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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.

Tong
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SGKw
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2 Answers2

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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!

am_11235...
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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,

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

An_Elephant
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    I don't understand. Doesn't the property you linked to keep the bounds of integration? – Akiva Weinberger Jan 15 '23 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. – Li Kwok Keung Jan 15 '23 at 14:10
  • @LiKwokKeung It is $2\int_1^3x(x-3)^2dx$, not just $\int_1^3x(x-3)^2dx$ – Kamal Saleh Jan 15 '23 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. – An_Elephant Jan 15 '23 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. – An_Elephant Jan 15 '23 at 16:13
  • @An_Elephant The first integral has bounds $\int_0^4$ and the second has bounds $\int_1^3$. With the King's Property, it seems that both integrals have bounds $\int_a^b$. – Akiva Weinberger Jan 15 '23 at 16:30
  • Oops I did a blunder. I'll correct it soon. I am a total nonsense. – An_Elephant Jan 15 '23 at 16:30