# Test1 date unknown+sol ```MATH 1014 3.00 PW – APPLIED CALCULUS II – Prof. Madras
Solutions for Practice Test #1
1. (a) Integrate by parts: Let u = ln x and dv = x−1/2 dx. Then du = (1/x) dx and
v = 2x1/2 , and we get
Z
Z
ln x
1
1/2
√ dx = 2x ln x − 2x1/2 dx
x
x
= 2x1/2 ln x − 4x1/2 + C .
(b) Use partial fractions. Write
3x + 4
(x − 2)(x + 3)
=
=
A
B
+
x−2 x+3
Ax + 3A + Bx − 2B
(x − 2)(x + 3)
Matching coefficients gives A + B = 3 and 3A − 2B = 4. The solution of this pair
of equations is A = 2 and B = 1. So
Z
Z
3x + 4
2
1
dx =
+
dx = 2 ln |x − 2| + ln |x + 3| + C.
(x − 2)(x + 3)
x−2 x+3
√
√
2. Let x = tan t. Then dx = sec2 t dt, and 1 + x2 = 1 + tan2 t = sec t. Thus
Z
Z
Z
sec t
1
1
2
√
sec t dt =
dx =
2
2 dt .
2
2
tan
t
sec
t
tan
t
x 1+x
It is not clear what to do with this, so try writing it in terms of sin and cos:
Z
Z
1/ cos t
cos t
dt .
=
dt
=
(sin t/ cos t)2
sin2 t
Now let w = sin t, so that dw = cos t dt:
Z
dw
1
=
= −w−1 + C = −
+ C.
w2
sin t
Finally, we need to express this in terms of x again. Since tan t = x, we can draw
the following triangle:
x
t
1
√
2
The length of the
√ hypoteneuse of this triangle must be 1 + x , and this tells us
that sin t = x/ 1 + x2 . So we can finish as follows:
√
Z
1 + x2
1
1
√
+C = −
+C.
dx = −
sin t
x
x2 1 + x2
3. The aquarium contains 4 cubic meters of water, so we only need to pump out
the top quarter of the water. Consider a layer of water of thickness ∆y at height y
meters above the bottom. The mass of this layer is
3
(density) &times; (volume) =
1000 kg./m &times; 4 &times; 1 &times; ∆y m3
=
4000∆y kg.
The force needed to raise this layer is 9.8 &times; 4000∆y newtons. This layer must be
raised a distance of 1 − y meters. Since we are only moving water from the top
quarter of the tank, y ranges from 3/4 to 1. Therefore, the work (in joules) required
for all of the layers is about
X
9.8 &times; 4000 &times; ∆y &times; (1 − yi )
i
(where yi is the height of the ith layer). This corresponds to the integral
Z 1
9.8 &times; 4000(1 − y) dy .
3/4
4. (This is similar to Example 4 on page 426 of the text; see the sketches there.)
The two curves intersect at
√ (0, 0) and (1, 1). Between these two points (i.e. for
0 ≤ x ≤ 1), the curve y = x lies above the curve y = x4 (and both lie above the
x-axis).
Consider the solid obtained by rotating this region about the x-axis. For each
number x between 0 and 1, let A(x) be the cross-section area of the slice through
the solid by the plane that is perpendicular to the x-axis and that intersects the
x-axis at the √
given value of x. The shape of the cross-section is a “washer” with
outer radius x and inner radius x4 (i.e., the radius of the hole in the washer is
x4 ). Therefore
√
A(x) = π( x)2 − π(x4 )2 = π(x − x8 )
and the volume of the solid is
2
Z 1
Z 1
x
x9
−
A(x) dx =
π(x − x8 ) dx = π
2
9
0
0
1
= π
0
1 1
−
2 9
=
7
π.
18
5. (a) We have
∆x =
2
4−2
= , x0 = 2,
3
3
x1 =
8
10
, x2 =
, x3 = 4
3
3
so the Trapezoidal Rule estimate is
T3
=
=
=
=
∆x
(f (x0 ) + 2f (x1 ) + 2f (x2 ) + f (x3 ))
2
2/3
(ln(2) + 2 ln(8/3) + 2 ln(10/3) + ln(4))
2
1
(.6931 + 2(.9808) + 2(1.2040) + 1.3863)
3
2.1493.
(b) The absolute value of the error is less than or equal to
|4 − 2|3 K2
,
12n2
where K2 is a number that satisfies
d2
ln x ≤ K2
dx2
for every x in [2, 4].
Since the second derivative of ln x is −x−2 , we need to make sure that x−2 ≤ K2
for every x in [2, 4]. Since the largest that x−2 gets (for x in [2, 4]) is 2−2 , we can
take K2 = 2−2 = 0.25. So to guarantee that Tn is accurate to 4 decimal places, we
need to choose n large enough so that
|4 − 2|3 (0.25)
≤ 10−4
12n2
i.e.
i.e.
2 &times; 104
≤ n2
12
40.82 ≤ n
Therefore, 41 subintervals suffice.
(c) The function ln x is concave down (note that its second derivative is negative).
A sketch of a positive concave-down function shows that the trapezoids would all lie
below the graph of the function. Therefore the sum of the areas of the trapezoids
is smaller than the area under the curve. For this reason, we expect the estimate
from (a) to be a bit smaller than the true value of the integral.
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