Question

If ${\int }_{-2}^{3}f\left(x\right)dx=5$ and ${\int }_1^3\{2-f(x\left)\right\}dx=6$

then the value of ${\int }_{-2}^{1}f\left(x\right)dx$ is?

• A. $-5$
• B. $3$
• C. $-7$
• D. $-3$

Answer 1

Answer: (A)

$-5$

${\int }_{-2}^{3}f\left(x\right)dx=5$

$\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx+{\int }_1^{3}f\left(x\right)dx=5$

$\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx+{\int }_1^3(4-f\left(0\right)dx)=5$

$\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx+{\int }_1^5(2-f\left(x\right))dx=5$

$\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx+2{\left[x\right]}_1^3+6=5$

$\Rightarrow {\int }_{-2}^{1}f\left(x\right)dx=-5$