Q.Show that $\int_{0}^{a}f(x)g(x)\,dx=2\int_{0}^{a}f(x)\,dx$, if $f$ and $g$ are defined as $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$

Concept: Even Function Property — The condition $f(x)=f(a-x)$ makes $f$ symmetric about $x=a/2$, and the relation $g(x)+g(a-x)=4$ allows a substitution trick.

Step 1: Let $I = \int_{0}^{a} f(x)g(x)\,dx$. Substitute $x \to a-x$:

since $f(a-x)=f(x)$.

Step 2: Add the two expressions for $I$: …

We start with two conditions:

$f(x) = f(a-x)$ for all $x$ — this is a symmetry about the midpoint $a/2$.

$g(x) + g(a-x) = 4$ — this tells us that the values of $g$ at symmetric points add to a constant 4.

The goal is to show that the product integral simplifies to twice the integral of $f$ alone. The key insight: split the integration interval $[0,a]$ into two symmetric halves, use a substitution on one half, and then apply both given conditions to collapse the expression.

1. Split the integral at the midpoint Write

2. Substitute in the second half In the second integral, let $t = a - x$. Then $x = a - t$, $dx = -dt$, and when $x = a/2$, $t = a/2$; when $x = a$, $t = 0$. So

3. Apply the symmetry of $f$ Since $f(a-t) = f(t)$ by the given condition, this becomes

4. Combine the two halves Now the original integral is

5. Use the condition on $g$ The given $g(x) + g(a-x) = 4$ simplifies the bracket:

6. Relate to the full integral of $f$ Because $f(x) = f(a-x)$, the integral of $f$ over $[0,a]$ is twice the integral over $[0, a/2]$: …

Method: Proving an integral identity via reflection and a symmetric condition

To prove a relation like $\displaystyle\int_0^a f(x)g(x)\,dx=2\int_0^a f(x)\,dx$, reflect with $x\to a-x$ and feed in the given symmetry conditions on $f$ and $g$.

Steps

Step 1: Write $I$ and reflect.

$I=\displaystyle\int_0^a f(x)g(x)\,dx=\int_0^a f(a-x)g(a-x)\,dx.$

Step 2: Apply the condition on $f$.

Given $f(a-x)=f(x)$, the reflected integrand becomes $f(x)g(a-x)$.

Step 3: Add the two expressions and use the condition on $g$. …

Common Mistakes

Mistake 1: Misreading $f(x)=f(a-x)$ as "even about $x=0$".

Why it's wrong: this symmetry is about the midpoint $x=\tfrac a2$, not the origin; it does not by itself let you write $\int_{-a}^{a}$-style rules. Correct approach: use it precisely as $f(a-x)=f(x)$ inside the reflected integral.

Mistake 2: Not using the condition $g(x)+g(a-x)=4$ after adding. …