Indefinite Integration 3 Question 7
7. If $f(x)=\int _{x^{2}}^{x^{2}+1} e^{-t^{2}} d t$, then $f(x)$ increases in
(2003, 1M)
(a) $(2,2)$
(b) no value of $x$
(c) $(0, \infty)$
(d) $(-\infty, 0)$
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Answer:
Correct Answer: 7. (d)
Solution:
- Given, $f(x)=\int _{x^{2}}^{x^{2}+1} e^{-t^{2}} d t$
On differentiating both sides using Newton’s Leibnitz’s formula, we get
$$ \begin{aligned} f^{\prime}(x)= & e^{-\left(x^{2}+1\right)^{2}} \frac{d}{d x}\left(x^{2}+1\right)-e^{-\left(x^{2}\right)^{2}} \frac{d}{d x}\left(x^{2}\right) \\ & =e^{-\left(x^{2}+1\right)^{2}} \cdot 2 x-e^{-\left(x^{2}\right)^{2}} \cdot 2 x \\ & =2 x e^{-\left(x^{4}+2 x^{2}+1\right)}\left(1-e^{2 x^{2}+1}\right) \\ & \quad\left[\text { where, } e^{2 x^{2}+1}>1, \forall x \text { and } e^{-\left(x^{4}+2 x^{2}+1\right)}>0, \forall x\right] \\ \therefore \quad & \quad f^{\prime}(x)>0 \end{aligned} $$
which shows $2 x<0$ or $x<0 \quad \Rightarrow \quad x \in(-\infty, 0)$
