Area between a curve and the x-axis
To calculate the area between a curve and the \(x\)-axis we must evaluate using definite integrals.
Example
Calculate the shaded area below:
Solution
\(\int\limits_0^5 {({x^2}} - 4x + 5)dx\)
\(= \left[ {\frac{{{x^3}}}{3} - 2{x^2} + 5x} \right]_0^5\)
\(= \left( {\frac{{125}}{3} - 50 + 25} \right) - \left( 0 \right)\)
\(= \frac{{125}}{3} - 25\)
\(= \frac{{50}}{3}unit{s^2}\)
You may have to work out the limits of integration before calculating the area under a curve.
Example
Find the area under the curve \(y = 4x - {x^2}\)
Solution
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:
\(\int\limits_0^4 {(4x - {x^2}} )dx\)
\(= \left[ {2{x^2} - \frac{{{x^3}}}{3}} \right]_0^4\)
\(= \left( {32 - \frac{{64}}{3}} \right) - \left( 0 \right)\)
\(= \frac{{32}}{3}unit{s^2}\)
