+136 Find \(\int_{0}^{\,5} {f(x)}\, dx\, \) if
\(f(x)=\begin{cases} 3 & \textrm{for } x<3 \\ x & \textrm{for } x\ge 3. \end{cases}\)
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+9673 Generally, for piecewise functions, you break the integral at the boundary points of the cases. In this problem, you break the integral at x = 3.
\(\newcommand{\dint}{\displaystyle\int} \dint_0^5 f(x)\,dx = \left(\dint_0^3 + \dint_3^5\right)f(x)\,dx = \dint_0^3 f(x) \,dx + \dint_3^5 f(x)\,dx\)
Now, you can use the fact that f(x) = 3 for 0 < x < 3 and f(x) = x for 3 < x < 5 to continue. Use the power rule of integration.
