Question

If ${\int }_0^1{e}^{x^2}(x-\alpha )dx=0,$ then

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

Answer 1

The correct option is A $1<\alpha <2$

We have ${\int }_0^1{e}^{x^2}(x-\alpha )dx=0$

$\Rightarrow {\int }_0^1{e}^{x^2}xdx={\int }_0^1{e}^{x^2}\alpha dx$

$\Rightarrow \frac{1}{2}{\int }_0^1{e}^{t}dt=\alpha {\int }_0^1{e}^{x^2}dx,$ where $t=x^2$

$\Rightarrow \frac{1}{2}(e-1)=\alpha {\int }_0^1{e}^{x^2}dx$

since, $e^{x^2}$ is an increasing function for $0\le x\le 1$, therefore,$1\le e^{x^2}\le e$
when $0\le x\le 1$

$\Rightarrow 1(1-0)\le {\int }_0^1{e}^{x^2}dx\le e(1-0)$

$\Rightarrow 1\le {\int }_0^1{e}^{x^2}dx\le e$

From equations (1) and $\left(2\right),$ we find that $L.H.$ S. of equation(1) is positive and ${\int }_0^1{e}^{x^2}dx$ lies between 1 and $e$ Therefore, $\alpha$ is a positive real number. Now, from equation $\left(1\right),\alpha =\frac{\frac{1}{2}(e-1)}{{\int }_0^1{e}^{x^2}dx}$

The denominator of equation (3) is greater than unity and the numerator lies between $0$ and $1$. Therefore, $0<\alpha <1$