Integration

Question 1

Improper integral

Answer 1

Integral evaluated as b approaches infinty

Question 2

Main techniques used to solve

Answer 2

Inspection
U-sub
By parts
Partial fractions

Question 3

Inspection

Answer 3

If numerator is derivative of the denominator then integral is simply ln(denominator) + c

Question 4

How can the fraction be manipulated so it fits for inspection

Answer 4

We can multiply by n/n and remove some constants from within the integral, alternatively we can use partial fractions or simply splitting the numerator

Question 5

U Substitution

Answer 5

Aim to simplify integral with the definition of du and get it into a standard integral form that is easily solved.

Question 6

Integration by parts equation

Answer 6

∫u’v = uv -∫uv’

Question 7

Integration by parts for cyclic functions

Answer 7

Set integral = to I, eventually our -∫uv’ will equal our original so we resub in I and rearrange to solve.
(May need to show that we’ve changed c)

Question 8

Alternative approaches related to integration by parts

Answer 8

May have to create a u’ = 1 or alternatively vary which is u’ and which is v as may not always be obvioius/ straightforward

Question 9

Partial fractions used when:

Answer 9

Need to split up an integral into more easily analysed components:

Question 10

3 forms of partial fractions

Answer 10

no repeated root, repeated root and irreducible quadratic

Question 11

no repeated root

Answer 11

= A/(x-n1) + B/(x-n2)

Question 12

repeated root

Answer 12

= A/(x-n1) + B/(x-n2) + C/(x-n2)^2

Question 13

quadratric

Answer 13

= A/(x-n1) + (Bx+C)/(x^2-n2x + n3)

Question 14

How to solve partial fractions

Answer 14

Multiply equation by denominator and use logical values of x to set some constants to 0 and then solve

Question 15

What if order of numerator is greater than that of the denominator

Answer 15

We use long algebraic divion

Question 16

Process of long algebraic division

Answer 16

Set up division with denominator as the divisor. Then multiply divisor by values that will help solve the equation when we have numerator - denominator and values will = 0. These values go to the top of the division and as bottom slowly simplifies to a point until it cant be simplified further, then our answer will be top line + remainder/divisor

Question 17

Length of curve equation

Answer 17

L =∫ (1+(f’(x))^2)^1/2 dx

Question 18

Integral of parametric equations

Answer 18

L = ∫ ((y(t))^2 + (x(t))^2)^1/2

Question 19

How to integrate absolute values:

Answer 19

Create a piecewise function that has one part when |inside| when inside is greater than 0 and one where inside is less than 0 (with this one being multiplied by -1)
We can then split the integral and determine what the limits will be then solve normally.

Question 20

Technique to solve integral with difficult bottom

Answer 20

Multiply by conjugate to try simplify (a^2 - b^2)