2
$\begingroup$

I have the following question before me: Find the number of roots of the equation $f(x)= \int_0^x (t-1)(t-2)(t-3)(t-4)dt =0$ in the interval $[0,5]$. $0$ is clearly one of the roots. But how can I find other roots, if any? I tried evaluating the integral and came up with a fifth degree polynomial in $x$ having no constant term. The equation seemed quite daunting to me. How can I get the roots quicker? Please suggest.

$\endgroup$
4
  • $\begingroup$ The derivative of $f(x)$ is $(x-1)(x-2)(x-3)(x-4)$. Find its extrema and variation table. $\endgroup$ – Jean-Claude Arbaut Jan 17 at 14:06
  • $\begingroup$ Find the value of the quintic at $1,2,3,4$ and using that its derivative is $0$ at these points, can you sketch the quintic? $\endgroup$ – Tavish Jan 17 at 14:10
  • $\begingroup$ @Jean-Claude Arbaut I have come up with $1,3$ as the points of local Maxima and $2,4$ as the points of local minima. But how are they of help in finding the roots of $f(x)=0$? $\endgroup$ – HARVEER RAWAT Jan 17 at 14:12
  • $\begingroup$ Apply the IVT. What's the sign of $f$ at the extrema? You will also need $f(0)$ and $f(5)$. Note you are not looking for the roots, but for the number of roots. $\endgroup$ – Jean-Claude Arbaut Jan 17 at 14:14
1
$\begingroup$

$f(x)=\dfrac{x^5}{5}-\dfrac{5x^4}{2} +\dfrac{35x^3}{3}-25x^2+24x$ so $x=0$ is a root in $[0,5]$

On the other hand $$\dfrac{df}{dx}=\dfrac{d}{dx}\int_0^x((t-1)(t-2)(t-3)(t-4))dt=(x-1)(x-2)(x-3)(x-4)$$ which gives two maximums $(1,\dfrac{251}{30})$ and $(3,\dfrac{81}{10})$ and two minimums $(2,\dfrac{116}{15})$ and $(4,\dfrac{112}{15})$ so all of the four extremums being positive there are not real solutions in $[1,5]$. We can not say that a non-real solution is in $[0,5]$ because it would implies an order in $\mathbb C$ where there is not any structure of order.

Consequently the only root is $x=0$-

$\endgroup$
3
  • $\begingroup$ At $x = 4$, as I have said, there is a minimum from which the function shoots up to infinity. What do you want me to add? $\endgroup$ – Piquito Jan 18 at 12:48
  • $\begingroup$ The last extremum being a minimum and positive, no problem at all. $\endgroup$ – Piquito Jan 18 at 17:10
  • $\begingroup$ I'm surprised you respond immediately. You must know things in the computer that I do not know (actually I do not know many things about calculators and computers). Regards. $\endgroup$ – Piquito Jan 18 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.