Question

If ${\int }_0^1{e}^{x^2}(x-\alpha )dx=0$ ,then which of the following is correct?

• 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 ${\int }_0^1{e}^{x^2}(x-\alpha )dx=0$
or ${\int }_0^1{e}^{x^2}xdx={\int }_0^1{e}^{x^2}\alpha$
Put $t=x^2$ in first integral
$\Rightarrow dt=2xdx$
$\Rightarrow \frac{1}{2}{\int }_0^1{e}^{t}dt=\alpha {\int }_0^1{e}^{x^2}dx$
or $\frac{1}{2}(e-1)=\alpha {\int }_0^1{e}^{x^2}dx$..............(1)
Since,$e^{x^2}$ is an increasing function for $0\le x\le 1$,$1\le e^{x^2}\le e$
$\Rightarrow {\int }_0^{1}1dx\le {\int }_0^1{e}^{x^2}dx\le {\int }_0^{1}edx$
$\Rightarrow 1(1-0)\le {\int }_0^1{e}^{x^2}dx\le e(1-0)$
$1\le {\int }_0^1{e}^{x^2}dx\le e$.............(2)
From equations (I) and (2), we find that L.H.S. of equation (I) 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 (I),
$\alpha =\frac{\frac{1}{2}(e-1)}{{\int }_0^1{e}^{x^2}dx}$......(3)
The denominator of equation (3) is greater than unity and the numerator lies between 0 and 1.Therefore, $0<\alpha <1$.
Hence, option 'C' is correct.