Question

Calculate the left Riemann sum for f(x)=1/√x over [1, 4] with n = 6(6 rectangles). When rounding, round your answers to four decimal places.

Answer 1

Calculate the left Riemann sum for $f(x) = \frac{1}{\sqrt{x}}$ over [1, 4] with $n = 6(6 \text{ rectangles})$. When rounding, round your answers to four decimal places.

Solution:

We have

$$\int_{1}^{4} \frac{1}{\sqrt{x}} dx \approx \sum_{i=0}^{5} h \frac{1}{\sqrt{x_i}} = h \sum_{i=0}^{5} \frac{1}{\sqrt{x_i}}$$

where $h = \frac{4 - 1}{n} = \frac{3}{6} = \frac{1}{2}; x_i = 1 + i \cdot h = 1 + \frac{i}{2}$. So

$$\begin{aligned} \int_{1}^{4} \frac{1}{\sqrt{x}} dx &\approx h \sum_{i=0}^{5} \frac{1}{\sqrt{x_i}} = \frac{1}{2} \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{1 + 1 \cdot \frac{1}{2}}} + \frac{1}{\sqrt{1 + 2 \cdot \frac{1}{2}}} + \frac{1}{\sqrt{1 + 3 \cdot \frac{1}{2}}} + \frac{1}{\sqrt{1 + 4 \cdot \frac{1}{2}}} + \right. \\ &\left. + \frac{1}{\sqrt{1 + 5 \cdot \frac{1}{2}}} \right) = \frac{1}{2} \left( 1 + \sqrt{\frac{2}{3}} + \frac{1}{\sqrt{2}} + \sqrt{\frac{2}{5}} + \frac{1}{\sqrt{3}} \right) \approx 1.8667 \end{aligned}$$

The exact solution is

$$\int_{1}^{4} \frac{1}{\sqrt{x}} dx = 2 \sqrt{x} \Big|_{1}^{4} = 2(\sqrt{4} - \sqrt{1}) = 2(2 - 1) = 2.$$

Answer:

$$\approx 1.8667$$