Q. 33 - 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.

${\int }_{-2}^{3}f\left(x\right)g\left(x\right)dx$.

Step-by-Step Solution

Answer

Since, ${\int }_{-2}^3 f\left(x\right)g\left(x\right)dx\ne {\int }_{-2}^3 f\left(x\right)dx\times {\int }_{-2}^3 g\left(x\right)dx$.

The definite integral cannot be found out as there is not enough information available. Therefore, the value is not possible to find.

Step 1 . Given information

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

${\int }_{-2}^{3}f\left(x\right)g\left(x\right)dx$.

Step 2 . The value is not possible to find.

Since, ${\int }_{-2}^3 f\left(x\right)g\left(x\right)dx\ne {\int }_{-2}^3 f\left(x\right)dx\times {\int }_{-2}^3 g\left(x\right)dx$.

The definite integral cannot be found out as there is not enough information available.