# Prove the equation $\int_0^4 f(x(x-3)^2) \,dx=2\int_1^3 f(x(x-3)^2) \,dx$

Let $$f:\mathbb{R}→\mathbb{R}$$ be a continuous function. Prove the equation $$\int_0^4 f(x(x-3)^2) \,dx=2\int_1^3 f(x(x-3)^2) \,dx.$$ I have tried substituting $$x(x-3)^2 = u.$$ But after that I couldn't figure out.

• Quick Tip: Think of it graphically. (desmos.com/calculator/ff5jtx5tmz) Commented Jan 15, 2023 at 12:48
• @mrtechtroid I think I have almost figured it out... thanks
– SGKw
Commented Jan 15, 2023 at 13:28

Consider the function $$g(x)-t=x(x-3)^2-t$$ and consider the respective inverse functions as $$z=z(t),y=y(t),x=x(t)$$ . Then using Vieta's formula on the equation $$g(x)-t=0$$ we have $$x(t)+y(t)+z(t)=6$$ . Therefore \begin{align*}\int_0^4f(x(x-3)^2)\,dx&=\left(\int_0^1+\int_1^3+\int_3^4\right)f(x(x-3)^2)\,dx\\&=\int_0^4f(t)\,dy(t)+\int_1^3f(x(x-3)^2)\,dx+\int_0^4f(t)\,dz(t)\\&=\int_0^4f(t)(y'(t)+z'(t))\,dt+\int_1^3f(x(x-3)^2)\,dx\\&=-\int_0^4f(t)\,dx(t)+\int_1^3f(x(x-3)^2)\,dx\\&=-\int_3^1f(x(x-3)^2)\,dx+\int_1^3f(x(x-3)^2)\,dx\\&=2\int_1^3f(x(x-3)^2)\,dx\end{align*} Done!

By using Kings Property (Equivalently substitute $$4-t$$ to $$x$$) We see that both LHS and RHS are equal except the factor of $$2$$ on the RHS. Hence, there's a typo in your question. There will not be $$2$$ and the equality holds.

Hence,

$$\int_0^4 f(x(x-3)^2) \,dx≠2\int_1^3 f(x(x-3)^2) \,dx$$

But,

$$\int_0^4 f(x(x-3)^2) \,dx=\int_1^3 f(x(x-3)^2) \,dx$$

• I don't understand. Doesn't the property you linked to keep the bounds of integration? Commented Jan 15, 2023 at 13:48
• @AkivaWeinberger Let $f(x)=x$, then $$\int_0^4f(x(x-3)^2) dx =\int_0^4x(x-3)^2 dx=8$$ and $$\int_1^3f(x(x-3)^2) dx =\int_1^3x(x-3)^2 dx=4.$$ Your assertion is incorrect in this case. Commented Jan 15, 2023 at 14:10
• @LiKwokKeung It is $2\int_1^3x(x-3)^2dx$, not just $\int_1^3x(x-3)^2dx$ Commented Jan 15, 2023 at 15:17
• @AkivaWeinberger I am afraid that I don't understand. Please elaborate more. I am just in high school and have used this property many times. Commented Jan 15, 2023 at 16:11
• @LiKwokKeung Yes and I'm shocked to see this. Right now, I can't find out that why it didn't worked. I'll be keeping to edit and correct it if I figure it out. But If you already know that what's wrong here or why this property can't be used , I would be grateful if you tell this. Commented Jan 15, 2023 at 16:13