If $\smallint f\left( x \right)\,\,dx = F\left( x \right),\;\,\,{\rm{then}}\,\,\smallint \,{x^3}\,f\left( {{x^2}} \right)\,\,dx$ is equal to

$\frac{1}{2}\,\,\{ \,{x^2}\,F\,\,\left( {{x^2}} \right)\,\, - \,\,\smallint F\,\,\left( {{x^2}} \right)\,\,d{x^2}\} \;$

$\frac{1}{2}\,\,\{ {x^2}\,F\,\,\left( {{x^2}} \right)\,\, - \,\,\smallint F\,\,\left( {{x^2}} \right)\,\,dx\} \;$

$\frac{1}{2}\,\,\left( {{x^2}\,F\,\,\left( x \right)\,\, - \,\,\frac{1}{2}\,\,\smallint F\,\,\left( {{x^2}} \right)\,\,dx} \right)$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

\int f\left( x \right)\,\,dx = F\left( x \right),\;\,\,\mathrm{then}\,\,\int \,{x^3}\,f\left( {{x^2}} \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}\,\,\{ \,{x^2}\,F\,\,\left( {{x^2}} \right)\,\, - \,\,\int F\,\,\left( {{x^2}} \right)\,\,d{x^2}\} \;

Option A (

\frac{1}{2}\\ {\x^2}\\ F\\,\left( {\x^2} \right)\\ - \\ \int F\\,\left( {\x^2} \right)\\ d{x^2}\\}\;

) is correct because it is derived using integration by parts with appropriate substitution.