Q. 37 - 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.

${\int }_6^3\left(f\right(x)+g(x\left)\right)dx$.

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 ${\int }_6^3\left(f\right(x)+g(x\left)\right)dx$ is, $-8$.

Step 1 . Given information

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

${\int }_6^3\left(f\right(x)+g(x\left)\right)dx$.

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

$$\begin{gathered} {\int }_6^3\left(f\right(x)+g(x\left)\right)dx={\int }_6^{3}f\left(x\right)dx+{\int }_6^{3}g\left(x\right)dx \\ =-{\int }_6^{-2}f\left(x\right)dx+{\int }_{-2}^{3}f\left(x\right)dx-{\int }_6^{3}g\left(x\right)dx \\ =-9+4-3 \\ =-8 \end{gathered}$$

Therefore, the exact value is, $-8$.