Q. 32 - 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.

${\int }_3^6\left(2f\right(x)-g(x\left)\right)dx$.

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 ${\int }_3^6\left(2f\right(x)-g(x\left)\right)dx$ is, $7$.

Step 1 . Given information

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

${\int }_3^6\left(2f\right(x)-g(x\left)\right)dx$.

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

$$\begin{gathered} {\int }_3^6\left(2f\right(x)-g(x\left)\right)dx=2{\int }_3^{6}f\left(x\right)dx-{\int }_3^{6}g\left(x\right)dx \\ = 2\left({\int }_{-2}^{6}f\left(x\right)dx-{\int }_3^{-2}f\left(x\right)dx\right)-3 \\ =2\left(9-4\right)-3 \\ =2\left(5\right)-3 \\ =10-3 \\ =7 \end{gathered}$$

Therefore, the required value is, $7$.