Riemann sum asked Dan and editied by me in many math groups

Exercise:

Find $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{\sqrt{2{{n}^{2}}-{{k}^{2}}}+\sqrt{2{{n}^{2}}+{{k}^{2}}}}{\sqrt{4{{n}^{4}}-{{k}^{4}}}}}$

Solution: we have $\sqrt{4{{n}^{4}}-{{k}^{4}}}=\sqrt{{{n}^{4}}\left( 4-\frac{{{k}^{4}}}{{{n}^{4}}} \right)}={{n}^{2}}\sqrt{4-{{\left( \frac{{{k}^{2}}}{{{n}^{2}}} \right)}^{2}}}={{n}^{2}}\sqrt{4-{{\left( \frac{k}{n} \right)}^{4}}}$

Also $\sqrt{2{{n}^{2}}-{{k}^{2}}}=\sqrt{{{n}^{2}}\left( 2-\frac{{{k}^{2}}}{{{n}^{2}}} \right)}=n\sqrt{2-{{\left( \frac{k}{n} \right)}^{2}}}$ and $\sqrt{2{{n}^{2}}+{{k}^{2}}}=n\sqrt{2+{{\left( \frac{k}{n} \right)}^{2}}}$

So $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{\sqrt{2{{n}^{2}}-{{k}^{2}}}+\sqrt{2{{n}^{2}}+{{k}^{2}}}}{\sqrt{4{{n}^{4}}-{{k}^{4}}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{\sqrt{2{{n}^{2}}-{{k}^{2}}}+\sqrt{2{{n}^{2}}+{{k}^{2}}}}{\sqrt{4{{n}^{4}}-{{k}^{4}}}}}$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{\sqrt{2-{{\left( \frac{k}{n} \right)}^{2}}}+\sqrt{2+{{\left( \frac{k}{n} \right)}^{2}}}}{\sqrt{4-{{\left( \frac{k}{n} \right)}^{4}}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1-0}{n}\sum\limits_{k=1}^{n}{\frac{\sqrt{2-{{\left( x_{k}^{*} \right)}^{2}}}+\sqrt{2+{{\left( x_{k}^{*} \right)}^{2}}}}{\sqrt{4-{{\left( x_{k}^{*} \right)}^{4}}}}}$

Where ${{h}_{k}}=\frac{b-a}{n}=\frac{1-0}{n}$ and $\Delta x_{k}^{*}=\frac{k}{n}$

So $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{\sqrt{2-{{\left( x_{k}^{*} \right)}^{2}}}+\sqrt{2+{{\left( x_{k}^{*} \right)}^{2}}}}{\sqrt{4-{{\left( x_{k}^{*} \right)}^{4}}}}}=\int_{0}^{1}{\frac{\sqrt{2-{{x}^{2}}}+\sqrt{2+{{x}^{2}}}}{\sqrt{4-{{x}^{4}}}}dx}$

Observe that $\sqrt{4-{{x}^{4}}}=\sqrt{{{2}^{2}}-{{\left( {{x}^{2}} \right)}^{2}}}=\sqrt{\left( 2-{{x}^{2}} \right)\left( 2+{{x}^{2}} \right)}$

So $\int_{0}^{1}{\frac{\sqrt{2-{{x}^{2}}}+\sqrt{2+{{x}^{2}}}}{\sqrt{4-{{x}^{4}}}}dx}=\int_{0}^{1}{\frac{1}{\sqrt{2+{{x}^{2}}}}dx}+\int_{0}^{1}{\frac{1}{\sqrt{2-{{x}^{2}}}}dx}$

But \[\int_{0}^{1}{\frac{dx}{\sqrt{2+{{x}^{2}}}}=arc\sinh \left( \frac{x}{\sqrt{2}} \right)_{0}^{1}=arc\sinh \left( \frac{1}{\sqrt{2}} \right)}\] and

$\int_{0}^{1}{\frac{dx}{\sqrt{2-{{x}^{2}}}}=\arcsin \left( \frac{x}{\sqrt{2}} \right)_{0}^{1}=\arcsin \left( \frac{1}{\sqrt{2}} \right)-\arcsin \left( 0 \right)=\frac{\pi }{4}}$

Thus $\int_{0}^{1}{\frac{\sqrt{2-{{x}^{2}}}+\sqrt{2+{{x}^{2}}}}{\sqrt{4-{{x}^{4}}}}dx}=\frac{\pi }{4}+arc\sinh \left( \frac{1}{\sqrt{2}} \right)$