1)

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


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

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

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

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

Answer:

Option D

Explanation:

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