Q. 31 - Calculus

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.

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

Step-by-Step Solution

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 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, $14$.