Answer: Â b) 23.94
Step-by-step explanation:
You are trying to find the area under the curve. Â Area = height x width.
Height is the y-value at the given coordinate --> f(x)
Width is the distance between the x-values --> dx
First, let's figure out dx: Â the distance from 1 to 4 is 3 units. We need to divide that into 6 sections because n = 6 --> dx = 1/2
So the intervals are when x = {1, 3/2, 2, 5/2, 3, 7/2, 4}
For midpoint sum, we need to find the midpoint of the intervals
[tex]\text{midpoint}\ \bigg\{1, \dfrac{3}{2}\bigg\}=\dfrac{1+\frac{3}{2}}{2}=\dfrac{5}{4}\\\\\\\text{midpoint}\ \bigg\{\dfrac{3}{2}, 2\bigg\}=\dfrac{\frac{3}{2}+2}{2}=\dfrac{7}{4}\\\\\\\text{midpoint}\ \bigg\{2, \dfrac{5}{2}\bigg\}=\dfrac{2+\frac{5}{2}}{2}=\dfrac{9}{4}\\\\\\\text{midpoint}\ \{\dfrac{5}{2}+3\}=\dfrac{\frac{5}{2}+3}{2}=\dfrac{11}{4}\\\\\\\text{midpoint}\ \{3, \dfrac{7}{2}\}=\dfrac{3+\frac{7}{2}}{2}=\dfrac{13}{4}\\\\\\\text{midpoint}\ \{\dfrac{7}{2},4\}=\dfrac{\frac{7}{2}+4}{2}=\dfrac{15}{4}[/tex]
Next, let's find the height for each of the midpoints:
f(x) = x² + 1
[tex]f\bigg(\dfrac{5}{4}\bigg)=\bigg(\dfrac{5}{4}\bigg)^2+1=\dfrac{41}{6}\\\\\\f\bigg(\dfrac{7}{4}\bigg)=\bigg(\dfrac{7}{4}\bigg)^2+1=\dfrac{65}{6}\\\\\\f\bigg(\dfrac{9}{4}\bigg)=\bigg(\dfrac{9}{4}\bigg)^2+1=\dfrac{97}{6}\\\\\\f\bigg(\dfrac{11}{4}\bigg)=\bigg(\dfrac{11}{4}\bigg)^2+1=\dfrac{137}{6}\\\\\\f\bigg(\dfrac{13}{4}\bigg)=\bigg(\dfrac{13}{4}\bigg)^2+1=\dfrac{185}{6}\\\\\\f\bigg(\dfrac{15}{4}\bigg)=\bigg(\dfrac{15}{4}\bigg)^2+1=\dfrac{241}{6}[/tex]
Now, let's find the Area: A = f(x) dx:
[tex]\text{Midpoint Sum:}\quad A=\dfrac{1}{2}\bigg(\dfrac{41}{16}+\dfrac{65}{16}+\dfrac{97}{16}+\dfrac{137}{16}+\dfrac{185}{16}+\dfrac{241}{6}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{1}{2}\bigg(\dfrac{766}{16}\bigg)\\\\\\.\qquad \qquad \qquad \qquad =\dfrac{383}{16}\\\\\\.\qquad \qquad \qquad \qquad =23.9375[/tex]
