How can one prove this property of integrals?$\int_0^bf(x)(b-x)\,dx=\int_0^b\left(\int _0^xf(t)\,dt\right)\,dx$

Question

$$\int_0^bf(x)(b-x)\,dx=\int_0^b\left(\int _0^xf(t)\,dt\right)\,dx$$

I can't understand how this property can be proven but it has held true for everything I have tried. How do you even approach this? I have tried substitution but that gets you no where.

Answer 1

Define functions $F$ and $G$ as follows: $$F(b):=\int_0^b f(x)(b-x)\,dx=b \int_0^b f(x)\,dx - \int_0^b xf(x)\,dx \\ G(b):=\int_0^b \left( \int_0^x f(t)\,dt \right) dx.$$ Then, differentiating "with respect to $b$", and using the product rule together with the FTC we have that $$F'(b)=1 \cdot \int_0^b f(x)\,dx + b \cdot f(b) - bf(b) = \int_0^b f(x)\,dx$$ and that $$G'(b)=\int_0^b f(t)\,dt$$ for all $b$. This means that $F' = G'$, and then $F=G+c$ for some constant $c$.

Since $F(0)=G(0)=0$, we conclude that $c=0$, and therefore $F = G$, meaning that $F(b)=G(b)$ for all $b$, as desired.

Answer 2

For the R.H.S. we have $$\int_0^b\left(\int^x_0f(t)dt\right)dx=\int_0^b\left(F(x)-F(0)\right)dx=\int_0^bF(x)dx-\int_0^bF(0)dx$$

Integrate by parts on the first integral so that $u=F(x)\implies du =f(x) dx$ and $dv=dx\implies v=x$ so we have that $$\int_0^bF(x)dx=xF(x)\bigg|^b_0-\int_0^bxf(x)dx=bF(b)-\int_o^bxf(x)dx$$ so the R.H.S. is equal to $$bF(b)-\int_0^bxf(x)dx-F(0)x\bigg|^b_0=b(F(b)-F(0))-\int_0^bxf(x)dx=$$

$$=\int_0^bbf(x)dx-\int_0^bxf(x)dx=\int_0^bf(x)(b-x)dx$$

Answer 3

The left-hand side equals $\int_0^b bf(x)\ dx-\int_0^b xf(x)\ dx$

Now use integration by parts where $u=\int_0^x f(t)dt$ and $dv=dx$. Then $du=f(x)dx$ and $v=x$. We see $uv|_0^b=\int_0^bbf(t)\ dt$ and $\int_0^b v\ du=xf(x)\ dx$, so the right-hand side equals

$$\int_0^bbf(t)dt-\int_0^bxf(x)\ dx$$

which is the same as the left-hand side.

Answer 4

Well, assuming you can change the order of integration $$ \int_0^b \int_0^x f(t) \ \text{d}t \ \text{d}x = \int_0^b \int_t^b f(t) \ \text{d}x \ \text{d}t = \int_0^b f(t) \left(\int_t^b \text{d}x \right) \ \text{d}t = \int_0^bf(t)(b-t) \ \text{d}t $$

Answer 5

\begin{align} & \int_0^b\left(\int _0^xf(t)\,dt\right)\,dx \\[10pt] = {} & \iint\limits_{(t,x)\,:\,0\,<\,t\,<\,x\,<\,b} f(t)\, d(t,x) \\[10pt] = {} & \int_0^b \left( \int_t^b f(t) \, dx \right) dt \\[10pt] = {} & \int_0^b f(t)(b-t) \, dt. \end{align}