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
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