Question

If ${\int }_{-2}^{3}f\left(x\right)dx=4,{\int }_{-2}^{6}f\left(x\right)dx=9,{\int }_{-2}^{3}g\left(x\right)dx=2$and ${\int }_3^{6}g\left(x\right)dx=3$,

then find the values of each definite integral in Exercises $29-40$. If there is not enough information, explain why.

${\int }_{-2}^6\left(g\right(x)+x)dx$.

Short answer

If ${\int }_{-2}^{3}f\left(x\right)dx=4,{\int }_{-2}^{6}f\left(x\right)dx=9,{\int }_{-2}^{3}g\left(x\right)dx=2$and ${\int }_3^{6}g\left(x\right)dx=3$, then the exact value of${\int }_{-2}^6\left(g\right(x)+x)dx$is,$21.$

Step-by-step solution

Step 1. Given information

${\int }_{-2}^{3}f\left(x\right)dx=4,{\int }_{-2}^{6}f\left(x\right)dx=9,{\int }_{-2}^{3}g\left(x\right)dx=2,{\int }_3^{6}g\left(x\right)dx=3$.

${\int }_{-2}^6\left(g\right(x)+x)dx$.

Step 2. The definite integrals can be found out as,

$$\begin{gathered} {\int }_{-2}^6\left(g\right(x)+x)dx={\int }_{-2}^{6}g\left(x\right)+{\int }_{-2}^{6}xdx \\ =\left({\int }_{-2}^{3}g\left(x\right)dx+{\int }_3^{6}g\left(x\right)dx\right)+{\left[\frac{x^2}{2}\right]}_{-2}^6 \\ =2+3+\left(\frac{36}{2}-\frac{4}{2}\right) \\ =2+3+18-2 \\ =21 \end{gathered}$$

Therefore, the required value is,$21$.