Area above and below the x-axis
Watch this video to learn about calculating the area above and below the x-axis.
To work out the area below the \(x\)-axis uses the same technique as for above the \(x\)-axis.
Example
Calculate the shaded area between the curve and the \(x\)-axis as shown below.
Solution
First find where the curve cuts the \(x\)-axis. Remember the curve cuts the \(x\)-axis when \(y = 0\).
\({x^2} - 6x = x(x - 6) = 0\)
\(\Rightarrow x = 0,\,6\)
\(\Rightarrow (0,0)\) and \((6,0)\)
Because the curve is below the \(x\)-axis, the integral from \(0\) to \(6\) will be negative. Thus we have:
\(- \int\limits_0^6 {({x^2} - 6x} )dx\)
\(= - \left[ {\frac{{{x^3}}}{3} - 3{x^2}} \right]_0^6\)
-\(= (0) - (72 - 108)\)
\(= 36unit{s^2}\)
Example: Extension
Calculate the total area of the curve \(y = {x^3}\) and the \(x\)-axis between \(x = 2\) and \(x = - 2\).
Solution
When you have a problem where part of the area is above the \(x\)-axis and part is below, then you should calculate each area separately and then total them at the end.
Area above curve
\(\int\limits_0^2 {{x^3}} dx\)
\(= \left[ {\frac{{{x^4}}}{4}} \right]_0^2\)
\(= \left( {\frac{{{{(2)}^4}}}{4}} \right) - \left( {\frac{{{{(0)}^4}}}{4}} \right)\)
\(= \frac{{16}}{4} - 0\)
\(= 4\,unit{s^2}\)
Area below curve
\(- \int\limits_{ - 2}^0 {{x^3}} dx\)
\(= - \left[ {\frac{{{x^4}}}{4}} \right]_{ - 2}^0\)
\(= - \left( {\frac{{{{(0)}^4}}}{4}} \right) - \left( {\frac{{{{( - 2)}^4}}}{4}} \right)\)
\(- \left( {0 - \frac{{16}}{4}} \right)\)
\(= 4\,unit{s^2}\)
\(Total\,area = 4 + 4\)
\(= 8\,unit{s^2}\)
