Prove that $2\int_a^b \int_a^x f(x)f(y) \, dy \, dx = \left[ \int_a^b f(x) \, dx \right]^2$ Suppose $f$ is a continuous single-variable function, prove that:
$$2\int_a^b \int_a^x f(x)f(y) \, dy \, dx = \left[ \int_a^b f(x) \, dx \right]^2$$
This question was just on my Calculus III final exam and I couldn't figure it out, could someone provide some insight, please?
 A: Let $F(x) = \int_a^x f(y) dy$, then
  \begin{align*}
    F(b)^2 - F(a)^2 & = \left[ \int_a^b f(x) dx \right]^2 - 0
  \end{align*}
  and
  \begin{align*}
   F(b)^2 - F(a)^2 & = \int_a^b \frac{d}{dx}(F(x))^2 dx \\ & = \int_a^b 2F(x)F'(x) \\& = 2\int_a^b f(x)F(x)\\ &= 2 \int_a^b\int_a^x f(y)f(x) dy dx
  \end{align*}
  Therefore,
  $$ 2 \int_a^b\int_a^x f(y)f(x) dy dx = \left[ \int_a^b f(x) dx \right]^2 $$
A: Hint: Let
$$F(x) = \int_a^x f(y)\,dy.$$
Apply the fundamental theorem of calculus to $F(b)^2 - F(a)^2$.
A: First look at
$$
\int_a^b \int_a^b\cdots\cdots.\tag1
$$
Part of the problem is to prove that  the given integral $\displaystyle\int_a^b\int_a^x \cdots\cdots$ is just half of that, i.e. half of the integral in $(1)$.
First look at the integral in $(1)$:
$$
\int_a^b \left(\int_a^b f(x) f(y)\,dy\right)\,dx
$$
Look at the inside integral first:
$$
\int_a^b f(x) f(y)\,dy
$$
As $y$ goes from $a$ to $b$, $x$ remains constant, so $f(x)$ is a constant factor and can be pulled out, getting
$$
f(x)\int_a^b f(y)\,dy.
$$
Now we have
$$
\int_a^b\left(f(x)\int_a^b f(y)\,dy\right)\,dx.
$$
In the integral with respect to $x$, as $x$ goes from $a$ to $b$, the entire inside integral does not change because no "$x$" appears within it, so that is a "constant" and may be pulled out, getting:
$$
\left(\int_a^b f(x)\,dx\right)\cdot\int_a^b f(y)\,dy.
$$
These two integrals are both the same thing, since we've merely re-named the variable that goes from $a$ to $b$.  So the value of this last expression is
$$
\left(\int_a^b f(x)\,dx\right)^2.
$$
Now look at
$$
\int_a^b \left(\int_a^x \cdots\cdots\,dy\right)\,dx
$$
You have $x$ going from $a$ to $b$, and for any fixed value of $x$ in that range, $y$ goes from $a$ to $x$.  Draw the picture and see that that's just half the square.  How do you know that the integral over that half of the square is equal to the integral over they other half?  The answer to that question is that the only difference between the two is that the roles of $x$ and $y$ are interchanged, and the whole expression is symmetric in $x$ and $y$.
A: HINT f(x) and f(y) are independent. Does this mean...you can separate the integrals in some way? (look in your book.)
A: The integral can be re-written as follows.
$$\int_a^b\int_a^x f(x)f(y) \,dy\,dx = \int_a^b f(x)\left(\int_a^x f(y)\,dy\right)\,dx $$
Let $$F(x) = \int_a^x f(y) dy$$
Then the integral becomes 
$$\int_a^b f(x)F(x) \,dx = \int_a^b F(x)f(x) \,dx$$
Since $f$ is continuous, we know from FTC II that $F$ is differentiable and
$$F'(x)=f(x)$$
So we can now use the substitution $u=F(x)$. To get
$$\int_a^b F(x)f(x) \,dx = \int_{x=a}^{x=b} u \,du = \frac{u^2}{2}|_{x=a}^{x=b}=\frac{F(b)^2-F(a)^2}{2}$$.
But $F(a)^2=0$ and $\displaystyle \frac{F(b)^2}{2}=\frac{1}{2}\left[ \int_a^b f(x) \,dx\right]^2$ by definition. 
The result follows.
