Using Property (4) of the Definite Integral

Find the area \(A\) under the graph of

\[ f(x) =\left\{ \begin{array}{c@{ }c@{ }ccc} x^{2} & \hbox{if} & 0 & \leq x &\lt10 \\ 100 & \hbox{if} & 10 &\leq x & \leq 15 \end{array} \right. \]

from \(0\) to \(15.\)

Solution See Figure 25. Since \(f\) is nonnegative on the closed interval \([0,15] ,\) then \(\int_{0}^{15}f(x)\, dx\) equals the area \(A\) under the graph of \(f\) from \(0\) to \(15.\) Since \(f\) is continuous on \([0,15],\) \[ \begin{eqnarray*} \int_{0}^{15}f(x)\, dx &=& \int_{0}^{10}f(x)\, dx+\int_{10}^{15}f(x)\, dx=\int_{0}^{10}x^{2}dx+\int_{10}^{15}100\,dx \\[4.5pt] &=& \left[\dfrac{x^{3}}{3}\right] _{0}^{10}+ \Big[100x\Big] _{10}^{15}=\dfrac{1000}{3}+500=\dfrac{2500}{3} \end{eqnarray*} \]

The area under the graph of \(f\) is approximately 833.33 square units.