The area above and below the x axis and the area between two curves is found by integrating, then evaluating from the limits of integration. Integration is also used to solve differential equations.

Part of

To calculate the area between a curve and the \(x\)-axis we must evaluate using definite integrals.

Calculate the shaded area below:

\[\int\limits_0^5 {({x^2}} - 4x + 5)dx\]

\[= \left[ {\frac{{{x^3}}}{3} - 2{x^2} + 5x} \right]_0^5\]

\[= \left( {\frac{{125}}{3} - 50 + 25} \right) - \left( 0 \right)\]

\[= \frac{{125}}{3} - 25\]

\[= \frac{{50}}{3}unit{s^2}\]

You may have to work out the limits of integration before calculating the area under a curve.

Find the area under the curve \(y = 4x - {x^2}\)

First, we need to find out where the curve cuts the \(x\)-axis. Remember, a curve cuts the \(x\)-axis when \(y = 0\).

\[4x - {x^2} = 0\]

\(x(4 - x) = 0\) - factorise taking out a common factor of \(x\).

\(x = 0\) or \(4 - x = 0\)

Therefore \(x = 0\) or \(x = 4\)

The curve cuts the \(x\)-axis at \((0,0)\) and \((4,0)\)

\(0\) and \(4\) are the limits of integration.

Hence the area is:

\[\int\limits_0^4 {(4x - {x^2}} )dx\]

\[= \left[ {2{x^2} - \frac{{{x^3}}}{3}} \right]_0^4\]

\[= \left( {32 - \frac{{64}}{3}} \right) - \left( 0 \right)\]

\[= \frac{{32}}{3}unit{s^2}\]