${\int }_{-1}^4 (2\mathrm{x}-3)\mathrm{dx}$ is equal to

Let $\mathrm{I}={\int }_{-1}^4 (2\mathrm{x}-3)\mathrm{dx}\text{ and }\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}-3$

then,

$\begin{array}{c}\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int (2\mathrm{x}-3)\mathrm{dx}\\ =\frac{2{\mathrm{x}}^2}{2}-3\mathrm{x}\\ ={\mathrm{x}}^2-3\mathrm{x}=\mathrm{F}\left(\mathrm{x}\right)\left(\text{ say }\right)\end{array}$

Now, by the fundamental theorem of integral calculus, we get

$\begin{array}{c}\mathrm{I}=\left[\mathrm{F}\right(\mathrm{x}){]}_{-1}^4=\mathrm{F}(4)-\mathrm{F}(-1)\\ =\left[(4{)}^2-3(4)\right]-\left[(-1{)}^2-3(-1)\right]\\ =(16-12)-(1+3)\\ =4-4=0\end{array}$