Question

If $\underset{0}{\overset{1}{\int }}{e}^{x^2}(x-\alpha )dx=0,$ then which of the following is true for $\alpha$?

• A. $1<\alpha <2$
• B. $\alpha <0$
• C. $0<\alpha <1$
• D. $\alpha =0$

Answer 1

The correct option is C $0<\alpha <1$

We have,
$\underset{0}{\overset{1}{\int }}{e}^{x^2}(x-\alpha )dx=0$
$\Rightarrow \underset{0}{\overset{1}{\int }}{e}^{x^2}xdx=\alpha \underset{0}{\overset{1}{\int }}{e}^{x^2}dx$
$\Rightarrow \frac{1}{2}\underset{0}{\overset{1}{\int }}{e}^{x^2}2x.dx=\alpha \underset{0}{\overset{1}{\int }}{e}^{x^2}dx$
$\Rightarrow \frac{1}{2}{\left|e^{x^2}\right|}_0^1=\alpha \underset{0}{\overset{1}{\int }}{e}^{x^2}dx$
$\Rightarrow \frac{1}{2}(e-1)=\alpha \underset{0}{\overset{1}{\int }}{e}^{x^2}dx$
$\Rightarrow \alpha =\frac{\frac{1}{2}(e-1)}{\underset{0}{\overset{1}{\int }}{e}^{x^2}dx}>0\text{but}<1$
$\therefore 0<\alpha <1$