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 approximating 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 interested 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
f (x) =
ln x
,
x
3 ≤ x ≤ 10
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
n
n
X
X
ln x∗i 7
∗
lim
f (xi )∆x = lim
n→∞
n→∞
x∗i n
i=1
i=1
2
for any choice of sample points x∗i , where x∗i is in the ith subinterval. Choosing, say, the right
endpoint of each as the sample point, we can see that
7
x∗i = 3 + i ,
n
so the above limit becomes
n
X
ln 3 + i n7 7
lim
.
7
n→∞
n
3
+
i
n
i=1
§5.2
18. Express the limit
lim
n→∞
n
X
cos xi
i=1
xi
∆x
as a definite integral on [π, 2π].
Answer: This is simply the definition of the definite integral
Z 2π
cos x
dx.
x
π
22. Use the form of the definition of the integral given in Theorem 4 to evaluate the integral
Z 4
(x2 + 2x − 5) dx.
1
Answer: Breaking the interval [1, 4] into n subintervals of equal width, each will be of width
∆x =
4−1
3
= .
n
n
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
3
3 2
= 1+i
+2 1+i
−5
n
n
6
9
6
= 1 + i + i2 2 + 2 + i − 5
n
n
n
9
12
= i2 2 + i − 2.
n
n
3
Therefore,
Z
n
X
4
(x2 + 2x − 5) dx = lim
n→∞
1
f (xi )∆x
i=1
n X
9
12
3
i 2 +i −2
= limn→∞
n
n
n
i=1
n X
36
6
2 27
= lim
i 3 +i 2 −
n→∞
n
n
n
i=1
" n
#
n
n
X 27 X
X
36
6
= lim
i2 3 +
i 2−
n→∞
n
n
n
i=1
i=1
i=1
"
#
n
n
n
27 X 2 36 X
6X
= lim
i + 2
i−
1
n→∞ n3
n
n
2
i=1
i=1
i=1
Therefore, since
n
X
i=1
n
X
1=1
i=
i=1
n
X
i=1
i2 =
n(n + 1)
2
n(n + 1)(2n + 1)
,
6
we see that the above limit is equal to
54n3 + 81n2 + 27n 36n2 + 36n
27 n(n + 1)(2n + 1) 36 n(n + 1) 6
+ 2
− n = lim
+
−6
lim
n→∞
n→∞ n3
6
n
2
n
6n3
2n2
54 36
+
−6
=
6
2
= 9 + 18 − 6
= 21.
Therefore,
Z
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
R2
(a) 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
4
R2
0
g(x)dx = 4.
(b)
R6
(c)
R7
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
π(2)2 = 2π.
2
Since it lies below the axis, the integral is negative, so
Z 6
g(x)dx = −2π.
2
2
0 g(x)dx
Answer: Since
Z 7
Z
g(x)dx =
0
2
6
Z
7
Z
g(x)dx +
0
Z
g(x)dx = 4 − 2π +
g(x)dx +
2
6
7
g(x)dx,
6
R7
we just need to determine 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
R7
Since this area lies above the axis,
Z 7
Z
g(x)dx = 4 − 2π +
0
6
g(x)dx = 1/2, so
7
g(x)dx = 4 − 2π +
6
44. Use the result of Example 3 to compute
Z
1
9
= − 2π ≈ −1.78.
2
2
3
(2ex − 1)dx.
1
R3
Answer: Example 3 says that 1 ex dx = e3 − e, we need to use the properties of the definite
R3
integral to express the given integral in terms of 1 ex dx.
Now, by Property 4,
3
Z
Z
x
3
x
(2e − 1)dx =
2e −
1
1
Z
3
1dx.
1
In turn, by Property 1,
3
Z
1dx = 1(3 − 1) = 2.
1
By Property 3,
3
Z
3
Z
x
ex dx.
2e dx = 2
1
1
Putting these together, then,
Z
3
Z
x
(2e − 1)dx = 2
1
3
ex dx − 2.
1
R3
Plugging in the value we know for 1 ex dx, we see that
Z 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
Z x2 p
h(x) =
1 + r3 dr.
0
Answer: Make the change of variables u = x2 . Then
!
Z u p
Z x2 p
d
d
0
3
3
1 + r dr =
1 + r dr .
h (x) =
dx
dx
0
0
By the Chain Rule, this is equal to
d
du
up
Z
1+
r3 dr
0
Using the Fundamental Theorem and the fact that
h0 (x) =
du
dx
du
.
dx
= 2x, we see that
p
p
p
1 + u3 (2x) = 1 + (x2 )3 (2x) = 2x 1 + x6 .
26. Evaluate the integral
2π
Z
cos θ dθ.
π
Answer: Since sin θ is an antiderivative of cos θ, the second part of the Fundamental Theorem
says that
Z 2π
h
i2π
cos θ dθ = sin θ
= sin 2π − sin π = 0 − 0 = 0.
π
π
36. Evaluate the integral
1
Z
10x dx.
0
Answer: Since
d
(10x ) = 10x ln 10,
dx
we see that
10x
ln 10
is an antiderivative of 10x . Therefore,
Z
0
1
10x
10 dx =
ln 10
x
1
=
0
6
10
1
9
−
=
.
ln 10 ln 10
ln 10
40. Evaluate the integral
Z
1
2
4 + u2
du.
u3
Answer: Re-write the integral as
Z 2
Z 2
Z 2
4
u2
−3
u−1 du.
4u
du
+
du
=
+
3
3
u
u
1
1
1
−2
Then, since u−2 = − 2u1 2 is an antiderivative for u−3 and since ln u is an antiderivative for
u−1 , we see that the above is equal to
i2 2 2 −1 2 h
3
4 2 + ln u = − +
+ (ln 2 − ln 1) = + ln 2.
2u 1
4 1
2
1
7