Find the number of roots of $F(x)= \int_0^x e^t(t^2-3t-5)\mathrm dt , x>0$ in the interval $(0,4)$

Question

Let $$F(x)= \int_0^x e^t(t^2-3t-5)\mathrm dt , x>0$$ Find the number of roots of $F(x)=0$ in the interval $(0,4)$.

My attempt: I simply integrated it and got $F(x)=e^x(x(x-5))$ which has roots $0$ and $5$ and none is between $0$ and $4$. So the answer is $0$, which is correct.

My question: Is there any other way to do this question in comparatively less time?

Answer 1

Hint: note that $$x^2-3x-5<0$$ On the interval $(0, 4)$

Answer 2

You have $F(0)=0, F'(x)=e^x(x^2-3x-5)<0$ in $(0,4)$. This means $F(x)$ is strictly decreasing in $(0,4)$, so that $F(x)<0$ in $(0,4)$.

Answer 3

Yes: it's easy to see, without computing the roots, that $t^2-3t-5$ has a negative and a positive root, and the latter is $>4$. Using the First Fundamental Theorem of Integral Calculus, $F(x)$ decreases on the interval $(0,4)$. Now $F(0)=0$, hence $F(x)<0$ for $0<x<4$.