Problem.

Solution

1.

Note that $\sin(x^2)$ has constant sign on any interval of the form $[\sqrt{k\pi},\sqrt{(k+1)\pi}]$. Therefore, we note

$$ \int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}}\vert \sin(x^2)\vert dx=\vert\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}} \sin(x^2) dx\vert. $$

Using a change in variables $\mu=x^2$, the above integral is equal to

$$ \vert\int_{k\pi}^{(k+1)\pi}\frac{1}{2\sqrt{\mu}}\sin(\mu)d\mu\vert. $$

Since $\frac{1}{\sqrt{(k+1)\pi}}\leq\frac{1}{\mu}\leq\frac{1}{\sqrt{k\pi}}$ on $[\sqrt{k\pi},\sqrt{(k+1)\pi}]$, we have

$$ \vert\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}}\frac{1}{2\sqrt{(k+1)\pi}}\sin(\mu)d\mu\vert\leq A_k\leq\vert\int_{\sqrt{k\pi}}^{\sqrt{(k+1)\pi}}\frac{1}{2\sqrt{k\pi}}\sin(\mu)d\mu\vert. $$

The upper and lower bounds are precisely $\frac{1}{\sqrt{k\pi}}$ and $\frac{1}{\sqrt{(k+1)\pi}}$, respectively. We are done.

2.

Note that

$$ B_n=\frac{1}{n}\Sigma_{k=n}^{2n-1}A_k. $$

From our result in (1), we have the upper and lower bounds

$$ \frac{1}{\sqrt{n\pi}}\Sigma_{k=n}^{2n-1}\frac{1}{\sqrt{(k+1)\pi}}\leq B_n \leq \frac{1}{\sqrt{n\pi}}\Sigma_{k=n}^{2n-1}\frac{1}{\sqrt{k\pi}} $$

Note the Riemann sum boundings via the curve $y=1/\sqrt{x}$, that

$$ 2(\sqrt{2n}-\sqrt{n+1})=\int_{n+1}^{2n}\frac{1}{\sqrt{x}}\leq\Sigma_{k=n}^{2n-1}\frac{1}{\sqrt{(k+1)\pi}}\leq\int_n^{2n-1}\frac{1}{\sqrt{x}}dx=2(\sqrt{2n-1}-\sqrt{n}). $$