Question

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

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

Answer 1

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

${\int }_{a-n}^n{e}^{x(a-x)}dx=\lambda ⟶1$
Let $I={\int }_{a-n}^{n}x{e}^{x(a-x)}dx⟶2$
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_{a-n}^n(a-x)e^{(a-x)(a-a+x)}dx={\int }_{a-n}^n(a-x)e^{x(a-x)}$
$I={\int }_{a-n}^{n}a{e}^{x(a-x)}dx-{\int }_{a-n}^{n}x{e}^{x(a-x)}dx$
$I=a{\int }_{a-n}^n{e}^{x(a-x)}dx-I$ (from equation 2)
$I=a\lambda -I$ (from equation 1)
$\therefore 2I=a\lambda$
$I=\frac{a\lambda }{2}$