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

If f(x)=∫x2x2+1e−t2dt,f(x) = \int_{x^{2}}^{x^{2} + 1}e^{- t^{2}}dt, then f(x)f(x) increases in

A

(2,2)(2,2)

B

No value of xx

C

(0,∞)(0,\infty)

D

(−∞,0)( - \infty,0)

Answer

(−∞,0)( - \infty,0)

Explanation

Solution

f′(x)=e−(x2+1)2.2x−e−(x2)2.2x=2xe−(x4+1+2x2)(1−e2x2+1)f'(x) = e^{- (x^{2} + 1)^{2}}.2x - e^{- (x^{2})^{2}}.2x = 2xe^{- (x^{4} + 1 + 2x^{2})}\left( 1 - e^{2x^{2} + 1} \right)

⇒ f′(x)>0,∀x∈(−∞,0).f'(x) > 0,\forall x \in ( - \infty,0).