Assume that we have functions f(x) and g(x) so that ∫−13f(x)dx=−1,∫35f(x)dx=4,∫−11g(x)dx=3,∫15g(x)dx=2 Determine ∫−15(2f(x)−5g(x))dx.

Step 1
Given:
${\displaystyle {\int }_{-1}^{3}f\left(x\right)dx=-1}$
${\displaystyle {\int }_3^{5}f\left(x\right)dx=4}$
${\displaystyle {\int }_{-1}^{1}g\left(x\right)dx=3}$
${\displaystyle {\int }_1^{5}g\left(x\right)dx=2}$
To find:
${\displaystyle {\int }_{-1}^5(2f\left(x\right)-5g\left(x\right))dx}$
Step 2
Using difference rule, we get:
${\displaystyle \Rightarrow {\int }_{-1}^5(2f\left(x\right)-5g\left(x\right))dx={\int }_{-1}^5\left(2f\left(x\right)\right)dx-{\int }_{-1}^5\left(5g\left(x\right)\right)dx}$
the constant can be taken out of the integral, hence we get:
${\displaystyle \Rightarrow {\int }_{-1}^5(2f\left(x\right)-5g\left(x\right))dx=2{\int }_{-1}^{5}f\left(x\right)dx-5{\int }_{-1}^{5}g\left(x\right)dx}$
Step 3
splitting the above integrals, we get:
${\displaystyle \Rightarrow {\int }_{-1}^5(2f\left(x\right)-5g\left(x\right))dx=2({\int }_{-1}^{3}f\left(x\right)dx+{\int }_3^{5}f\left(x\right)dx)}$
${\displaystyle -5({\int }_{-1}^{1}g\left(x\right)dx+{\int }_1^{5}g\left(x\right)dx)}$
plugging in the given values of these integrals, we get:
${\displaystyle \Rightarrow {\int }_{-1}^5(2f\left(x\right)-5g\left(x\right))dx=2(-1+4)-5(3+2)}$