# Proof using Fundamental Theorem of Calculus (Showing RHS = LHS)

Question: Prove that $$\int^x_0\big[\int^u_0 f(t) dt\big] du = \int^x_0f(u)(x-u) du$$ where $$f$$ is a continuous function.

Attempt: My lecturer hinted that it'd be helpful to apply the fundamental theorem of calculus to $$F(u) = u \int^u_0f(t) dt$$.

I know that I can apply the FTC to $$F(u)$$ since $$f$$ is continuous, meaning it is also a Riemann integrable function and thus the conditions of the FTC are met.

To find $$F'(u)$$, I will let $$G(u) = u$$ and $$H(u)=\int^u_0f(t) dt$$ be functions such that $$F(u)=G(u)H(u)$$. Thus, $$G'(u)=1$$ and, applying the FTC, $$H'(u)=f(u)$$.

Using the product rule, $$F'(u)=G(u)H'(u)+G'(u)H(u)=uf(u)+\int^u_0f(t) dt$$.

However, this is where I get stuck and am not sure how to use this to prove the initial equation.

Any help would be greatly appreciated.

## 3 Answers

Continuing from where you left off, integrate both sides wrt $$u$$ from $$0$$ to $$x$$ to get

\begin{equation}\begin{aligned} \int_{0}^{x}F'(u)du & = \int_{0}^{x}\left(uf(u) + \int_{0}^{u} f(t) dt\right) du \\ F(x) - F(0) & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ x\int_{0}^{x}f(t)dt - 0 & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ \int_{0}^{x}xf(u)du & = \int_{0}^{x}uf(u)du + \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du \\ \int_{0}^{x}\left(\int_{0}^{u}\left(f(t) dt\right)\right) du & = \int_{0}^{x}xf(u)du - \int_{0}^{x}uf(u)du \\ \int_0^x\left(\int_0^u f(t) dt\right) du & = \int_0^x f(u)(x-u) du \end{aligned}\end{equation}\tag{1}\label{eq1A}

Note that with going from the third to the fourth lines, since $$t$$ and $$u$$ are just dummy integration variables then, I used a substitution of $$u = t$$ in the LHS integral. This was so you can more easily see how that integral can be combined with the other integral in the second last line to then get the last line which is what you're trying to show.

• Going from the second to the third line (in the LHS), how did you go from $F(x)-F(0)=x\int^x_0 f(t)dt - 0$? – Viv4660 May 17 '20 at 2:53
• @Viv4660 You've defined $F(u) = u\int_{0}^{u}f(t)dt$. Thus, $F(x) = x\int_{0}^{x}f(t)dt$ and $F(0) = 0\int_{0}^{0}f(t)dt = 0$. – John Omielan May 17 '20 at 2:55
• Ahh I get it thank you so much for your help! – Viv4660 May 17 '20 at 3:07

$$\newcommand{\d}{\; \mathrm{d} #1}$$ $$\newcommand{\bb}{\left( #1 \right)}$$ $$\newcommand{\sb}{\left[ #1 \right]}$$ We apply integration by parts and FTOC: \begin{align*} \int_0^x \bb{\int_0^u f(t) \d{t}} \d{u} &= \sb{u\int_0^u f(t) \d{t}}_{u=0}^{u=x} - \int_0^x u\underbrace{\bb{\int_0^u f(t) \d{t}}'}_{=f(u)\text{ by FTOC}}\d{u} \\ &= x\int_0^x f(t) \d{t} - \int_0^x uf(u)\d{u} \\ &= \int_0^x xf(u) \d{u} - \int_0^x uf(u)\d{u} \\ &= \int_0^x (x - u)f(u) \d{u} \end{align*}

• This is my preferred approach. +1 – Paramanand Singh May 17 '20 at 4:23

hint

The derivative of the LHS is

$$\int_0^xf(u)du$$

the RHS can be put as

$$x\int_0^xf(u)du-\int_0^xuf(u)du$$

its derivative is $$\int_0^xf(u)du+xf(x)-xf(x)=$$ $$\int_0 ^xf(u)du$$ they have the same derivative, so theire difference is a constant $$C$$. for $$x=0$$, we find that $$C=0$$.

• Can you please explain further for how, in the RHS, you differentiated the term $\int^x_0 uf(u) du$? – Viv4660 May 17 '20 at 2:46
• @Viv4660: by fundamental theorem of calculus the derivative of this term is $xf(x)$. – Paramanand Singh May 17 '20 at 4:21
• @ParamanandSingh Ah yes I realise that now! For some reason the $uf(u)$ being a product of two functions kind of threw me and I thought the process would be different, but I realise now that doesn't matter. As always, thanks for your help :) – Viv4660 May 17 '20 at 5:27