 2021-02-14

Assume that we have functions f(x) and g(x) so that
${\int }_{-1}^{3}f\left(x\right)dx=-1,{\int }_{3}^{5}f\left(x\right)dx=4,{\int }_{-1}^{1}g\left(x\right)dx=3,{\int }_{1}^{5}g\left(x\right)dx=2$
Determine
${\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx$. Demi-Leigh Barrera

Step 1
Given:
${\int }_{-1}^{3}f\left(x\right)dx=-1$
${\int }_{3}^{5}f\left(x\right)dx=4$
${\int }_{-1}^{1}g\left(x\right)dx=3$
${\int }_{1}^{5}g\left(x\right)dx=2$
To find:
${\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx$
Step 2
Using difference rule, we get:
$⇒{\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx={\int }_{-1}^{5}\left(2f\left(x\right)\right)dx-{\int }_{-1}^{5}\left(5g\left(x\right)\right)dx$
the constant can be taken out of the integral, hence we get:
$⇒{\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx=2{\int }_{-1}^{5}f\left(x\right)dx-5{\int }_{-1}^{5}g\left(x\right)dx$
Step 3
splitting the above integrals, we get:
$⇒{\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx=2\left({\int }_{-1}^{3}f\left(x\right)dx+{\int }_{3}^{5}f\left(x\right)dx\right)$
$-5\left({\int }_{-1}^{1}g\left(x\right)dx+{\int }_{1}^{5}g\left(x\right)dx\right)$
plugging in the given values of these integrals, we get:
$⇒{\int }_{-1}^{5}\left(2f\left(x\right)-5g\left(x\right)\right)dx=2\left(-1+4\right)-5\left(3+2\right)$ Jeffrey Jordon