Proving that $0<\int_0^\sqrt{2}x\sin(\sqrt[3]{4x^2})\le1$

Question

I need to prove the inequality $0<\int_0^\sqrt{2}x\sin(\sqrt[3]{4x^2})\le1$

What I've done so far:

For $x\in[0, \sqrt{2}]$,

$0\le 4x^2\le8\Rightarrow 0\le \sqrt[3]{4x^2}\le2\lt\pi\Rightarrow0\le \sin(\sqrt[3]{4x^2})\le1 \Rightarrow0\le x\sin(\sqrt[3]{4x^2})\le\sqrt{2}$

And now, we know that if $m\le f(x)\le M$ for all $x \in [a,b]$, then $m(b-a)\le\int_a^bf(x)\le M(b-a)$, thus $0\le\int_0^\sqrt{2}x\sin(\sqrt[3]{4x^2})\le \sqrt2*(\sqrt2)=2$

But I don't know what to do from here. I need to prove that:

1) the integral is > then zero and not just $\ge$.

2) that it's less then 1 and not 2.

Any help will be appreciate!

Answer 1

We have $0\le x\sin(\sqrt[3]{4x^2})\le x$ for $x\in[0, \sqrt{2}]$. Hence

$0 \le \int_0^\sqrt{2}x\sin(\sqrt[3]{4x^2}) dx \le\ \int_0^\sqrt{2}x dx=1$.

Since $x\sin(\sqrt[3]{4x^2})$ is $ \ge 0$, continuous and not identically $0$, we have $0 < \int_0^\sqrt{2}x\sin(\sqrt[3]{4x^2}) dx$.

Answer 2

since $\vert \sin(\cdot)\vert\leq 1$ it follows $$ \int_{0}^{\sqrt{2}}x\sin{(\sqrt[3]{4x^2})}dx\leq \int_{0}^{\sqrt{2}}x dx=1 $$

Best regards