Answer
$$
f(x)=\left\{\begin{array}{ll}{2 x+3} & {\text { if } x \leq 0} \\ {-\frac{x}{4}-3} & {\text { if } x>0}\end{array}\right.
$$
To find $\int_{-1}^{4}f(x) d x $ by using Exercise 49 and the Fundamental Theorem as follows:
$$
\begin{aligned}
\int_{-1}^{4} f(x) d x&=\int_{-1}^{0}(2 x+3)d x+ \int_{0}^{4}\left(-\frac{x}{4}-3\right) d x\\
&=-12.
\end{aligned}
$$
Work Step by Step
$$
f(x)=\left\{\begin{array}{ll}{2 x+3} & {\text { if } x \leq 0} \\ {-\frac{x}{4}-3} & {\text { if } x>0}\end{array}\right.
$$
To find $\int_{-1}^{4}f(x) d x $ by using Exercise 49 and the Fundamental Theorem as follows:
$$
\begin{aligned}
\int_{-1}^{4} f(x) d x&=\int_{-1}^{0}(2 x+3)d x+ \int_{0}^{4}\left(-\frac{x}{4}-3\right) d x\\
&=\left.\left(x^{2}+3 x\right)\right|_{-1} ^{0}+\left.\left(-\frac{x^{2}}{8}-3 x\right)\right|_{0} ^{4}\\
&=-(1-3)+(-2-12)\\
&=2-14\\
&=-12
\end{aligned}
$$
