$\frac{d}{{dx}}\,\,\mathop \smallint \nolimits_{f\left( x \right)}^{{\rm{g}}\left( x \right)} \,\,{\rm{h}}\,\,\left( t \right)\,\,dt$ is equal to

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]$

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]$

$h\,\,\left[ {\,{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,h\left[ {\,f\,\,\left( x \right)} \right]$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

\frac{d}{{dx}}\,\,\int \nolimits_{f\left( x \right)}^{\mathrm{g}\left( x \right)} \,\,\mathrm{h}\,\,\left( t \right)\,\,dt

is equal to

View answer and explanation

Correct option: C

g'\left( x \right)\,\,h\,\,\left[ {\mathrm{g}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]

Option C (

g'\left( x \right)\\,,h\\,\left[ {\mathrm{g}\\,,\left( x \right)} \right]\\, - \\,f'\left( x \right)\\,,h\\,\left[ {f\left( x \right)} \right]

) is correct because it represents the Leibniz rule for differentiating an integral with variable limits.