We are asked to evaluate the definite integral \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\). This is a standard type of integral that can be solved using a property of definite integrals.

Evaluating the Definite Integral

Let the given integral be denoted by \(I\). \(I = \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx \quad \cdots (1)\)

We will use the property of definite integrals which states: \(\int_0^a f(x)\ dx = \int_0^a f(a-x)\ dx\)

Applying this property to the integral \(I\), we replace \(x\) with \((a-x)\) in the integrand. Note that \((a-(a-x)) = a-a+x = x\). So, the integrand \(\frac{f(a-x)}{f(x)+f(a-x)}\) becomes: \(\frac{f(a-(a-x))}{f(a-x)+f(a-(a-x))} = \frac{f(x)}{f(a-x)+f(x)}\)

Thus, the integral \(I\) can also be written as: \(I = \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx \quad \cdots (2)\)

Now, we add the two expressions for \(I\) from equations (1) and (2): \(I + I = \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx + \int_0^a \frac{f(x)}{f(a-x)+f(x)}\ dx\) \(2I = \int_0^a \left( \frac{f(a-x)}{f(x)+f(a-x)} + \frac{f(x)}{f(x)+f(a-x)} \right)\ dx\)

Since the denominators are the same, we can combine the numerators: \(2I = \int_0^a \frac{f(a-x) + f(x)}{f(x)+f(a-x)}\ dx\)

The numerator and the denominator are identical. Assuming \(f(x)+f(a-x) \neq 0\) over the interval \([0, a]\), the fraction simplifies to 1: \(2I = \int_0^a 1\ dx\)

Now, we evaluate the simple integral: \(\int_0^a 1\ dx = [x]_0^a = a - 0 = a\)

So, we have: \(2I = a\)

Solving for \(I\): \(I = \frac{a}{2}\)

Thus, the value of the definite integral \(\rm \int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx \) is \(\rm \frac{a}{2}\).

Revision Table: Key Integral Properties

Here is a summary of the property used in solving this integral:

Additional Information: Applying the Property

The property \(\int_0^a f(x)\ dx = \int_0^a f(a-x)\ dx\) is very useful for definite integrals with limits from \(0\) to \(a\). It is often applied when the integrand has a form that relates \(f(x)\) and \(f(a-x)\).

Common forms where this property simplifies the integral include:

• \(\int_0^a \frac{f(x)}{f(x)+f(a-x)}\ dx\)
• \(\int_0^a \frac{f(a-x)}{f(x)+f(a-x)}\ dx\) (as seen in this problem)
• \(\int_0^a f(x)g(a-x)\ dx\) where \(\int_0^a f(a-x)g(x)\ dx\) is also calculated

In many cases, adding the original integral \(I = \int_0^a f(x)\ dx\) and the transformed integral \(I = \int_0^a f(a-x)\ dx\) leads to a significant simplification of the integrand, often resulting in a constant or a much simpler function to integrate.