Question

If $f(x)=\int_{x^2}^{x^2+1} e^{-t {^2}}dt$ then f(x) increases in

• A. (2, 2)
• B. no value of x
• C. (0,$\infty$)
• D. $(-\infty,0)$

Answer 1

Answer: (D)

$(-\infty,0)$

Answer 2

Given, $f(x)=\int_{x^2}^{x^2+1} e^{-t {^2}}dt$
On differentiating both sides using Newton's Leibnitz's
formula, we get
f'(x)=e^{-(x^2+1^2)} \bigg \\{\frac{d}{dx}(x^2+1)\bigg \\}-e^{-(x^2)^2} \bigg\\{\frac{d}{dx}(x^2)\bigg\\}
$\, \, \, \, \, \, \, \, =e^{-(x^2+1)^2} .2x-e^{-(x^2)^2}.2x$
$\, \, \, \, \, \, \, \, =2xe^{-(x^4+2x^2+1)}(1-e^{2x^2}+1)$
$[where, e^{2x^2}+1 >1, \forall x \, and \, e^{-(x^4+2x^2+1)}> 0, \forall x]$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, f'(x)>0$
which shows 2x < 0 or x < 0 $\Rightarrow \, \, x \in(-\infty ,0)$