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.

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

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 ${\int }_{-2}^6\left(f\right(x)+g(x\left)\right)dx$is,$14$.

Step-by-step solution

Step 1. Given information

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

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

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

$$\begin{gathered} {\int }_{-2}^6\left(f\right(x)+g(x\left)\right)dx={\int }_{-2}^{6}f\left(x\right)dx+{\int }_{-2}^{6}g\left(x\right)dx \\ =9+{\int }_{-2}^{3}g\left(x\right)dx+{\int }_3^{6}g\left(x\right)dx \\ =9+2+3 \\ =14 \end{gathered}$$

Therefore, the required value is, $14$.