Question

If $ƒ(a+b-x)=ƒ(x)$,then $∫^b_axƒ(x)dx$ is equal to

• A. $\frac{a+b}{2}∫^b_aƒ(b-x)dx$
• B. $\frac{a+b}{2}∫^b_aƒ(b+x)dx$
• C. $\frac{b-a}{2}∫^b_aƒ(x)dx$
• D. $\frac{a+b}{2}∫^b_aƒ(x)dx$

Answer 1

Answer: (D)

$\frac{a+b}{2}∫^b_aƒ(x)dx$

Answer 2

The correct answer is D:$(\frac{a+b}{2})∫^b_aƒ(x)dx$
Let $I=∫^b_axƒ(x)dx...(1)$
$I=∫^b_a(a+b-x)ƒ(a+b-x)dx\,\,\,\, (∫^b_aƒ(x)dx=∫^b_aƒ(a+b-x)dx)$
$⇒I=∫^b_a(a+b-x)ƒ(x)dx$
$⇒I=(a+b)∫^b_aƒ(x)dx-I [Using(1)]$
$⇒I+I=(a+b)∫^b_aƒ(x)dx$
$⇒2I=(a+b)∫^b_aƒ(x)dx$
$⇒I=(\frac{a+b}{2})∫^b_aƒ(x)dx$
Hence,the correct Answer is D.