Answer
Estimating the area under the graph
Since the function is increasing on [0, 1], we use left endpoints to
obtain lower sums and right endpoints to obtain upper
sums.
Work Step by Step
Using finite approximations to estimate the area under
the graph of the function using
$\text{Genera Rule:}$
Area = sum of rectangles' area under the curve
The Area of a Rectangle = Width * Height
Width is $\Delta x$ = $\frac{\text{final point - initial point}}{\text{number of rectangles}}$
Height is $f(x_i)$ where $x_i$ is the height at each corresponding point $i$
The difference between lower and upper sums
Area (Lower Sum) = $\sum_{i=0}^{n-1} f(x) \dot\ \Delta x$ where n is the number of rectangles
Area (Upper Sum) = $\sum_{i=1}^{n} f(x) \dot\ \Delta x$ where n is the number of rectangles
a. a lower sum with two rectangles of equal width.
$\Delta x=\frac{1-0}{2}=\frac{1}{2}\ \text{and}\ x_i = i\Delta x=\frac{i}{2}\Rightarrow\ \sum_{i=0}^{1} \Delta x \dot\ x_i^2 =\sum_{i=0}^{1} \frac{1}{2}(\frac{i}{2})^2 = \frac{1}{2}(\frac{0}{2})^2 + \frac{1}{2}(\frac{1}{2})^2=\frac{1}{2}(\frac{1}{4})=\frac{1}{8}$
b. a lower sum with four rectangles of equal width.
$\Delta x=\frac{1-0}{4}=\frac{1}{4}\ \text{and}\ x_i = i\Delta x=\frac{i}{4}\Rightarrow\ \sum_{i=0}^{3} \Delta x \dot\ x_i^2 =\sum_{i=0}^{3} \frac{1}{4}(\frac{i}{4})^2 = \frac{1}{4}(\frac{0}{4})^2 + \frac{1}{4}(\frac{1}{4})^2+\frac{1}{4}(\frac{2}{4})^2+\frac{1}{4}(\frac{3}{4})^2=\frac{7}{32}$
c. an upper sum with two rectangles of equal width.
$\Delta x=\frac{1-0}{2}=\frac{1}{2}\ \text{and}\ x_i = i\Delta x=\frac{i}{2}\Rightarrow\ \sum_{i=1}^{2} \Delta x \dot\ x_i^2 =\sum_{i=1}^{2} \frac{1}{2}(\frac{i}{2})^2 = \frac{1}{2}(\frac{1}{2})^2 + \frac{1}{2}(\frac{2}{2})^2=\frac{1}{2}(\frac{1}{4})+\frac{1}{2}(1)=\frac{3}{8}$
d. an upper sum with four rectangles of equal width
$\Delta x=\frac{1-0}{4}=\frac{1}{4}\ \text{and}\ x_i = i\Delta x=\frac{i}{4}\Rightarrow\ \sum_{i=1}^{4} \Delta x \dot\ x_i^2 =\sum_{i=1}^{4} \frac{1}{4}(\frac{i}{4})^2 = \frac{1}{4}(\frac{1}{4})^2 + \frac{1}{4}(\frac{2}{4})^2+\frac{1}{4}(\frac{3}{4})^2+\frac{1}{4}(\frac{4}{4})^2=\frac{15}{32}$
