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.

  • $\begingroup$ 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

$f(x)=\dfrac{x^5}{5}-\dfrac{5x^4}{2} +\dfrac{35x^3}{3}-25x^2+24x$ 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 $x=0$-

  • $\begingroup$ 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

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