Math 113 HW #11 Solutions

§5.14. (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 yourestimate 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.

0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4

0.5

1

1.5

2

Since each rectangle has width 1, the area of the first rectangle is 1 · 1 = 1, the area ofthe 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.

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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 fourrectangles are

f(0) =√

0 = 0

f(1) =√

1 = 1

f(2) =√

2 ≈ 1.414

f(3) =√

3 ≈ 1.732.

0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4

0.5

1

1.5

2

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) =lnx

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 subintervalsof equal width, each will have width ∆x = 7/n. Then the area under the graph will be givenby

limn→∞

n∑i=1

f(x∗i )∆x = limn→∞

n∑i=1

lnx∗ix∗i

7n

2

for any choice of sample points x∗i , where x∗i is in the ith subinterval. Choosing, say, the rightendpoint of each as the sample point, we can see that

x∗i = 3 + i7n

,

so the above limit becomes

limn→∞

n∑i=1

ln(3 + i 7

n

)3 + i 7

n

7n

.

§5.218. Express the limit

limn→∞

n∑i=1

cos xi

xi∆x

as a definite integral on [π, 2π].

Answer: This is simply the definition of the definite integral∫ 2π

π

cos x

xdx.

22. Use the form of the definition of the integral given in Theorem 4 to evaluate the integral∫ 4

1(x2 + 2x− 5) dx.

Answer: Breaking the interval [1, 4] into n subintervals of equal width, each will be of width

∆x =4− 1

n=

3n

.

Moreover, the right endpoint of the ith subinterval will be

xi = 1 + i3n

.

Therefore, the height of the ith rectangle will be (since we’re using right endpoints),

f(xi) = x2i + 2xi − 5

=(

1 + i3n

)2

+ 2(

1 + i3n

)− 5

= 1 + i6n

+ i29n2

+ 2 + i6n− 5

= i29n2

+ i12n− 2.

3

Therefore, ∫ 4

1(x2 + 2x− 5) dx = lim

n→∞

n∑i=1

f(xi)∆x

= limn→∞

n∑i=1

(i2

9n2

+ i12n− 2)

3n

= limn→∞

n∑i=1

[i2

27n3

+ i36n2− 6

n

]

= limn→∞

[n∑

i=1

i227n3

+n∑

i=1

i36n2−

n∑i=1

6n

]

= limn→∞

[27n3

n∑i=1

i2 +36n2

n∑i=1

i− 6n

n∑i=1

1

]

Therefore, sincen∑

i=1

1 = 1

n∑i=1

i =n(n + 1)

2n∑

i=1

i2 =n(n + 1)(2n + 1)

6,

we see that the above limit is equal to

limn→∞

[27n3

n(n + 1)(2n + 1)6

+36n2

n(n + 1)2

− 6n

n

]= lim

n→∞

[54n3 + 81n2 + 27n

6n3+

36n2 + 36n

2n2− 6]

=546

+362− 6

= 9 + 18− 6= 21.

Therefore, ∫ 4

1(x2 + 2x− 5) dx = 21.

34. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral

(a)∫ 20 g(x)dx

Answer: Since on [0, 2] the graph of g(x) is just a straight line of slope −2 coming downfrom y = 4 to y = 0, the area is just the area of the triangle

122 · 4 = 4.

Since this area is above the x-axis, definite integral equals the area, so∫ 20 g(x)dx = 4.

4

(b)∫ 62 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

12π(2)2 = 2π.

Since it lies below the axis, the integral is negative, so∫ 6

2g(x)dx = −2π.

(c)∫ 70 g(x)dx

Answer: Since∫ 7

0g(x)dx =

∫ 2

0g(x)dx +

∫ 6

2g(x)dx +

∫ 7

6g(x)dx = 4− 2π +

∫ 7

6g(x)dx,

we just need to determine∫ 76 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

121 · 1 =

12.

Since this area lies above the axis,∫ 76 g(x)dx = 1/2, so∫ 7

0g(x)dx = 4− 2π +

∫ 7

6g(x)dx = 4− 2π +

12

=92− 2π ≈ −1.78.

44. Use the result of Example 3 to compute∫ 3

1(2ex − 1)dx.

Answer: Example 3 says that∫ 31 exdx = e3− e, we need to use the properties of the definite

integral to express the given integral in terms of∫ 31 exdx.

Now, by Property 4, ∫ 3

1(2ex − 1)dx =

∫ 3

12ex −

∫ 3

11dx.

In turn, by Property 1, ∫ 3

11dx = 1(3− 1) = 2.

By Property 3, ∫ 3

12exdx = 2

∫ 3

1exdx.

Putting these together, then, ∫ 3

1(2ex − 1)dx = 2

∫ 3

1exdx− 2.

Plugging in the value we know for∫ 31 exdx, we see that∫ 3

1(2ex − 1)dx = 2(e3 − e)− 2 = 2(e3 − e− 1) ≈ 32.73.

5

§5.314. Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function

h(x) =∫ x2

0

√1 + r3 dr.

Answer: Make the change of variables u = x2. Then

h′(x) =d

dx

(∫ x2

0

√1 + r3 dr

)=

d

dx

(∫ u

0

√1 + r3 dr

).

By the Chain Rule, this is equal to

d

du

(∫ u

0

√1 + r3 dr

)du

dx.

Using the Fundamental Theorem and the fact that dudx = 2x, we see that

h′(x) =√

1 + u3 (2x) =√

1 + (x2)3 (2x) = 2x√

1 + x6.

26. Evaluate the integral ∫ 2π

πcos θ dθ.

Answer: Since sin θ is an antiderivative of cos θ, the second part of the Fundamental Theoremsays that ∫ 2π

πcos θ dθ =

[sin θ

]2π

π= sin 2π − sinπ = 0− 0 = 0.

36. Evaluate the integral ∫ 1

010x dx.

Answer: Sinced

dx(10x) = 10x ln 10,

we see that10x

ln 10is an antiderivative of 10x. Therefore,∫ 1

010xdx =

[10x

ln 10

]1

0

=10

ln 10− 1

ln 10=

9ln 10

.

6

40. Evaluate the integral ∫ 2

1

4 + u2

u3du.

Answer: Re-write the integral as∫ 2

1

(4u3

+u2

u3

)du =

∫ 2

14u−3du +

∫ 2

1u−1du.

Then, since u−2

−2 = − 12u2 is an antiderivative for u−3 and since lnu is an antiderivative for

u−1, we see that the above is equal to[4−12u2

]2

1

+[lnu]21

=(−2

4+

21

)+ (ln 2− ln 1) =

32

+ ln 2.

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