Question

If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then

\( {1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx \) is equal to:

• A. \( \int_{a - 1}^{b - 1} {f(x+1)dx} \)
• B. \( \int_{a + 1}^{b + 1} {f(x + 1)dx} \)
• C. \( \int_{a - 1}^{b - 1} {f(x)dx} \)
• D. \( \int_{a + 1}^{b + 1} {f(x)dx} \)

Hint:

Use the symmetry \(f(a+b-x)=f(x+1)\) (a King-type substitution \(x\to a+b-x\)) to simplify the integral.

Answer 1

The Correct Option is A

Let \(I=\int_a^b x\,(f(x)+f(x+1))\,dx\). Substitute \(x\to a+b-x\): since \(f(a+b-x)=f\big((a+b+1)-(x+1)\big)=f(x+1)\) and \(f(a+b-x+1)=f(x)\),
\(I=\int_a^b (a+b-x)\,(f(x+1)+f(x))\,dx\).
Adding the two forms: \(2I=(a+b)\int_a^b (f(x)+f(x+1))\,dx\), so \(\dfrac{I}{a+b}=\dfrac12\int_a^b (f(x)+f(x+1))\,dx\).
Now \(\int_a^b f(x)\,dx\) with \(x=a+b+1-t\) gives \(\int_{a+1}^{b+1}f(t)\,dt=\int_a^b f(x+1)\,dx\). So both parts are equal and
\(\dfrac{I}{a+b}=\int_a^b f(x+1)\,dx=\int_{a+1}^{b+1}f(t)\,dt\), which equals \(\int_{a-1}^{b-1} f(x+1)\,dx\) after shifting the variable (it also equals \(\int_a^b f(x)dx\)).
Answer: (A) \(\int_{a-1}^{b-1} f(x+1)\,dx\)