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If {f{{({x})}}}={_{{{x}^{{2}}}}^{{{x}^{{2}}+{1}}}}{e}^{{-{{t}}

Question

If \displaystyle{f{{\left({x}\right)}}}={\int_{{{x}^{{2}}}}^{{{x}^{{2}}+{1}}}}{e}^{{-{{t}}}}^{2}{\left.{d}{t}\right.} , then $f (x)$ increases in $(0, 2)$ (b) no value of $x$ $(0, \infty)$ (d) $(- \infty, 0)$

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Owen Discussed

If \displaystyle{f{{\left({x}\right)}}}={\int_{{{x}^{{2}}}}^{{{x}^{{2}}+{1}}}}{e}^{{-{{t}}}}^{2}{\left.{d}{t}\right.} , then

f (x)

increases in

(0, 2)

(b) no value of

x

(0, \infty)

(d)

(- \infty, 0)

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Question Text

If \displaystyle{f{{\left({x}\right)}}}={\int_{{{x}^{{2}}}}^{{{x}^{{2}}+{1}}}}{e}^{{-{{t}}}}^{2}{\left.{d}{t}\right.} , then

f (x)

increases in

(0, 2)

(b) no value of

x

(0, \infty)

(d)

(- \infty, 0)

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