If $f(a+b-x)=f(x)$ then $\int ^b_axf(x)dx$ is equal to

We are asked to evaluate

\(\displaystyle I = \int_a^b x f(x)\, dx \quad \text{given } f(a+b-x)=f(x).\)

Step 1: Put substitution \(t=a+b-x\). Then \(dx=-dt\).

When \(x=a \Rightarrow t=b\), when \(x=b \Rightarrow t=a\).

So, \[ I = \int_a^b x f(x)\, dx = \int_b^a (a+b-t) f(t)(-dt) = \int_a^b (a+b-t) f(t)\, dt. \]

Step 2: Add both forms of \(I\):

\[ 2I = \int_a^b [x f(x) + (a+b-x) f(x)] dx = \int_a^b (a+b) f(x)\, dx. \]

Step 3: Simplify:

\[ I = \frac{a+b}{2} \int_a^b f(x)\, dx. \]

Final Answer: \(\displaystyle \frac{a+b}{2}\int_a^b f(x)\, dx\) → matches Option 1.