Question

If ${\int }_{-1}^{-4}f\left(x\right)dx=4$ and ${\int }_2^{-4}(3-f(x\left)\right)dx=7$, then the value of ${\int }_{-2}^{1}f(-x)dx$, is

• A. 2
• B. -3
• C. 5
• D. none of these

Answer 1

The correct option is D none of these

We have,
${\int }_2^{-4}(3-f(x\left)\right)dx=7$
$\Rightarrow {\int }_2^{-4}3dx-{\int }_2^{-4}f\left(x\right)dx=7$
$\Rightarrow 3(-4-2)-{\int }_2^{-4}f\left(x\right)dx=7$
$\Rightarrow {\int }_2^{-4}f\left(x\right)dx=-25$
$\Rightarrow {\int }_{-4}^{2}f\left(x\right)dx=25$
$\Rightarrow {\int }_{-4}^{-1}f\left(x\right)dx+{\int }_{-1}^{2}f\left(x\right)dx=25$
$\Rightarrow -4+{\int }_{-1}^{2}f\left(x\right)dx=25$ $[\because {\int }_{-1}^{-4}f\left(x\right)dx=4]$
$\Rightarrow {\int }_{-1}^{2}f\left(x\right)dx=29$
$\Rightarrow -{\int }_1^{-2}f(-t)dt=29,$ where $t=-x$
$\therefore {\int }_{-2}^{1}f(-t)dt=-29$