Y2, C3 - Methods in Calculus

Question 1

When does a definite integral exist

Answer 1

When it converges

Question 2

When doesn’t a definite integral exist

Answer 2

When it diverges

Question 3

When is an integral improper

Answer 3

Either:
One or both limits are infinite
f(x) is undefined at x = a, x = b, or another point in the interval [a,b]

Question 4

How do you integrate with infinity

Answer 4

Take a limit as t –> infinity
Solve the integral normally

Question 5

Integrate 1 / x^2 dx from Infinity to 1

Answer 5

Infinity = t
lim t –> infinity
1 / x^2 = -x^-1
(-1 / t - (-1 / 1) = 1

Question 6

What do you do when integrating with f(x) undefined for some value

Answer 6

Set lim t –> (undefined value)
Solve normally, then plug t in as its undefined value

Question 7

What do you do when both limits are infinite (possible an undefined value in between)

Answer 7

Take integral as 2 separate integrals
Integral from infinity –> 0 +
Integral from 0 –> - infinity

Question 8

What is the formula for finding the mean value of a function
y = f(x) over [a,b]

Answer 8

1 / (b - a) * integral of f(x) dx from a to b

Question 9

The mean value of f(x) over an interval is 1/6 * ln(2) + pi/18,
what is the mean value of f(x) + ln(k) over the same interval

Answer 9

1/6 * ln(2k^6) + pi/18

Question 10

y = arcsin(x^2) find dy/dx

Answer 10

Use formula booklet (arcsin(x) = 1 / (root(1 - x^2))
Therefore 1 / (root(1 - x^2)) * 2x = 2x / (root(1 - x^4))

Question 11

What does the integral of d/dx arccos(x) = in terms of d dx arcsin(x)

Answer 11

• dx arcsin(x)

Question 12

What substitutions should you use when proving integrals of inverse trig functions
For: arcsin, arctan, arcosh, arcsinh

Answer 12

x = a * asinu
x = a * tanu
x = a * coshu
x = a * sinhu

Question 13

Find the integral of 1 / (25 + 9x^2) dx

Answer 13

Put in a form where the coefficient of x^2 = 1
= 1/9 integral [1 / (25/9 + x^2)]
a = 5/3
1/9 * 3/5 * arctan(3x/5) + c
1/15 * arctan(3x/5) + c

Question 14

How would you split integral [(x+4) / (root(1 - 4x^2))] dx

Answer 14

integral [x / (root(1 - 4x^2))] dx + integral [4 / (root(1 - 4x^2))] dx

Question 15

How would you integrate 4 / root(1 - 4x^2) dx

Answer 15

Take out factors to leave form 1 / root(a^2 - x^2)
(4 / root(4)) integral 1 / root(1/4 - x^2)dx
Use formula booklet

Question 16

How would you write 1 / (x(x^2 + 1)) in partial fractions

Answer 16

A/x + (Bx+C)/(x^2 + 1))
A = 1/9
B = -1/9
C = 1
ans = (1/9) / x + ((-1/9)x+1) / (x^2 + 1)

Question 17

What do you do if a fraction is top heavy? (avoiding doing algebraic long division)

Answer 17

Just write it out as A being the first constant term and then solving by comparing coefficients normally

Question 18

Express (x^4 + x) / (x^4 + 5x^2 + 6) as partial fractions (hence differentiate (optional))

Answer 18

(x^4 + x) / (x^4 + 5x^2 + 6) = A + (Bx + C) / (x^2 + 2) + (Dx + E) / (x^2 + 3)
x^4 + x = A(x^4 + 5x^2 + 6) + …
(comparing coefficients)
ans = 1 + (x+4) / (x^2 + 2) = (x-9) / (x^2 + 3)
A = 1, B = 1, C = 4, D = -1, E = -9

Question 19

How would you integrate 1 / (x^2 - 8x + 8) dx by completing the square

Answer 19

Denominator = (x - 4)^2 - 8
Use formula booklet:
x = (x-4)
a = root(8) = 2root(2)
Sub in x and a values:
ans = 1 / 4root(2) * ln((x-4-2root(2)) / (x-4+2root(2)) + c

Question 20

How would you integrate 1 / (root(12x + 2x^2)) dx

Answer 20

Complete the square then use formula booklet to sub in x and a values:
= 1 / root(2*((x+3)^2 - 9)) dx
Take the 1/root(2) out:
1 / root(2) * integral( 1 / root((x+3)^2 - 9)) dx
= 1 / root(2) * arcosh(x + 3 / 3) + c