int_0^{alpha / 3} frac{f(x)}{f(x)+fleft(frac{ | vedclass

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(D) Let $I = \int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x$  $(i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-

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$\frac{\alpha}{6}$

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(D) Let $I = \int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x$  $(i)$ Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we replace $x$ with $(\frac{\alpha}{3} - x)$: $I = \int_0^{\alpha / 3} \frac{f(\frac{\alpha}{3} - x)}{f(\frac{\alpha}{3} - x) + f(\frac{\alpha - 3(\frac{\alpha}{3} - x)}{3})} dx$ $I = \int_0^{\alpha / 3} \frac{f(\frac{\alpha}{3} - x)}{f(\frac{\alpha}{3} - x) + f(x)} dx$  $(ii)$ Adding $(i)$ and $(ii)$: $2I = \int_0^{\alpha / 3} \frac{f(x) + f(\frac{\alpha}{3} - x)}{f(x) + f(\frac{\alpha}{3} - x)} dx$ $2I = \int_0^{\alpha / 3} 1 dx$ $2I = [x]_0^{\alpha / 3} = \frac{\alpha}{3}$ $I = \frac{\alpha}{6}$

$\int_0^{\alpha / 3} \frac{f(x)}{f(x)+f\left(\frac{\alpha-3 x}{3}\right)} d x=$

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