Area between curves

A slightly more complex process is finding the area between curves. In this case it doesn't matter whether the shaded area is above, below or cut into pieces by the x-axis. Again, we'll take you through a couple of worked examples.

The first step is to find where the curves meet, i.e., where

f(x) = g(x)

$\eqalign{ \Rightarrow x^2 & = & 2x - x^2 \cr \Rightarrow 2x^2 - 2x = 2x\left( {x - 1} \right) & = & 0 \cr \Rightarrow x & = & 0,1 \cr}$

So, 0 and 1 are the limits of integration

Area $= \int_0^1 {\left( {g(x) - f(x)} \right)dx}$ ; that is, the integrand is the upper curve minus the lower curve.

Simplify before integrating (this saves having to integrate two terms with the same power.)

$= \int_0^1 ((2x - x^2) - (x^2)) dx = \int_0^1 (2x - 2x^2) dx$

$\eqalign{ & = & \left[ {x^2 - 2{{x^3 } \over 3}} \right]_0^1 \cr\cr & = & \left( {1 - {2 \over 3}} \right) - \left( 0 \right) \cr\cr & = & {1 \over 3}\;unit^2 \cr}$

Question: Find the area between the curve and straight line:

Answer

The first step is to find where the curves meet.

$\eqalign{ f(x) & = & g(x) \cr \Rightarrow x^2 & = & x \cr \Rightarrow x^2 - x & = & x\left( {x - 1} \right) = 0 \cr \Rightarrow x & = & 0,1 \cr}$

$\eqalign{ Area & = & \int_0^1 {(x - x^2 )dx} \cr\cr & = & \left[ {{{x^2 } \over 2} - {{x^3 } \over 3}} \right]_0^1 \cr\cr & = & \left( {{1 \over 2} - {1 \over 3}} \right) - \left( 0 \right) \cr\cr & = & {1 \over 6}\;units^2 \cr}$

Now try a Test Bite

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Area between curves

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