Finding the Area Between a Graph and the \(x\)-Axis

Find the area of the region enclosed by the graph of \(f(x)=x^{2}-4\), the \(x\)-axis, and the lines \(x=0\) and \(x=5\).

Solution First we graph the function \(f\). On the interval \([0,5]\), the graph of \(f\) intersects the \(x\)-axis at \(x=2\). As shown in Figure 7, \(f(x)\leq 0\) on the interval \([0, 2]\), and \(f(x)\geq 0\) on the interval \([2,5]\). So, the area \(A\) is the sum of the areas \(A_{1}\) and \(A_{2},\) where \begin{eqnarray*} A_{1} &=&\int_{0}^{2}\left[ ~0-f(x) \right] ~{\it dx}=\int_{0}^{2}-~ ( x^{2}-4) ~{\it dx}=\left[ -\dfrac{x^{3}}{3}+4x\right] _{0}^{2}=- \dfrac{8}{3}+8=\dfrac{16}{3} \\[4pt] A_{2} &=&\int_{2}^{5}f(x) ~{\it dx}=\int_{2}^{5}( x^{2}-4) ~{\it dx}=\left[ \dfrac{x^{3}}{3}-4x\right] _{2}^{5}=\left( \dfrac{125}{3} -20\right) -\left( \dfrac{8}{3}-8\right)\\[4pt] &=&\dfrac{81}{3}=27 \end{eqnarray*}

The area \(A\) we seek is \(A=A_{1}+A_{2}=\dfrac{16}{3}+\dfrac{81}{3}=\dfrac{97}{3}\) square units.