Label any extreme or inflection points by their x-coordinate; also use the fact that g(0) ≈ 0.37.
(6 points) From (a) we know g(−1) = 0. From (b) we know g(1) = 2g(0) ≈ 2 × 0.37 = 0.74.
From (c) we know g is decreasing near the left of x = − √12 and increasing near the right of
x = − √12 . Hence by first derivative test g has a local (actually absolute, but we do not ask for
it) minimum at x = − √12 . Similarly, g has a local (actually absolute) maximum at x = √12 .
Recall that a point of inflection is a point at which the function changes its concavity. From
(d) we know x = 0 is an inflection point. By this definition, x = 1 and x = −1 are inflection
point since the function g is defined only on [−1, 1]. However, we can still interpret x = ±1 as
inflection points by extending g beyond [−1, 1]. Therefore, we accept answers with or without
±1 as inflection points.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 8 of 16
6. (12 points)
(a) Give a precise statement of the Mean Value Theorem.
(2 points) The Mean Value Theorem states that if f is a function that is continuous on the closed
interval [a, b] and differentiable on the open interval (a, b) then there is some number c in (a, b)
so that
f (b) − f (a)
f 0 (c) =
b−a
It is very important to say that c is in the open interval from a to b!
(b) Let f (x) = x − e ln x. Use the Mean Value Theorem to show that if b is any number larger than e,
then f (b) > 0.
(7 points) We want to show that f (b) = b − e ln b > 0 for all b > e. To do this, note that for
any number b > e, the function f (x) is continuous on the closed interval [e, b] and differentiable
on the open interval (e, b). So we can use the mean value theorem with a = e, b = b, and the
function f (x). From part a, the mean value theorem says that for some c between e and b,
f 0 (c) =
f (b) − f (e)
b−e
Now, f (e) = e − e ln(e), where ln(e) = 1. So f (e) = 0. Also, we computed that f (b) =
b − e ln(b).Thus, we get that for some c between e and b,
f 0 (c) =
b − e ln(b)
b−e
Next, f 0 (x) = 1 − xe . Thus, f 0 (c) = 1 − ec . We know that c is between e and b. That means, in
particular, that c is bigger than e. Therefore, ec is less than 1, so f 0 (c) = 1 − ec is positive.
We combine the fact that f 0 (c) is greater than 0 with the result of the mean value theorem to
get that
b − e ln(b)
>0
b−e
We also know that b > e, so b − e > 0. Thus we can multiply both sides by b − e to get that
b − e ln(b) > 0. So, we are done.
(c) Which is larger, eπ or π e ? Justify your answer. (Hint: you can use the result of part (b) even if
you did not prove it.)
(3 points) The trick to this part is to figure out what to plug in to f (x). We showed that for
b > e, b − e ln(b) > 0. That is,
b > e ln b
so
b > ln be
raising e to both sides,
e
eb > eln(b ) = be
Since π > e, we can set b = π and get that eπ > π e .
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 9 of 16
7. (12 points) A particle moving along a line has position at time t (in minutes) given by
2
s(t) = t3 − 3t2 + 4t
3
meters.
(a) Find the initial velocity and acceleration; that is, the velocity and acceleration at t = 0.
(2 points) Since the position is s(t) = 32 t3 − 3t2 + 4t, the velocity is v(t) = s0 (t) = 2t2 − 6t + 4
and the acceleration is a(t) = s00 (t) = 4t − 6. At time t = 0, the velocity is v(0) = 4 and the
acceleration is a(0) = −6.
(b) At what time, after t = 0, does the particle first come to a stop?
(3 points) The particle is stopped when v(t) = 0. We solve
2t2 − 6t + 4 = 0
2(t2 − 3t + 2) = 0
2(t − 1)(t − 2) = 0,
so v(t) = 0 at t = 1 and t = 2. The first time the particle comes to a stop is t = 1.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 10 of 16
For easy reference, the particle’s position function is s(t) = 23 t3 − 3t2 + 4t.
(c) Find the position of the particle at the second time the particle comes to a stop after t = 0.
(2 points) From part b, the second time the particle comes to a stop is t = 2. The position at
that time is s(2) = 23 (2)3 − 3(2)2 + 4(2) = 4/3.
(d) What is the total distance traveled by the particle between t = 0 and t = 3? Give all the steps in
your reasoning.
(5 points) The velocity v(t) = 2(t − 1)(t − 2) is positive when t is between 0 and 1, negative
when t is between 1 and 2, and positive when t is between 2 and 3. The total distance traveled
from time 0 to time 1 is s(1) − s(0) = 5/3 − 0 = 5/3. The total distance traveled from time 1
to time 2 is |s(2) − s(1)| = |4/3 − 5/3| = 1/3. The total distance traveled from time 2 to time 3
is s(3) − s(2) = 3 − 4/3 = 5/3. All told, from time 0 to time 3, the particle travels a distance
of 5/3 + 1/3 + 5/3 = 11/3. Note that the overall displacement (i.e., net change in position) is
5/3−1/3+5/3 = 3, which is simply s(3)−s(0); but the correct answer to the question as written
is 11/3.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 11 of 16
8. (15 points) Let f (x) = x2 − 1.
(a) On the axes below, sketch a graph of f over the domain [1, 3], and then draw the approximating
rectangles that are used to estimate the area under the curve (and above the x-axis) between
x = 1 and x = 3 according to the Midpoint Rule; use n = 4 rectangles.
(3 points)
(b) Write an expression involving only numbers that represents the area estimate using the rectangles
described above. (You do not have to expand or simplify the expression!)
(3 points) The midpoints of the intervals are x = 1.25, 1.75, 2.25, 2.75. The heights of the
rectangles are f (x) at those x-values. The widths of the rectangles are all 0.5. The total area is
0.5 × f (1.25) + 0.5 × f (1.75) + 0.5 × f (2.25) + 0.5 × f (2.75)
= 0.5[(1.252 − 1) + (1.752 − 1) + (2.252 − 1) + (2.752 − 1)].
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 12 of 16
(c) Find the exact area of the same region by evaluating the limit of a Riemann sum that uses the
Right Endpoint Rule. (That is, do not use the Fundamental Theorem of Calculus.) Show all
reasoning.
(9 points) The general form for the limit of a Riemann sum using the Right Endpoint Rule is
Z
b
f (x)dx = lim
a
n→∞
n
X
f (xi )∆x,
i=1
2
where ∆x = b−a
n and xi = a + i∆x. We have a = 1, b = 3, and f (x) = x − 1, meaning that
2
2i
∆x = n and xi = 1 + n . Now,
#
"
Z 3
n
X
2
2i 2
2
−1 ·
(x − 1)dx = lim
1+
n→∞
n
n
1
i=1
n X
4i 4i2
2
= lim
+ 2 ·
n→∞
n
n
n
i=1
!
n
n
8 X
8 X 2
i+ 3
i
= lim
n→∞ n2
n
i=1
i=1
8 n(n + 1)
8 n(n + 1)(2n + 1)
= lim
·
+
·
n→∞ n2
2
n3
6
!
8 · 1(1 + n1 ) 8 · 1(1 + n1 )(2 + n1 )
= lim
+
n→∞
2
6
=
8 8·2
20
+
= .
2
6
3
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 13 of 16
9. (5 points) Verify the following indefinite integral expression by differentiation, showing your steps:
Z
1
1
x
2 dx = x + 1 + ex − ln(e + 1) + C
x
(1 + e )
All we need to do for this problem is to differentiate the right hand side and show that it is equal to
1
.
(1+ex )2
0
1
x
x+
− ln(e + 1)
1 + ex
0
=x0 + (1 + ex )−1 − (ln(ex + 1))0
1
=1 + (−1)(1 + ex )−2 × (ex )0 − x
× (ex )0
e +1
ex
ex
=1 −
−
.
(1 + ex )2 ex + 1
Now we combine the fractions by making the denominators common:
ex
ex
−
(1 + ex )2 ex + 1
(ex + 1)2
ex
ex (ex + 1)
= x
−
−
(e + 1)2 (1 + ex )2
(ex + 1)2
(ex + 1)2 − ex − (ex + 1)ex
=
(ex + 1)2
x
2
(e ) + 2ex + 1 − ex − (ex )2 − ex
=
(ex + 1)2
1
,
= x
(e + 1)2
1−
as desired.
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 14 of 16
10. (9 points) Suppose r(t) is a continuous function which gives the instantaneous rate of change of North
America’s pigeon population, measured in thousands of pigeons per year, where t is measured in years
since January 1, 1900.
Z 80
r(t) dt represent? Express your answer in terms relevant to this
(a) What does the quantity
20
situation, and make it understandable to someone who does not know any calculus; be sure to
use any units that are appropriate, and also explain what the sign of this quantity would signify.
Z 80
(4 points) The number
r(t) dt is the net change in the North American pigeon population, in
20
thousands of pigeons, over the period between January 1, 1920 and January 1, 1980. A negative
quantity signifies a net population decrease, and a positive quantity signifies a net increase.
Z 80
r(t) dt is equal to the difference
Alternate solution: The number
20
The North American pigeon population
on January 1, 1980, in thousands
The North American pigeon population
−
on January 1, 1920, in thousands
.
(Note that the meaning of a negative or positive quantity is built implicitly into this formulation.)
Z x+10
(b) Define a function Q by Q(x) =
r(t) dt. In the context of this application, what does the
x
value of Q(x) represent in terms of x? Follow the same guidelines as in part (a) in expressing
your answer.
(2 points) Q(x) gives the net change in the North American pigeon population, in thousands of
pigeons, over a ten-year period that starts exactly x years after January 1, 1900. (Note that x
need not be a positive integer; if x negative, then we begin our ten-year period exactly |x| years
before January 1, 1900.) If Q(x) is positive, then the population experienced a net increase over
this decade; if Q(x) is negative, then the population experienced a net decrease instead.
(c) Suppose Q has a critical number at x = c. Give a simple condition involving c that must hold for
the function r; your answer should not involve integrals.
(3 points) If Q has a critical number at c, then either Q0 (c) = 0 or Q is not differentiable at c.
But note that Q(x) is differentiable for all x: we have
Z 0
Z x+10
Z x
Z x+10
r(t) dt +
r(t) dt = − r(t) dt +
r(t) dt
Q(x) =
x
0
0
0
and this is a difference of two “area” functions, each of which is differentiable by the Fundamental
Theorem of Calculus (since we are given that r(t) is continuous). Thus, Q0 (c) = 0 for any critical
number c of Q. We compute Q0 (x) below, using the chain rule with u = x + 10:
Z x+10
Z x
d
Q0 (x) =
r(t) dt −
r(t) dt
dx
0
Z0 u
Z x
d
d
=
r(t) dt −
r(t) dt
dx
dx
0
0
Z u
Z x
d
du
d
=
r(t) dt ·
−
r(t) dt
du 0
dx dx
0
= r(u) · 1 − r(x)
= r(x + 10) − r(x)
Thus, c is a critical number of Q if and only if 0 = Q0 (c) = r(c + 10) − r(c), or equivalently
r(c + 10) = r(c) .
Math 41, Autumn 2012
Solutions to Final Exam — December 10, 2012
Page 15 of 16
11. (22 points) Evaluate each of the following integrals, showing all reasoning.
Z x−1
1
2
√
+ sec x tan x + x dx
(a)
+√
3
2
x
1 − x2
(5 points) First, note that
Z
Z
x−1
√
dx = (x − 1)x−1/3 dx
3
x
Z
= x2/3 − x−1/3 dx
3 5 3 2
= x3 − x3 + C
5
2
Next, we can see by inspection that an antiderivative of
Z
1
2x
= 2−x is
So we get
1
x−1
2
√
+ sec(x) tan(x) + x dx
+√
3
2
2
x
1−x
Z
Z
Z
Z
x−1
1
2
√
=
dx + sec(x) tan(x)dx +
dx
dx + √
3
2
2x
x
1−x
3 5 3 2
2−x
= x 3 − x 3 + 2 arcsin(x) + sec(x) −
+C
5
2
ln 2
Z
(b)
−2−x
ln(2) .
sin θ
dθ
1 + cos2 θ
(5 points)
Z
sin θ
dθ =
1 + cos2 θ
Z
−1
du
1 + u2
using u-substitution with u = cos θ and du = − sin θdθ
= − arctan u + C
= − arctan(cos(θ)) + C
Math 41, Autumn 2012
Z
(c)
Solutions to Final Exam — December 10, 2012
Page 16 of 16
t2 ln(2t) dt
(6 points)
Z
1
1
t2 ln(2t)dt = t3 ln(2t) −
3
3
Z
t2 dt
using integration by parts with u = ln(2t), du = 1t dt, v = 13 t3 and dv = t2 dt.
1
1
= t3 ln(2t) − t3 + C
3
9
Z
(d)
π2
√
cos( x) dx
0
(6 points)
Z
π2
√
Z
π
cos( x)dx =
0
using u-substitution with u =
dx = 2udu.
2u cos(u)du
0
√
x, du =
1
√
dx.
2 x
√
We solve for dx to get dx = 2 xdu, so
Z
π
= 2u sin(u) − 2
0
π
sin(u)du
0
using integration by parts with v = 2u, dv = 2du, w = sin(u) and dw = cos(u)du. Note that the
limits of integration are the same.
π
π
= 2u sin(u) − (−2 cos(u))
0
= 2 cos(π) − 2 cos(0)
= −4
since sin(π) = sin(0) = 0, and cos(π) = −1 while cos(0) = 1.
0