Math homework!!

MATH 1310 Homework Assignment 3 Winter 2022

1. Let Jn(a) =

∫ a

0

1

(x2 + a2)n

dx where a is a real number and n is a non-negative integer. Evaluate

J0(a) and J1(a). As well, for a 6= 0 and n ≥ 1, find a formula for Jn+1(a) in terms of Jn(a). As

part of your solution, state the formula for integration by parts.

2. Suppose f(x) has a continuous second derivative for all x ∈ R. For any x ∈ R, prove

f(x) = f(0) + f ′(0)x +

∫ x

0

(x−s)f ′′(s)ds.

3. Provide an example of an indefinite integral that when integrating requires sin(x) as a substitu-

tion. As well, evaluate your indefinite integral.

4. Let g be a continuous function. Prove

∫ b

a

g(x)dx =

∫ a+b

2

a

(g(x) + g(a + b−x))dx.

5. Find and explain the errors (if any) in the following solution.

We evaluate

∫

1

1 + t4

dt by first factorizing the denominator as follows. Notice 1 + t4 = (1 +

t2)2 − 2t2 and factoring we have

1 + t4 = (1 + t2)2 − 2t2 = (1 + t2 −

√

2t)(1 + t2 +

√

2t).

Considering the concept of partial fractions we have

1

1 + t4

=

At + B

t2 +

√

2t + 1

+

Ct + D

t2 −

√

2t + 1

(which implies 1 = (At + B)(t2 −

√

2t + 1) + (Ct + D)(t2 +

√

2t + 1)) where A, B, C and D are

determined by solving the following system of equations

1 = B + D,

0 = A−

√

2B + C +

√

2D,

0 = −

√

2A + B +

√

2C + D,

0 = A + C.

The solution to this system of equations is A = −C = 1

2

√

2

, B = D = 1

2

. We can now write

1

1 + t4

as

1

1 + t4

=

1

2

√

2

t + 1

2

t2 +

√

2t + 1

+

− 1

2

√

2

t + 1

2

t2 −

√

2t + 1

=

1

2

√

2

t +

√

2

(t2 +

√

2t + 1)

−

1

2

√

2

t−

√

2

(t2 −

√

2t + 1)

=

1

2

√

2

t +

√

2

((t +

√

2

2

)2 + 1

2

)

−

1

2

√

2

t−

√

2

((t−

√

2

2

)2 + 1

2

)

Copyright 2022 c© A. Chow and D. Yu 1

where in the last line we completed the square. It follows that∫

1

1 + t4

dt =

∫

1

2

√

2

t +

√

2

((t +

√

2

2

)2 + 1

2

)

dt−

∫

1

2

√

2

t−

√

2

((t−

√

2

2

)2 + 1

2

)

dt

let w = t +

√

2

2

and let u = t−

√

2

2

=

1

2

√

2

∫

w +

√

2

2

w2 + 1

2

dw −

1

2

√

2

∫

u−

√

2

2

u2 + 1

2

du

=

1

2

√

2

∫

w

w2 + 1

2

dw +

1

4

∫

1

w2 + 1

2

dw −

1

2

√

2

∫

u

u2 + 1

2

du +

1

4

∫

1

u2 + 1

2

du

let p = w2 +

1

2

and let q = u2 +

1

2

=

1

4

√

2

∫

1

p

dp +

1

4

∫

1

w2 + 1

2

dw −

1

4

√

2

∫

1

q

dq +

1

4

∫

1

u2 + 1

2

du

=

1

4

√

2

∫

1

p

dp +

1

2

∫

1

(

√

2w)2 + 1

dw −

1

4

√

2

∫

1

q

dq +

1

2

∫

1

(

√

2u)2 + 1

du.

We know the antiderivatives of these integrals and hence evaluating we have∫

1

1 + t4

dt =

1

4

√

2

ln |p| +

1

2

√

2

arctan(

√

2w) −

1

4

√

2

ln |q| +

1

2

√

2

arctan(

√

2u)

=

1

4

√

2

ln |w2 +

1

2

| +

1

2

√

2

arctan(

√

2w) −

1

4

√

2

ln |u2 +

1

2

| +

1

2

√

2

arctan(

√

2u)

=

1

4

√

2

ln

∣∣∣∣∣∣

(

t +

√

2

2

)2

+

1

2

∣∣∣∣∣∣ + 12√2 arctan

(

√

2

(

t +

√

2

2

))

−

1

4

√

2

ln

∣∣∣∣∣∣

(

t−

√

2

2

)2

+

1

2

∣∣∣∣∣∣ + 12√2 arctan

(

√

2

(

t−

√

2

2

))

=

1

4

√

2

ln

(t + √2

2

)2

+

1

2

+ 1

2

√

2

arctan

(

√

2

(

t +

√

2

2

))

−

1

4

√

2

ln

(t− √2

2

)2

+

1

2

+ 1

2

√

2

arctan

(

√

2

(

t−

√

2

2

))

.

In the last line, the absolute values have been removed because

(

t +

√

2

2

)2

+ 1

2

and

(

t−

√

2

2

)2

+ 1

2

are positive for any value of t.

6. Evaluate

∫

1

1 + t3

dt using a similar approach as in question 5 (but without the errors, if there

were any).

7. Evaluate

∫

4t4

(1 + t4)2

dt using integration by parts. Hint: question 5.

8. Using symmetry and an appropriate definite integral, prove the area of a circle with radius r is

πr2. (continued on the next page)

Copyright 2022 c© A. Chow and D. Yu 2

9. Sketch the region between y =

9

x2 + 9

, y = 0, x = 0 and x = 3. Using the washer method (also

known as the slicing method), find the volume of Q, where Q is the solid obtained by rotating

this region about the line y = 0. As part of your solution, explain what information from curve

sketching you used to determine the sketch of this region. Recall curve sketching was discussed

in Math 1300.

10. Evaluate

∫ π

−π

sin(mx) sin(nx)dx for all positive integers m and n. Hint: it is possible for m to

be equal to n.

11. Suppose there is a new function of x called York whose domain is R and it is a positive function.

As well, the derivative of York is the function Math, whose domain is also R. The two functions

satisfiy (York)2−(Math)2 = 1. An antiderivative of York is the function Calculus, whose domain

is also R. Compute the arc length of the curve y if y = York(x) for all x ∈ [0,b] where b is a

positive constant. As part of your solution, state the arc length formula.

12. Suppose f(x) is a differentiable function and a is a positive number. State the formula that

determines the area of a surface obtained by rotating y = f(x) for x ∈ [a,b] about the x-axis.

As well, find the surface area of S where S is the surface created by rotating the curve y = 1

3

x3

for x ∈ [0,a] about y = 0.

13. List four integration concepts you have thus far learned in either Math 1300 or Math 1310.

Each integration concept on your list is to have a one to two sentence description, and the

description must be mathematically correct and clearly stated. Here is an example of how one

of the integration concepts in your list could look like:

(i) Let a and b be real numbers, and let f(x) and g(x) be continuous functions for x ∈ [a,b].

The area between f(x) and g(x) for x ∈ [a,b] is equal to the definite integral

∫ b

a

(f(x) −g(x)) dx

if f(x) ≥ g(x) for all x ∈ [a,b].

Copyright 2022 c© A. Chow and D. Yu 3