Question

If ${\int }_{a-n}^n{e}^{x(a-x)}dx=\lambda$, then the value of ${\int }_{a-n}^{n}x{e}^{x(a-x)}dx,a\ne 2n,$ is

• A. $\frac{a\lambda }{2}$
• B. $a\lambda$
• C. $2a\lambda$
• D. none of these

Answer 1

The correct option is A $\frac{a\lambda }{2}$

$I={\int }_{a-n}^{n}x{e}^{x(a-x)}dx$

$={\int }_{a-n}^n[n+a-n-x]e^{(a-x)x}dx(\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx)$

$\Rightarrow I={\int }_{a-n}^a(a-x)e^{(a-x)x}dx$
So, $2I={\int }_{a-n}^{n}x{e}^{x(a-x)}dx+{\int }_{a-n}^a(a-x)e^{(a-x)x}dx={\int }_{a-n}^n(x+a-x)e^{x(a-x)}dx={\int }_{a-n}^{n}a{e}^{x(a-x)}dx$
$\Rightarrow I=\frac{a}{2}{\int }_{a-n}^n{e}^{x(a-x)}dx$

$\Rightarrow I=\frac{a}{2}\lambda$$(\because {\int }_{a-n}^n{e}^{x(a-x)}dx=\lambda )$

Hence, option A.