1. INTEGRATION
To find the area under a curve you could draw a series if trapezia and add their areas which would give an approximation to the area.
As the trapezia become thinner the approximation gets better
Therefore if the width of each trapezium is very small the area of the trapezia will be the same as the area under the curve

2. Add 1 to the power then Divide by the new power
To differentiate we multiplied by the power and then we subtracted 1 from the power
If we exactly reverse the process then we
Add 1 to the power then
Divide by the new power

3.  x dx = x2 dx = x3 + c 3 xn dx = xn+1 + c n +1
Means Integrate with respect to x
 x dx =
X2 + c 2
x2 dx = x3 + c 3
xn dx = xn+1 + c
n +1

4. [ ] = [ ] = 42 – 22  x dx = 2 2 = 8 – 2 = 6 -13 – -33 3x2 dx = x3 4 2
[ ] = X2 2
 x dx = 4 2
= 8 – 2 = 6
[ ] = -1 -3
-13 – -33
3x2 dx = x3 -1 -3
= 26

5. Find  (x – 2)2 dx =  x2 – 4x +2 dx = [⅓x3 – 2x2 + 2x]-1 -3
Find  (x – 2)2 dx -1 -3
=  x2 – 4x +2 dx -1 -3
= [⅓x3 – 2x2 + 2x]
= [⅓(-1)3 – 2(-1)2 + 2(-1)] -
[⅓(-3)3 – 2(-3)2 + 2(-3)]
= [-⅓ – 2 - 2] -
[-9 – 18 -6]
= -4⅓ + 33
= 28⅔

6. WHY? Find  x2 – 9 dx = [⅓x3 – 9x] = [⅓(3)3 – 9(3)] – [⅓(1)3 – 9(1)]
[⅓(27) – 9(3)] - [ ⅓ - 9]
= [9 - 27] -
[8⅔]
= - 26⅔
WHY?

7. Because the area is negative it means that the area is under the curve
The easiest solution is to swap the limits round 1 3
So find  x2 – 9 dx

8. Beware of things like  2x -4 dx1 -1
Beware of things like  2x -4 dx 1 -1
= [x2 – 4x]
= [12 – 4(1)] – [(-1)2 - 4(-1)]
= 1 – 4 –1 + 4 = 0
Half the area is above and half below so the area = 0