AREA BETWEEN TWO CURVES USING INTEGRATION

Problem 1 :

Find the area bounded between the functions y = x3 - x & y = 3x.

Solution:

Let us find the point of intersection,

y = x3 - x ---> (1)

y = 3x ---> (2)

(1) = (2)

x3 - x = 3x

x3 - 4x = 0

x(x2 - 4) = 0

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

x = 0, x = 2 and x = -2

=20(3x)-x3-x dx+0-2x3-x-(3x)=3x22-x44+x2220+x44-x22-3x220-2=3(2)22-(2)44+(2)22-[0]+[0]-(-2)44-(-2)22-3(-2)22=4+4=8

Using graphing calculator, find the area between the curves.

Problem 2 :

Find the area bounded between the functions y = x2, y = 1/x & y = 4

Solution:

y = x2 ----(1)  and y = 1/x ----(2)

x2 = 1/x

(x3 - 1) / x = 0

x3 - 1 = 0

x3 = 1

x = 1

area-between-two-curvesq2.png
=41y-1y dy=23y32-ln(y)41=23(4)32-ln(4)-23=143-ln(4)

Problem 3 :

Find the area bounded by y = 2x, y = 4x and y = 1/x.

Solution:                                                                             

area-between-two-curvesq3.png

So, the required area is 0.63 square units.

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