1. INTEGRATION & AREA NB: we should think of a line as a very simple curve! Consider X Y y = 2x + 3 1 5 A B A is (1,5) & B is (5,13) (I) (II) 5 8 4 Area(I) = 8 X 4  2= 16 Area(II) = 5 X 4 = 20 Total = 36 Now consider  1 5 (2x + 3) dx= [ ] 1 5 x 2 + 3x= (25 + 15) – (1 + 3)= 36 Comparing answers we should see that the area can also be obtained by integration so we can use this for curves as follows….

2. Area Under a Curve X Y y = f(x) a b The area under the curve y = f(x) from x = a to x = b is given by  a b f(x) dx MUST BE LEARNED !!! Ex13 X Y y = p(x) Shaded area =  -2 7 -2 7 p(x) dx

3. Ex14 X Y y = x 2 – 4x + 5 2 5 Area =  2 5 (x 2 – 4x + 5) dx = [ ] 2 5 1 / 3 x 3 – 2x 2 + 5x = ( 125 / 3 – 50 + 25) – ( 8 / 3 – 8 + 10) = 12 units 2

4. Ex15 X Y y = 2x(6 – x) NB: need limits! Curve cuts X-axis when 2x(6 – x) = 0so x = 0 or x = 6 Area =  0 6 2x(6 – x) dx = =  0 6 (12x – 2x 2 ) dx = [ ] 0 6 6x 2 – 2 / 3 x 3 = (216 – 144) - 0 = 72 units 2

5. Ex16 X Y y = (x – 1)(x – 6) NB: need limits! Curve cuts X-axis when (x – 1)(x – 6) = 0so x = 1 or x = 6 Area =  1 6 (x – 1)(x – 6) dx = =  1 6 (x 2 - 7x + 6) dx = [ ] 1 6 1 / 3 x 3 – 7 / 2 x 2 + 6x = (72 – 126 + 36) - ( 1 / 3 – 7 / 2 + 6) = -20 5 / 6 units 2 (**) (**) Area can’t be negative. Negative sign indicates area is below X-axis. Actual area = 20 5 / 6 units 2

6. Ex17 X Y y = x(x – 4) 0 4 6 partA partB Need to find each section separately ! Area A =  = =  0 4 x(x – 4) dx 0 4 (x 2 – 4x) dx = [ ] 1 / 3 x 3 – 2x 2 0 4 = (21 1 / 3 – 32) - 0 = -10 2 / 3 (really 10 2 / 3 ) Area B =  6 4 x(x – 4) dx = [ ] 1 / 3 x 3 – 2x 2 4 6 = (72 – 72) - (21 1 / 3 – 32) = 10 2 / 3 Total = 10 2 / 3 + 10 2 / 3 = 21 1 / 3 units 2

7. NB: both areas are identical in size however the different signs indicate position above and below the X-axis! If we try to calculate the area in one step then the following happens Area =  0 6 x(x – 4) dx = [ ] 0 6 1 / 3 x 3 – 2x 2 = (72 – 72) – 0 = 0 It is obvious the total area is not zero but the equal magnitude positive and negative parts have cancelled each other out. Hence the need to do each bit separately.

8. Ex18 X Y y = x 3 – 4x 2 – x + 4 To find limits must solve x 3 – 4x 2 – x + 4 = 0 using polynomial method. Start with x = 1. 11 -4 -1 4 1 1 -3 -4 0 f(1) = 0 so (x – 1) a factor Other factor is x 2 – 3x – 4 = (x – 4 )(x + 1)

9. Solving (x + 1)(x – 1)(x – 4) = 0gives x = -1 or x = 1 or x = 4 so 1 st area =  1 (x 3 – 4x 2 – x + 4) dx = [ ] 1 1 / 4 x 4 - 4 / 3 x 3 – 1 / 2 x 2 + 4x = ( 1 / 4 - 4 / 3 – 1 / 2 + 4) – ( 1 / 4 + 4 / 3 – 1 / 2 - 4) = 5 1 / 3 2 nd area =  1 4 (x 3 – 4x 2 – x + 4) dx = [ ] 1 4 1 / 4 x 4 - 4 / 3 x 3 – 1 / 2 x 2 + 4x = (64 – 256 / 3 – 8 + 16) - ( 1 / 4 - 4 / 3 – 1 / 2 + 4) = - 15 3 / 4 (Really 15 3 / 4 ) So total area = 5 1 / 3 + 15 3 / 4 = 21 1 / 12 units 2