Chapter 5.3, Problem 2QQ

To determine

To find: the approximation of Riemann sum.

Answer to Problem 2QQ

Option B is correct. Reimann sum approximation is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x}}dx.$

Explanation of Solution

Given:

The function is $f(x)=\overline{)x}$ .

Formula used:

If $y=f(x)$ is defined over closed interval and interval is partitioned into n sub intervals, length of $k^{th}$ subinterval is $\Delta x_k$ . On each sub interval select some number. Denote the number chosen from the $k^{th}$ subinterval by $c_k$ .

Riemann sum $R={\displaystyle \sum _{k=1}^{n}f(c_k)\Delta x_k}$ .

Calculation:

The given function is $f(x)=\sqrt{x}$ .

  $\begin{array}{l}\frac{1}{20}\left(\sqrt{\frac{1}{20}}+\sqrt{\frac{2}{20}}+\sqrt{\frac{3}{20}}+\mathrm{.................}+\sqrt{\frac{20}{20}}\right)=\frac{1}{20}{\displaystyle \sum _{k=1}^n\sqrt{\frac{n}{20}}}\\ \frac{1}{20}\left(\sqrt{\frac{1}{20}}+\sqrt{\frac{2}{20}}+\sqrt{\frac{3}{20}}+\mathrm{.................}+\sqrt{\frac{20}{20}}\right)={\displaystyle \sum _{k=1}^n\sqrt{n}}\frac{1}{20\sqrt{20}}\end{array}$

For Riemann sum approximation to integral, length of sub-interval must be tending to zero. So take length of sub interval as small as possible.

  $\Delta x=\frac{\text{difference of limit of integration}}{\text{a large value tends to infinity}}$

  $\begin{array}{l}\Delta x=\frac{1-0}{20\sqrt{20}}\\ \Delta x=\frac{1}{20\sqrt{20}}\end{array}$

Riemann sum approximation is ${\displaystyle \underset{0}{\overset{1}{\int }}\sqrt{x}}dx.$

Option b is correct.