Question

Let f(x) =$\int_{x^{2}}^{x^{2} + 1}{e^{- t^{2}}dt}$, x Î (–¥, ¥) then the interval for which f(x) is increasing is –

• A. (–¥, 0]
• B. [0, ¥)
• C. [–2, 2)
• D. None

Answer 1

Answer: (A)

(–¥, 0]

Answer 2

By Newton – Leibnitz Formula

f '(x) = 2x $(e^{- (x^{2} + 1)^{2}} - e^{- x^{4}})$

2x $\left( \frac{1}{e^{(x^{2} + 1)^{2}}} - \frac{1}{e^{(x^{2})^{2}}} \right)$ > 0

̃ x < 0