If $f\,\,\left( x \right) = \mathop \smallint \limits_{{x^2}}^{{x^4}} \,\,\sin \,\,\sqrt {t\,} dt$, then $f'\,\,\left( x \right)$ equals

$\sin \,\,{x^2} - \,\,\sin \,\,x$

${x^{4\,}}\,\,\sin \,\,{x^2}\,\, - \,\,x\,\,\sin \,\,x$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f\,\,\left( x \right) = \int \limits_{{x^2}}^{{x^4}} \,\,\sin \,\,\sqrt {t\,} dt

, then

f'\,\,\left( x \right)

equals

View answer and explanation

Correct option: B

4\,{x^3}\,\,\sin \,\,{x^2}\,\, - \,\,2\,x\,\,\sin \,\,x

Option B (

4\,{x^3}\,\,\sin \,\,{x^2}\,\, - \,\,2\,x\,\,\sin \,\,x

) is correct because it is the result of applying the Leibniz rule for differentiating under the integral sign.

Practice All Chapter MCQs →

\int \nolimits_0^\infty \,\,\log \,\,\left( {x + \frac{1}{x}} \right)\,\,\frac{{dx}}{{1 + {x^2}}}

f\left( x \right)

\int \nolimits_0^{{t^2}} \,\,x\;\,f\left( x \right)\,\,dx = \frac{2}{5}\,\,{t^5}

f\left( {\frac{4}{{25}}} \right)

\int\limits_{{0}}^{11}[x]^{3} . \mathrm{dx}

[*]

\int\limits_{2}^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x

\int\limits_{{-2}}^0\left(x^3+3 x^2+3 x+3+(x+1) \cos (x+1)\right) d x

I\,\,\left( {m,\,\,n} \right) = \int \nolimits_0^1 \,\,{t^m}\,\,{\left( {1 + t} \right)^n}\,\,dt

I\left( {m,\,\,n} \right)

I\,\,\left( {m + 1,\,\,n + 1} \right)