Question

If $I={\int }_0^2{e}^{x^4}(x-a)dx=0$, then $a$ lies in the interval

• A. $(0,2)$
• B. $(-1,0)$
• C. $(2,3)$
• D. $(-2,-1)$

Answer 1

The correct option is A $(0,2)$

Given, $I={\int }_0^2{e}^{x^4}(x-\alpha )dx=0$
$\Rightarrow {\int }_0^2{e}^{x^4}xdx={\int }_0^2{e}^{x^4}\alpha dx$
Here, we see that ${\int }_0^2{e}^{x^4}xdx$ gives us an area between two curves $e^{x^4}$ and $x$ from $x=0$ to $x=2$.
Similarly, ${\int }_0^2{e}^{x^4}\alpha dx$ gives us a same area, so $\alpha$ lies between $0$ to $2$, i.e., $\alpha \in (0,2)$.