prove this: $\int_{2}^{4}\frac{\mathrm{d} x}{p(x)}\leq \frac{2}{7}$

Question

I need to prove this: $$\int_{2}^{4}\frac{\mathrm{d} x}{p(x)}\leq \frac{2}{7}$$ when the polynomial $p(x)$ is: $$p(x)=x^{3}+(x-1)^{2}-2$$

How shell I begin?

Answer 1

Hint:

Prove that $p(x)\geq 7$ on $[2,4]$.

Answer 2

Observe that for $p(x)=x^{3}+(x-1)^{2}-2$ we have $$p(2)=7$$ and the derivative $$p'(x)=3x^2+2x-2$$ has two zeros at $$x_{\pm}=-\frac{1}{3} \pm \frac{1}{3}\sqrt{7}<-\frac{1}{3} + \frac{1}{3}\sqrt{9}=\frac{2}{3}<2$$ so $p'(x)$ has the same sign to the right of $x_+$ such that $p(x)$ either only increases or only decreases for all $x\geq 2$ (and indeed for all $x>x_+$). As $p'(2)=14>0$ it follows that $p'(x)>0$ and thus $p(x)>7$ for all $x\geq2$.

Thus $$\int_{2}^{4}\frac{\mathrm{d} x}{p(x)}<\int_{2}^{4}\frac{1}{7}\mathrm{d} x = \frac{2}{7}$$ and there is even a strict inequality separating your original integral from $\frac{2}{7}.$