Q102 / 225 · JEE Main 2003 · Mathematics

If $\mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})=\mathrm{f}(\mathrm{x})$ then $\int_a^b \mathrm{xf}(\mathrm{x}) \mathrm{dx}$ is equal to

  • Correct answer
    A
    $\frac{a+b}{2} \int_a^b f(a+b-x) d x$
  • B
    $\frac{a+b}{2} \int_a^b f(b-x) d x$
  • C
    $\frac{a+b}{2} \int_a^b f(x) d x$
  • D
    $\frac{b-a}{2} \int_a^b f(x) d x$

Solution

$I=\int_a^b x f(x) d x=\int_a^b(a+b-x) f(a+b-x) d x$ $=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(a+b-x) d x$ $=(a+b) \int_a^b f(a+b-x) d x-\int_a^b x f(x) d x$ $2 I=(a+b) \int_a^b f(x) d x$ $I=\frac{(a+b)}{2} \int_a^b f(x) d x ; I=\frac{(a+b)}{2} \int_a^b f(a+b-x) d x$

Hence, the correct answer is option: 'A'.

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