Riemann Sum Exercise Credit to Dr. Yousef Abbas

Exercise:

Evaluate , $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{16i}{{{n}^{2}}}\sqrt{16-\frac{16{{i}^{2}}}{{{n}^{2}}}}}$

Solution: we know that $\frac{16{{i}^{2}}}{{{n}^{2}}}={{\left( \,\frac{4i}{n}\, \right)}^{2}}$ and $\frac{16i}{{{n}^{2}}}=\frac{4}{n}\times \frac{4i}{n}$

So $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{16i}{{{n}^{2}}}\sqrt{16-\frac{16{{i}^{2}}}{{{n}^{2}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{4}{n}\times \frac{4i}{n}\sqrt{16-{{\left( \frac{4i}{n} \right)}^{2}}}}}$

          So ${{x}^{*}}_{i}=\frac{4i}{n}$ and $\Delta x=\frac{4}{n}=\frac{4-0}{n}$

Thus $\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{16i}{{{n}^{2}}}\sqrt{16-\frac{16{{i}^{2}}}{{{n}^{2}}}}=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\left( \Delta x \right){{x}^{*}}_{i}\sqrt{16-{{\left( {{x}^{*}}_{i} \right)}^{2}}}}}=\int_{0}^{4}{x\sqrt{16-{{x}^{2}}}dx}$

Let $u=16-{{x}^{2}}\Rightarrow du=-2xdx\Rightarrow -\frac{du}{2}=xdx$

So $\int_{0}^{4}{x\sqrt{16-{{x}^{2}}}dx}=-\frac{1}{2}\int_{16}^{0}{\sqrt{u}\,du}=\frac{1}{2}\int_{0}^{16}{\sqrt{u}du}=\left. \frac{1}{3}{{u}^{\frac{3}{2}}} \right|_{0}^{16}=\frac{1}{3}{{\left( 16 \right)}^{3/2}}=\frac{64}{3}$