Question

If the function $f\left(x\right)$ is symmetric about the line $x=3$ , then the value of the integral $I={\int }_{-2}^8\frac{f\left(x\right)}{f\left(x\right)+f\left(6-x\right)}dx$ is

• A. $0$
• B. $5$
• C. $10$
• D. $16$

Answer 1

Answer: (B)

As $f\left(x\right)$ is symmetric about $x=3,$ we get,
$f\left(3+x\right)=f\left(3-x\right)$
Replacing $x$ by $3-x,$ we get,
$f\left(6-x\right)=f\left(x\right)$
Now, using $\left(a+b-x\right)$ property in $I,$ and adding, we get,
$\Rightarrow 2I={\int }_{-2}^8\frac{f\left(x\right)+f\left(6-x\right)}{f\left(x\right)+f\left(6-x\right)}dx$
$\Rightarrow 2I={\int }_{-2}^{8}1dx={\left[x\right]}_{-2}^8$
$\Rightarrow 2I=10$
$\Rightarrow I=5$