Solution: APPM 1350
Exam 3
Spring 2016
and so we have
Z ah
⇡a ⇡|x|+
a
1. The following problems are not related.
(a)(8 pts) Approximate the area of the region bounded by the function f (x) =
8p
< 4
x2 ,
if 0  x < 2
: 1 x 1,
if x 2
2
x-axis from x = 0 to x = 4 using a Riemann Sum with 4 subintervals of equal length and left endpoints.
(b)(8 pts) Suppose an object has velocity v(t) = t2
(c)(8 pts) If
Z
(a)(8 pts) Find dy/dx if y =
4
2g(x) for 0  x  4.
g(x) dx = 3.4, find the net area of the region bounded by the curve h(x) = 1
Solution: (a)(8 pts) Note that
x = (b
0)/4 = 1 and so xi = a + i x = 0 + i · 1 = i for i = 0, 1, 2, 3, 4.
a)/n = (4
Thus, using left endpoints, we have
L4 = f (x0 ) x + f (x1 ) x + f (x2 ) x + f (x3 ) x = f (0) · 1 + f (1) · 1 + f (2) · 1 + f (3) · 1 = 2 +
=

0
4t
t3
3
2
t3
+
3
0
(c)(8 pts) The net area is
Z 4
Z 4
h(x) dx =
(1
0

4
=
2
2g(x)) dx =
0
1
5 p
=
+ 3
2
2
4) dx
2
4t
3+0+
Z
✓
8
8
3
◆
4
dx
2
0
+
Z
✓
64
3
16
◆
✓
8
3
◆
16 16 16
8
=
+
+
= 16
3
3
3
4
g(x) dx = 4
2(3.4) = 4
6.8 =
1
3⇡ 2
⇡a2 + ⇡a2 =
a
2
2
✓Z
3x
sin(t4 ) dt
0
◆3
(Justify your answer and cite any theorems that you use.)
(b)(8 pts) The height H (in feet) of a palm tree after growing for t years is given by H(t) =
(c)(8 pts) Suppose F (x) is an antiderivative of f (x) = (sin x)/x, x > 0. Express
your answer and cite any theorems that you use.)
Z
3
1
sin(2x)
dx in terms of F . (Justify
x
d
dx
✓Z
3x
sin(t4 ) dt
0
◆3
=3
✓Z
3x
sin(t4 ) dt
0
◆2
· sin((3x)4 ) · 3 = 9
(b)(8 pts) We have, by definition,
Z 8
Z
p
1
1 9 1/2
5 3
Have =
( t + 1 + 5t1/3 ) =
u du + · · t4/3
8 0 0
8 1
8 4
| {z }
8
✓Z
1 2 3/2
· u
8 3
=
0
3x
sin(t4 ) dt
0
9
(b)(9 pts)
Z
5
1
x
p
2x
1
dx
◆2
sin(81x4 )
15
· 16
32
+
1
u=t+1
1
(27
12
=
1) +
15
26 90
116
29
=
+
=
=
2
12 12
12
3
thus the tree’s average height over the first 8 years was 29/3 ft.
0
(c)(8 pts) Using the u-substitution u = 2x ) x = u/2 and dx = du/2 and so dx/x =
d✓
p
(a)(9 pts)
cos2 (✓) 3 1 + tan(✓)
t + 1 + 5t1/3 , find the
Solution: (a)(8 pts) Using the chain rule and the Fundamental Theorem of Calculus (Part I) we have
2.8
Z
2. Evaluate the following integrals, remember to show all work and simplify your answer:
Z
p
tree’s average height for 0  t  8.
dy/dx =
(b)(8 pts) The total distance is given by
Z 4
Z 2
Z 4
|t2 4| dt =
(t2 4) dt +
(t2
0
p
i
a2 x2 dx = 2⇡a2
3. The following problems are not related, justify your answers and cite any theorems that you use.
and the
4, find the total distance travelled by the object for 0  t  4.
0
p
(c)(9 pts)
Z ah
a
⇡a ⇡|x|+
p
a2
x2
i
dx, for a > 0.
Solution: (a)(9 pts) Let u = 1 + tan(✓) then du = sec2 (✓)d✓ = d✓/ cos2 (✓), thus
Z
Z
d✓
du
3
3
p
=
= u2/3 + C = (1 + tan(✓))2/3 + C
3
2
2
2
u1/3
cos (✓) 1 + tan(✓)
(u+1)
(b)(9 pts) Let u = 2x 1 ) x = (u + 1)/2 and dx = du/2 thus xdx = (u+1)
· du
2
2 =
4 du and so
Z 5
Z 9
Z 9
xdx
1
u+1
1
1/2
1/2
p
p du =
=
(u + u
) du
4 1
4 1
u
2x 1
1
✓
◆ 9
✓
◆ ✓
◆

1 2 3/2
1
2
2
1
8
=
u + 2u1/2
=
· 27 + 2 · 3
+2 =
24
=6
4 3
4
3
3
4
3
1
(c)(9 pts) We can use symmetry of functions and geometry for this integral, note that
Z a
⇡a dx = ⇡a · [(a ( a)] = 2⇡a2 (area of a rectangle)
a
Z a
Z a
1
⇡|x| dx = 2⇡
|x| dx = 2⇡ · a · a = ⇡a2 (y = ⇡|x| is even and area of a triangle)
2
a
0
Z ap
1
a2 x2 dx = ⇡a2 (area of a semicircle of radius a)
2
a
3
1
sin(2x)
dx =
x
Z
6
2
sin(u)
du = F (u)
u
2
u
·
du
2
= du/u thus
6
= F (6)
F (2)
2
where the last equality follows by the Fundamental Theorem of Calculus (Part Deux).
4. The following problems are not related, remember to justify your answers.
(a)(10 pts) Ralphie has asked you to design an open-top stainless steel box. It is to have square base and a volume
of 32 ft3 , to be welded from thin stainless steel and to weigh no more than necessary. Find the dimensions (length,
width, and height) of such a box.
2
16
=
3
3
(b)(10 pts) Suppose f (x) = x4 /4 + x2
3x, set up (but do not evaluate) Newton’s method to approximate a critical
point of the function f (x). (In other words, give a formula for calculating the (n + 1)-st approximation, xn+1 , in terms
of the n-th approximation, xn .) Simpify your answer.
✓
◆
2⇡i
sin 2⇡ +
equivalent to? Clearly write
n
n
i=1
down your answer(s) in your blue book, no justification necessary - be sure to copy down the entire answer, don’t
(c)(5 pts) Which of the five choices given below is the limit lim
n!1
just write down the roman numeral of your choice(s).
Z 3⇡
Z 2⇡
Z ⇡
sin(2x)
(i)
dx (ii)
sin(2x) dx (iii)
sin(2x + ⇡) dx
2
⇡
⇡
0
n
X
⇡
(iv)
Z
⇡
0
⇡
sin(2⇡ + x) dx
x
(v)
Z
⇡
sin(2x + ⇡) dx
⇡
Solution: (a)(10 pts) For the vat to weigh the least we need to minimize the amount of material used to construct the
vat, so we need to minimize the surface area of the vat, thus if x represents the width of the base and if h represents
the height of the vat, then the optimization problem we are solving is
minimize SA = x2 + 4xh subject to V = x2 h = 32.
Using the constraint we see that h = 32/x2 , x > 0 and so we need to minimize
SA(x) = x2 + 4x ·
and so
d
dx SA
32
128
d
= x2 +
=)
SA = 2x
x2
x
dx
= 0 implies x3 = 64 ) x = 4 and note that
128
x2
d
dx SA
is undefined when x = 0 but this is not a feasible answer
256
d2
= 2+ 3
>0
dx2 SA
x x=4
x=4
Derivative Test and so we have a local minimum at x = 4, thus the dimensions of the vat that will meet
since this not in the domain of the problem (the volume is positive). Now note that
by the 2nd
Ralphie’s specifications are w ⇥ l ⇥ h = 4 ft ⇥ 4 ft ⇥ 2 ft.
(b)(10 pts) Note that if f (x) = x4 /4 + x2
3x then f 0 (x) = x3 + 2x
3 and so we wish to set up Newtons’s Method
to approximate a root of the equation f 0 (x) = 0, thus we have
xn+1 = xn
f 0 (xn )
= xn
f 00 (xn )
(c)(5 pts) Choice (ii). Consider the integral
Z
x3n + 2xn 3
2x3 + 3
= n2
for n = 0, 1, 2, . . .
3x2n + 2
3xn + 2
2⇡
sin(2x) dx then
x = (2⇡ ⇡)/n = ⇡/n and xi = a+i x = ⇡+i⇡/n,
⇡
now if we use right endpoints, i.e. x⇤i = xi then we have
lim
n!1
n
X
i=1
f (x⇤i ) x = lim
n!1
n
X
i=1
f (xi ) x = lim
n!1
n
X
i=1
sin(2(⇡ + i⇡/n))
◆
✓
n
X
⇡
⇡
2⇡i
= lim
sin 2⇡ +
n n!1 i=1 n
n