SATHEE: Definite Integration Question 7

Definite Integration Question 7

Question 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

$\begin{aligned} f^{'} (x) = & e^{- {(x^{2} + 1)}^{2}} \frac{d}{d x} (x^{2} + 1) - e^{- {(x^{2})}^{2}} \frac{d}{d x} (x^{2}) & = e^{- {(x^{2} + 1)}^{2}} \cdot 2 x - e^{- {(x^{2})}^{2}} \cdot 2 x & = 2 x e^{- (x^{4} + 2 x^{2} + 1)} (1 - e^{2 x^{2} + 1}) & [where, e^{2 x^{2} + 1} > 1, \forall x and e^{- (x^{4} + 2 x^{2} + 1)} > 0, \forall x] ∴ & f^{'} (x) > 0 \end{aligned}$

which shows $2 x < 0$ or $x < 0 \Rightarrow x \in (- \infty, 0)$

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Definite Integration Question 7

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