Let f and g be continuous functions on [0, a] such that $$f(x) = f(a-x)$$ and $$g(x) + g(a-x) = 4$$, then $$\int_0^a f(x)g(x)dx$$ is equal to

We have the integral

$$I=\int_0^{a}f(x)\,g(x)\,dx.$$

The question tells us two important facts:

$$f(x)=f(a-x)\qquad\text{and}\qquad g(x)+g(a-x)=4\;.$$

Because $$f$$ and $$g$$ are continuous on $$[0,a]$$, we are allowed to perform a change of variable without worrying about points of discontinuity.

Now, in the integral $$I$$, let us make the substitution

$$x=a-t\;.$$

Then $$dx=-dt$$. When $$x=0$$, we have $$t=a$$, and when $$x=a$$, we have $$t=0$$. Reversing the limits will remove the minus sign. Explicitly:

$$\begin{aligned} I&=\int_{x=0}^{x=a}f(x)\,g(x)\,dx\\ &=\int_{t=a}^{t=0}f(a-t)\,g(a-t)\,(-dt)\\ &=\int_{t=0}^{t=a}f(a-t)\,g(a-t)\,dt. \end{aligned}$$

Because $$f(x)=f(a-x)$$, we can replace $$f(a-t)$$ by $$f(t)$$:

$$I=\int_0^{a}f(t)\,g(a-t)\,dt.$$

The symbol of integration is a dummy variable, so we may rename $$t$$ back to $$x$$ for ease of comparison:

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

We now have two perfectly valid expressions for $$I$$:

$$I=\int_0^{a}f(x)\,g(x)\,dx\quad\text{and}\quad I=\int_0^{a}f(x)\,g(a-x)\,dx.$$

Adding these two equal quantities term-by-term, we obtain

$$\begin{aligned} 2I&=\int_0^{a}f(x)\,g(x)\,dx+\int_0^{a}f(x)\,g(a-x)\,dx\\ &=\int_0^{a}f(x)\bigl[g(x)+g(a-x)\bigr]\,dx. \end{aligned}$$

But the problem states that

$$g(x)+g(a-x)=4\;.$$

Substituting this constant into the integrand gives

$$2I=\int_0^{a}f(x)\,(4)\,dx=4\int_0^{a}f(x)\,dx.$$

Finally, dividing both sides by $$2$$ yields

$$I=2\int_0^{a}f(x)\,dx.$$

Thus the value of $$\displaystyle\int_0^{a}f(x)g(x)\,dx$$ is exactly twice the value of $$\displaystyle\int_0^{a}f(x)\,dx.$$

Hence, the correct answer is Option D.