AREAS BETWEEN CURVES

a b X

Yy = f(x)

Area A = a

bf(x) dx

A

a b X

Y

y = g(x)

B

Area B = a

bg(x) dx

a b X

Yy = f(x)

y = g(x)

C

Area C

= AreaA - AreaB

= a

bf(x) dx -

a

bg(x) dx

= a

b[ f(x) – g(x) ] dx

NB: In the interval a < x < b if f(x) > g(x) then the area enclosed between the curves

from x = a to x = b is given by

a

b[ f(x) – g(x) ] dx

ie

a b X

Y

MUST BE LEARNED !!

Ex 19 Find the finite between area between the curves.

y = x2 – 6x

y = 12x - 2x2

Upper curve is y = 12x - 2x2

Limits are 0 and 6.

Area = 0

6

12x - 2x2 - (x2 – 6x) dx

Brackets round 2nd formula

= 0

612x - 2x2 - x2 + 6x dx

= 0

618x - 3x2 dx

= [ ]

0

69x2 – x3

= (324 – 216) - 0 = 108units2

Ex20

X

Y y = x4

y = 8x

Find shaded area!

Limits:

x4 = 8xx4 - 8x = 0

x(x3 – 8) = 0

x = 0 or x3 = 8

x= 2

0 < x < 2

take x = 1

8x = 8

x4 = 1

So 8x > x4

Area = 0

28x - x4 dx

= [ ]0

24x2 – 1/5x5

= (16 – 32/5) - 0

= 93/5 units2

OK! Who hid my banana?

Ex21 Find the finite area enclosed by the parabolic curves y = 2x2 + x – 9 and y = x2 + 2x – 3.

*********

Roughly Limits

2x2 + x – 9 = x2 + 2x – 3

x2 - x – 6 = 0

(x - 3)(x + 2) = 0

x = -2 or x = 3

For -2 < x < 3

taking x = 0

2x2 + x – 9 = -9 & x2 + 2x – 3 = -3

so y = x2 + 2x – 3 is the upper curve

ctd

Area = -2

3x2 + 2x – 3 – (2x2 + x – 9 ) dx

= -2

3x2 + 2x – 3 – 2x2 - x + 9 dx

= -2

3–x2 + x + 6 dx

= [ ]-2

3–1/3x3 + 1/2x2 + 6x

= (-9 + 41/2 + 18) – (8/3 + 2 - 12)

= 205/6 units2