If \[f\,\,\left( {a + b - x} \right) = f\,\,\left( x \right),\,\,\;{\rm{then}}\,\,\mathop \smallint \nolimits_a^b \,\,xf\,\,\left( x \right)\,\,dx\] is equal to

$\frac{{a + b}}{2}\,\,\mathop \smallint \nolimits_a^b \,\,f\left( x \right)\,\,dx$

$\frac{{a + b}}{2}\,\,\mathop \smallint \nolimits_a^b \,\,f\left( {b - x} \right)\,\,dx$

$\frac{{b - a}}{2}\,\,\mathop \smallint \nolimits_a^b \,\,f\left( x \right)\,\,dx$

$\frac{{a + b}}{2}\,\,\mathop \smallint \nolimits_a^b \,\,f\left( {a + b + x} \right)\,\,dx$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

f\,\,\left( {a + b - x} \right) = f\,\,\left( x \right),\,\,\;\mathrm{then}\,\,\int \nolimits_a^b \,\,xf\,\,\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{{a + b}}{2}\,\,\int \nolimits_a^b \,\,f\left( x \right)\,\,dx

Option B (

\frac{{a + b}}{2}\,\int \nolimits_a^b \,f\left( x \right)\,dx

) is correct because it uses the property of definite integrals with the given functional relationship to simplify the integral.