Question
If $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt,$$ then$$f\left( x \right)$$ increases in-
A.
$$\left( { - 2,\,2} \right)$$
B.
no value of $$x$$
C.
$$\left( {0,\,\infty } \right)$$
D.
$$\left( { - \infty ,\,0} \right)$$
Answer :
$$\left( { - \infty ,\,0} \right)$$
Solution :
We have $$f\left( x \right) = \int\limits_{{x^2}}^{{x^2} + 1} {{e^{ - {t^2}}}} dt$$
Then $$f'\left( x \right) = {e^{ - {{\left( {{x^2} + 1} \right)}^2}}}.2x - {e^{ - {x^4}}}.2x$$
[Using Leibnitz theorem, ]
$$\eqalign{
& \frac{d}{{dx}}\int\limits_{\phi \left( x \right)}^{\psi \left( x \right)} {f\left( t \right)dt} \cr
& = f\left[ {\psi \left( x \right)} \right].\psi '\left( x \right) - f\left[ {\phi \left( x \right)} \right].\phi '\left( x \right) \cr
& = 2x\left[ {{e^{ - {{\left( {{x^2} + 1} \right)}^2}}} - {e^{ - {x^4}}}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because {\left( {{x^2} + 1} \right)^2} > {x^4} \cr
& \Rightarrow {e^{ + {{\left( {{x^2} + 1} \right)}^2}}} > {e^{{x^4}}} \Rightarrow {e^{ - {{\left( {{x^2} + 1} \right)}^2}}} < {e^{ - {x^4}}} \cr
& \therefore {e^{ - {{\left( {{x^2} + 1} \right)}^2}}} - {e^{ - {x^4}}} < 0 \cr
& \therefore f'\left( x \right) > 0,\,\forall \,x < 0 \cr} $$
$$\therefore f\left( x \right)$$ increases when $$x<0$$