Q. 39 - 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.

${\int }_3^{-2}\left(2x^2-3g\left(x\right)\right)dx$.

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 ${\int }_3^{-2}\left(2x^2-3g\left(x\right)\right)dx$ is, $-\frac{52}{3}$.

Step 1 . Given information

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

${\int }_3^{-2}\left(2x^2-3g\left(x\right)\right)dx$.

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

$$\begin{gathered} {\int }_3^{-2}\left(2x^2-3g\left(x\right)\right)dx=2{\int }_3^{-2}{x}^{2}dx-3{\int }_3^{-2}g\left(x\right)dx \\ =2\left[\frac{1}{3}\left(-2^3-3^3\right)\right]+3{\int }_{-2}^{3}g\left(x\right)dx \\ =2\left[\frac{1}{3}(-8-27)\right]+3\left(2\right) \\ =\frac{2\left(35\right)}{3}+6 \\ =\frac{-70+18}{3} \\ =-\frac{52}{3} \end{gathered}$$

Therefore, the exact value is, $-\frac{52}{3}$.