Math 113 HW #11 Solutions
[Pages:7]Math 113 HW #11 Solutions
?5.1
4. (a) Estimate the area under the graph of f (x) = x from x = 0 to x = 4 using four approx-
imating rectangles and right endpoints. Sketch the graph and the rectangles. Is your estimate an underestimate or an overestimate? Answer: Since [0, 4] has length 4, each of the four rectangles will have width 4/4 = 1, so the right endpoints are 1, 2, 3 and 4. Thus, the heights of the four rectangles are
f (1) = 1 = 1
f (2) = 2 1.414
f (3) = 3 1.732
f (4) = 4 = 2.
2
1.5
1
0.5
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
Since each rectangle has width 1, the area of the first rectangle is 1 ? 1 = 1, the area of the second is 2 ? 1 = 2, etc. Thus, we can estimate the area under the curve as
1 + 2 + 3 + 2 6.146. Since f (x) is an increasing function, this is an over-estimate of the actual area. (b) Repeat part (a) using left endpoints.
1
Answer: The endpoints of the four sub-intervals are the same, though now we're in-
terested in the left endpoints, which are 0, 1, 2, and 3. Thus, the heights of the four
rectangles are
f (0) = 0 = 0
f (1) = 1 = 1
f (2) = 2 1.414
f (3) = 3 1.732.
2
1.5
1
0.5
0
0.4
0.8
1.2
1.6
2
2.4
2.8
3.2
3.6
4
Thus, the area contained in these rectangles is
0 + 1 + 2 + 3 4.146,
which is an underestimate of the actual area.
18. Use Definition 2 to find an expression for the area under the graph of
ln x f (x) = , 3 x 10
x as a limit. Do not evaluate the limit.
Answer: Since [3, 10] has length 10 - 3 = 7, if we break this interval up into n subintervals
of equal width, each will have width x = 7/n. Then the area under the graph will be given
by
lim
n
n i=1
f (xi )x
=
lim
n
n i=1
ln xi xi
7 n
2
for any choice of sample points xi , where xi is in the ith subinterval. Choosing, say, the right endpoint of each as the sample point, we can see that
xi
=
3
+
i
7 n
,
so the above limit becomes
lim
n
n i=1
ln
3
+
i
7 n
3
+
i
7 n
7 .
n
?5.2
18. Express the limit
lim n cos xi x
n i=1
xi
as a definite integral on [, 2].
Answer: This is simply the definition of the definite integral
2 cos x dx.
x
22. Use the form of the definition of the integral given in Theorem 4 to evaluate the integral
4
(x2 + 2x - 5) dx.
1
Answer: Breaking the interval [1, 4] into n subintervals of equal width, each will be of width
4-1 3
x =
=.
nn
Moreover, the right endpoint of the ith subinterval will be
3
xi
=
1
+
i. n
Therefore, the height of the ith rectangle will be (since we're using right endpoints),
f (xi) = x2i + 2xi - 5
32
3
= 1+i +2 1+i -5
n
n
=
1
+
6 i n
+
i2
9 n2
+
2
+
6 i n
-
5
=
i2
9 n2
+
12 i
n
-
2.
3
Therefore,
4
n
1
(x2 + 2x - 5) dx
=
lim
n
f (xi)x
i=1
n
= limn
i2
9 n2
+
12 i
n
-
2
3 n
i=1
n
= lim
n
i2
27 n3
+
36 i n2
-
6 n
i=1
= lim
n
n
i2
27 n3
+
n
36 i n2 -
n
6 n
i=1
i=1
i=1
= lim
n
27 n3
n
i2
+
36 n2
n
6
i-
n
n
1
i=1
i=1
i=1
Therefore, since
n
1=1
i=1
n
n(n + 1)
i=
2
i=1
n
i2
=
n(n
+
1)(2n
+
1) ,
6
i=1
we see that the above limit is equal to
27 n(n + 1)(2n + 1) 36 n(n + 1) 6
54n3 + 81n2 + 27n 36n2 + 36n
lim
n
n3
6
+ n2
2
- n = lim
n
n
6n3
+ 2n2 - 6
54 36 = + -6
62
= 9 + 18 - 6
= 21.
Therefore,
4
(x2 + 2x - 5) dx = 21.
1
34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral
(a)
2 0
g(x)dx
Answer: Since on [0, 2] the graph of g(x) is just a straight line of slope -2 coming down
from y = 4 to y = 0, the area is just the area of the triangle
1 2 ? 4 = 4.
2
Since this area is above the x-axis, definite integral equals the area, so
2 0
g(x)dx
=
4.
4
(b)
6 2
g(x)dx
Answer: On [2, 6] the graph of g(x) is a semi-circle of radius 2 lying below the x-axis.
Its area is
1 (2)2 = 2. 2
Since it lies below the axis, the integral is negative, so
6
g(x)dx = -2.
2
(c)
7 0
g(x)dx
Answer: Since
7
2
6
7
7
g(x)dx = g(x)dx + g(x)dx + g(x)dx = 4 - 2 + g(x)dx,
0
0
2
6
6
we just need to determine
7 6
g(x)dx.
Since this is a straight line of slope 1 going up
from the x-axis (at x = 6) to y = 1 (at x = 7), it describes a triangle of area
1
1
1?1= .
2
2
Since this area lies above the axis,
7 6
g(x)dx
=
1/2,
so
7
7
19
g(x)dx = 4 - 2 + g(x)dx = 4 - 2 + = - 2 -1.78.
0
6
22
44. Use the result of Example 3 to compute
3
(2ex - 1)dx.
1
Answer: Example 3 says that
3 1
exdx
=
e3
-
e,
we
need
to
use
the
properties
of
the
definite
integral to express the given integral in terms of
3 1
ex
dx.
Now, by Property 4,
3
3
3
(2ex - 1)dx = 2ex - 1dx.
1
1
1
In turn, by Property 1,
3
1dx = 1(3 - 1) = 2.
1
By Property 3,
3
3
2exdx = 2 exdx.
1
1
Putting these together, then,
3
3
(2ex - 1)dx = 2 exdx - 2.
1
1
Plugging in the value we know for
3 1
exdx,
we
see
that
3
(2ex - 1)dx = 2(e3 - e) - 2 = 2(e3 - e - 1) 32.73.
1
5
?5.3
14. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function
x2
h(x) =
1 + r3 dr.
0
Answer: Make the change of variables u = x2. Then
d h (x) =
dx
x2
1 + r3 dr
d =
0
dx
u
1 + r3 dr .
0
By the Chain Rule, this is equal to
d
u
1 + r3 dr
du .
du 0
dx
Using
the
Fundamental
Theorem
and
the
fact
that
du dx
=
2x,
we
see
that
h (x) = 1 + u3 (2x) = 1 + (x2)3 (2x) = 2x 1 + x6.
26. Evaluate the integral
2
cos d.
Answer: Since sin is an antiderivative of cos , the second part of the Fundamental Theorem
says that
2
2
cos d = sin = sin 2 - sin = 0 - 0 = 0.
36. Evaluate the integral
1
10x dx.
0
Answer: Since
d (10x) = 10x ln 10, dx
we see that
10x
ln 10
is an antiderivative of 10x. Therefore,
1
10xdx =
10x 1 10
=
-
1
=
9 .
0
ln 10 0 ln 10 ln 10 ln 10
6
40. Evaluate the integral Answer: Re-write the integral as
2 4 + u2 1 u3 du.
2 1
4 u2 u3 + u3
2
2
du = 4u-3du + u-1du.
1
1
Then,
since
u-2 -2
=
-
1 2u2
is
an
antiderivative
for u-3
and
since
ln u
is
an
antiderivative for
u-1, we see that the above is equal to
-1 2
2
22
3
4 2u2
+
1
ln u =
1
-+ 41
+ (ln 2 - ln 1) = + ln 2. 2
7
................
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