MATH 172.507
Exam 1 Solutions
February 24, 2016
2
Z
1.
Evaluate
0
6x2
dx.
x3 + 1
(A) 2 ln(6)
(B) 2 ln(7)
(C) 2 ln(8)
(D) 2 ln(9)
(E) 2 ln(10)
Solution: D We perform a substitution. Let u = x3 + 1, so du = 3x2 dx. Therefore,
2
Z
0
2.
6x2
dx =
x3 + 1
Z
u(2)
u(0)
h
i9
2
du = 2 ln |u| = 2 ln 9 − 2 ln 1 = 2 ln 9 .
u
1
Consider the region in the first quadrant bounded by x = y 2 , y = 3, and the y-axis.
Which integral below gives the volume of the solid obtained by rotating this region
about the x-axis?
Z 9
(A) π
(9 − x) dx
0
Z
9
x2 dx
(B) π
0
9
Z
x(3 − x) dx
(C) 2π
0
Z
3
y(3 − y 2 ) dy
(D) 2π
0
Z
(E) 2π
3
y 4 dy
0
Solution: A We first find the intersection point between x = y 2 and y = 3, which
is easily seen to be (x, y) = (9, 3). The region
in question then sits above the interval
√
0 ≤ x ≤ 9 and between the curves y = x and y = 3. We then need to use washers
to find the volume of the solid obtained by rotating the region around the x-axis. We
see that for a given x with 0 ≤ x ≤ 9, the outer radius is 3 and the inner radius is
√
x. Thus,
Z 9
Z 9
√ 2
2
V =π
(3 − ( x) ) dx = π
(9 − x) dx .
0
0
1
Z
3.
Evaluate
π
x sin(x) dx.
0
(A) π + 1
(B) 2π
(C) π
(D) −π
(E) −π − 1
Solution: C We use integration by parts, picking u = x and dv = sin(x) dx. Then
du = dx and v = − cos(x). Thus,
Z
Z
x sin(x) dx = −x cos(x) + cos(x) dx = −x cos(x) + sin(x),
and so
Z
π
h
x sin(x) dx = −x cos(x) + sin(x)
= (−pi(−1) + 0) − (0 + 0) = π .
0
0
4.
iπ
A 10 meter cable has a total mass of 4/(9.8) kg and a constant mass density. It is
initially hanging from the top of a tall building. How much work (in Joules) is done
against gravity when lifting the cable to the roof of the building? (Assume that the
acceleration of gravity is 9.8 m/s2 .)
(A) 10
(B) 20
(C) 30
(D) 40
(E) 50
Solution: B We see right away that the mass density δ of the cable is
δ=
4
kg/m.
(9.8)(10)
A piece of the cable of infinitesimal width dx that is a distance x from the top of the
building, then has a mass of
m=
4
dx kg,
(9.8)(10)
and so the force of gravity acting on this piece is
F = mg =
(4)(9.8)
2
dx = dx N.
(9.8)(10)
5
Our piece at x will travel a distance x to get to the top of the building, so
2 10
Z 10
2
x
100
W =
x dx =
=
= 20 J .
5
5 0
5
0
2
Z
5.
Let F (x) =
x
√
t3 + t dt. What is F 0 (2)?
1
18
(A) √
8
√
(B) 8
√
(C) 10
√
√
(D) 8 − 2
√
√
(E) 10 − 2
Solution: C From the Fundamental Theorem of Calculus we know that F 0 (x) =
√
√
√
x3 + x. Therefore, F 0 (2) = 8 + 2 = 10 .
6.
Consider the following limit of Riemann sums:
r
r
r
2
2
4
2(n − 1)
lim
1 + 1 + + 1 + + ··· + 1 +
.
n→∞ n
n
n
n
Which definite integral below is equal to this limit?
Z 2
(A)
x dx
0
3
Z
1
dx
1+x
1
Z 2
√
(1 + x) dx
(C)
√
(B)
0
Z
(D)
3
√
x dx
1
(E) None of the above
Solution: D When we look at the expressions under the square-roots, we see that
each is n2 more than√the preceding one, so the length of the interval of integration is
2. If we let f (x) = x, then the limit is
2
2
4
2(n − 1)
lim
f (1) + f 1 +
+f 1+
+ ··· + f 1 +
,
n→∞ n
n
n
n
and so we are integrating f (x) over the interval that starts at 1 and ends at 1 + 2 = 3.
Z 3
√
Thus the limit is
x dx .
1
3
7.
Let R be the region in the first quadrant that is below the graph of y = −x2 − 4x + 16
and above the graph of y = x2 . Find the area of R.
Solution: We find the intersection points between the two parabolas: x2 = −x2 −
4x + 16 yields the equation
2x2 + 4x − 16 = 0 ⇔ x2 + 2x − 8 = 0 ⇔ (x + 4)(x − 2) = 0.
Therefore the parabolas intersect when x = −4 and x = 2. Since the region R is in the
first quadrant, it sits above the interval 0 ≤ x ≤ 2. The top curve is y = −x2 −4x+16,
and the bottom curve is y = x2 , so the area of R is
Z 2
Z 2
2
2
(−2x2 − 4x + 16) dx
(−x − 4x + 16 − x ) dx =
A=
0
0
2
2 3
2
= − x − 2x + 16x
3
0
16
= − − 8 + 32
3
56
=
.
3
8.
Find the average value of the function f (x) = sin(3x) cos(x) over the interval [0, π2 ].
Solution: The average value of f on [0, π2 ] is
2
π
π/2
Z
sin(3x) cos(x) dx.
0
We use the identity
1
sin(A) cos(B) = (sin(A − B) + sin(A + B)).
2
so
2
π
Z
0
π/2
2
sin(3x) cos(x) dx =
π
Z
π/2
1
(sin(2x) + sin(4x)) dx
2
0
π/2
1
1
1
=
− cos(2x) − cos(4x)
π
2
4
0
1
1
1
1
1
=
− cos(π) − cos(2π) + cos(0) + cos(0)
π
2
4
2
4
1 1 1 1 1
=
− + +
π 2 4 2 4
=
1
.
π
4
9.
Evaluate the following integrals.
Z
√
(a) 2x3 x2 + 4 dx
Z
(b) cos2 (2x) dx
Z
(c) ex sin(x) dx
Solution: (a) We use substitution. Let u = x2 + 4, so du = 2x dx. Also x2 = u − 4.
Therefore,
Z
Z
√
√
3
2x x2 + 4 dx = x2 · 2x · x2 + 4 dx
Z
√
= (u − 4) u du
Z
= (u3/2 − 4u1/2 ) du
2
8
= u5/2 − u3/2 + C
5
3
2
8
= (x2 + 4)5/2 − (x2 + 4)3/2 + C .
5
3
(b) We use the identity that cos2 (θ) = 21 (1 + cos(2θ)), and so
Z
2
Z
cos (2x) dx =
1
x 1
(1 + cos(4x)) dx =
+ sin(4x) + C .
2
2 8
(c) We use integration by parts with u = ex and dv = sin(x) dx, so du = ex dx and
v = − cos(x). Therefore,
Z
Z
x
x
e sin(x) dx = −e cos(x) + ex cos(x) dx.
We need to perform integration by parts again, with u = ex and dv = cos(x) dx (so
du = ex dx and v = sin(x)), and we obtain
Z
Z
x
x
x
e sin(x) dx = −e cos(x) + e sin(x) − ex sin(x) dx.
Moving the integral on the right to the left side, we have
Z
2 ex sin(x) dx = −ex cos(x) + ex sin(x) + C,
and so
Z
1
1
ex sin(x) dx = − ex cos(x) + ex sin(x) + C .
2
2
5
10.
Let R be the first region in the first quadrant to the right of the y-axis that is bounded
1
above by the graph of y = cos(x) and below by the graph of y = √ .
2
In each part below, set up an integral for volume of the given solid (but do not
evaluate). It may help to first sketch the region R.
(a) The solid obtained by rotating R about the x-axis:
(b) The solid obtained by rotating R about the line x = −1:
(c) The solid whose base is R and whose cross-sections perpendicular to the x-axis
are squares:
√
2 is a horizontal line, and it intersects y = cos(x)
Solution: The graph
of
y
=
1/
√
when cos(x) = 1/ 2, so when x = π4 . Thus R is the region that is above the interval
√
0 ≤ x ≤ π4 and is bounded on the bottom by y = 1/ 2 and on the top by y = cos(x).
(a) Rotating about the
√ x-axis we need to use washers. The outer radius is cos(x) and
the inner radius is 1/ 2, so
π/4 Z
V = π
0
1
cos (x) −
2
2
dx .
(b) Rotating about the vertical line x = −1, we need to use cylindrical shells. For
π
a shell starting
√ at some x with 0 ≤ x ≤ 4 , the radius is x + 1 and the height is
cos(x) − 1/ x. Therefore,
π/4
Z
V = 2π
0
1
(x + 1) cos(x) − √ dx .
2
(c) We need to integrate the
√ areas of the cross-sections, which at a given x is a square
of side-length cos(x) − 1/ 2. Thus,
Z
V =
0
π/4 1
cos(x) − √
2
6
2
dx .