Question

If the function $f\left(x\right)$ is symmetric about the line $x=3$ , then the value of the integral $I=\displaystyle \int _{- 2}^{8} \frac{f \left(x\right)}{f \left(x\right) + f \left(6 - x\right)}dx$ 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,
$\Rightarrow 2I=\displaystyle \int _{- 2}^{8}\frac{f \left(x\right) + f \left(6 - x\right)}{f \left(x\right) + f \left(6 - x\right)}dx$
$\Rightarrow 2I=\displaystyle \int _{- 2}^{8} 1dx=\left[x\right]_{- 2}^{8}$
$\Rightarrow 2I=10$
$\Rightarrow I=5$