Question

If $f(x)=f(a−x)$ and $g(x)+g(a−x)=2$ then the value of $\underset{0}{\overset{a}{∫}}f(x)g(x)dx$ is

• A. $\underset{0}{\overset{a}{∫}}f(x)dx$
• B. $\underset{0}{\overset{a}{∫}}g(x)dx$
• C. $\underset{0}{\overset{a}{∫}}[g(x)−f(x)]dx$
• D. $\underset{0}{\overset{a}{∫}}[g(x)+f(x)]dx$

Answer 1

Answer: (A)

Given that $f(x)=f(a−x)...$(i)
and $g(x)+g(a−x)=2....$(ii)
Now, let $I=\underset{0}{\overset{a}{∫}}f(x)g(x)dx$
$=\underset{0}{\overset{a}{∫}}f(a−x)g(a−x)dx$
$⇒I=\underset{0}{\overset{a}{∫}}f(x)[2−g(x)]dx$
[using (i) and (ii)]
$=\underset{0}{\overset{a}{∫}}2f(x)dx−\underset{0}{\overset{a}{∫}}f(x)g(x)dx$
$⇒I=\underset{0}{\overset{a}{∫}}2f(x)dx−I$
$⇒2I=\underset{0}{\overset{a}{∫}}2f(x)dx$
$⇒I=\underset{0}{\overset{a}{∫}}f(x)dx$
Note: $\underset{0}{\overset{a}{∫}}f(x)dx=\underset{0}{\overset{a}{∫}}f(a−x)dx$.