When the function to be integrated is not continuous, but bounded, we will separate the integration interval using the additive property at the points where the function is not continuous.
$$\displaystyle \int_0^5 f(x) \ dx$$ , where $$f(x) =\left\{ \begin{array} {rcl} x+1 & \mbox{if} & x < 3 \\ 7x^2-5 & \mbox{if} & x \geqslant 3 \end{array}\right.$$
(This function is not continuous in $$x=3$$)
$$\displaystyle \int_0 ^5 f(x) \ dx = \int_0^3 f(x) \ dx + \int_3^5 f(x) \ dx =\int_0^3 x+1 \ dx + \int_3^5 7x^2-5 \ dx =$$
$$=\displaystyle \Big[ \frac{x^2}{2}+x\Big]_0^3+\Big[\frac{7x^3}{3}-5x \Big]_3^5=\frac{1745}{6}$$