Mathematics | VITEEE 2016 | Discussion Page

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

$\frac{a+b}{2}\int_{a}^{b} f(b-x) dx$

$\frac{a+b}{2}\int_{a}^{b} f(b+x) dx$

$\frac{b-a}{2}\int_{a}^{b} f(x) dx$

$\frac{a+b}{2}\int_{a}^{b} f(x) dx$

Option D

Let I= $\int_{a}^{b} x f(x) dx$

Let a+b-x= z $\Rightarrow$ -dx=dz

when x=a,z=b and when x=b,z=a

$\therefore$ $I= -\int_{b}^{a} (a+b-z)f(z) dz$

$I= (a+b)\int_{b}^{a} f(x) dx-\int_{a}^{b} x f(x) dx$

$I= (a+b)\int_{b}^{a} f(x) dx-I;$

$2I= (a+b)\int_{a}^{b} f(x) dx$

Hence , I= $\left(\frac{a+b}{2}\right)\int_{a}^{b} f(x) dx$

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