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= 16Area(II) = 5 X 4 = 20

Total = 36

Now consider

1

5

(2x + 3) dx = [ ]1

5

x2 + 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….

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

Yy = p(x)

Shaded area =

-2 7

-2

7

p(x) dx

Ex14

X

Y

y = x2 – 4x + 5

2 5

Area = 2

5

(x2 – 4x + 5) dx = [ ]2

51/3x3 – 2x2 + 5x

= (125/3 – 50 + 25) – (8/3 – 8 + 10)

= 12 units2

Ex15

X

Y

y = 2x(6 – x)

NB: need limits!

Curve cuts X-axis when 2x(6 – x) = 0 so x = 0 or x = 6

Area = 0

6

2x(6 – x) dx = 0

6

(12x – 2x2) dx

= [ ]0

6

6x2 – 2/3x3

= (216 – 144) - 0

= 72 units2

Ex16

X

Yy = (x – 1)(x – 6)

NB: need limits!

Curve cuts X-axis when (x – 1)(x – 6) = 0 so x = 1 or x = 6

Area = 1

6

(x – 1)(x – 6) dx = 1

6

(x2 - 7x + 6) dx

= [ ]1

61/3x3 – 7/2x2 + 6x

= (72 – 126 + 36) - (1/3 – 7/2 + 6)

= -205/6 units2 (**)

(**) Area can’t be negative.

Negative sign indicates area is below X-axis.

Actual area = 205/6 units2

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

(x2 – 4x) dx

= [ ]1/3x3 – 2x20

4

= (211/3 – 32) - 0

= -102/3 (really 102/3)

Area B = 6

4x(x – 4) dx

= [ ]1/3x3 – 2x24

6

= (72 – 72) - (211/3 – 32)

= 102/3

Total = 102/3 + 102/3 = 211/3 units2

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

61/3x3 – 2x2

= (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.

Ex18

X

Y

y = x3 – 4x2 – x + 4

To find limits must solve x3 – 4x2 – x + 4 = 0 using polynomial method. Start with x = 1.

1 1 -4 -1 4

11

-3-3-4

-40

f(1) = 0 so (x – 1) a factor

Other factor is x2 – 3x – 4 = (x – 4 )(x + 1)

Solving (x + 1)(x – 1)(x – 4) = 0 gives x = -1 or x = 1 or x = 4

so

1st area = -1

1

(x3 – 4x2 – x + 4) dx = [ ] -1

11/4x4 - 4/3x3 – 1/2x2 + 4x

= (1/4 - 4/3 – 1/2 + 4) – (1/4 + 4/3 – 1/2 - 4)

= 51/3

2nd area = 1

4

(x3 – 4x2 – x + 4) dx = [ ]1

41/4x4 - 4/3x3 – 1/2x2 + 4x

= (64 – 256/3 – 8 + 16) - (1/4 - 4/3 – 1/2 + 4)

= -153/4 (Really 153/4)

So total area = 51/3 + 153/4 = 211/12 units2