# Number of roots of the equation $f(x)= \int_0^x (t-1)(t-2)(t-3)(t-4)dt =0$

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.

• The derivative of $f(x)$ is $(x-1)(x-2)(x-3)(x-4)$. Find its extrema and variation table. – Jean-Claude Arbaut Jan 17 at 14:06
• 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? – Tavish Jan 17 at 14:10
• @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 $f(x)=0$? – HARVEER RAWAT Jan 17 at 14:12
• 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. – 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 $$x=0$$-
• 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? – Piquito Jan 18 at 12:48