Question

When rounded to the nearest hundredth, the area under the graph of f(x) = x2 + 1 from x = 1 to x = 4 with 6 subintervals using the midpoint sum is a)20.38. b)23.94. c)27.88. d)33.47. My learning platform is rather bad at explaining math so if anyone sees this, I would take learning how to do this over an answer any day

Answer 1

Answer: b) 23.94

Explanation:

You are trying to find the area under the curve. 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: 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

`\text{midpoint}\ \bigg\{1, (3)/(2)\bigg\}=(1+(3)/(2))/(2)=(5)/(4)\\\\\\\text{midpoint}\ \bigg\{(3)/(2), 2\bigg\}=((3)/(2)+2)/(2)=(7)/(4)\\\\\\\text{midpoint}\ \bigg\{2, (5)/(2)\bigg\}=(2+(5)/(2))/(2)=(9)/(4)\\\\\\\text{midpoint}\ \{(5)/(2)+3\}=((5)/(2)+3)/(2)=(11)/(4)\\\\\\\text{midpoint}\ \{3, (7)/(2)\}=(3+(7)/(2))/(2)=(13)/(4)\\\\\\\text{midpoint}\ \{(7)/(2),4\}=((7)/(2)+4)/(2)=(15)/(4)`

Next, let's find the height for each of the midpoints:

f(x) = x² + 1

`f\bigg((5)/(4)\bigg)=\bigg((5)/(4)\bigg)^2+1=(41)/(6)\\\\\\f\bigg((7)/(4)\bigg)=\bigg((7)/(4)\bigg)^2+1=(65)/(6)\\\\\\f\bigg((9)/(4)\bigg)=\bigg((9)/(4)\bigg)^2+1=(97)/(6)\\\\\\f\bigg((11)/(4)\bigg)=\bigg((11)/(4)\bigg)^2+1=(137)/(6)\\\\\\f\bigg((13)/(4)\bigg)=\bigg((13)/(4)\bigg)^2+1=(185)/(6)\\\\\\f\bigg((15)/(4)\bigg)=\bigg((15)/(4)\bigg)^2+1=(241)/(6)`

Now, let's find the Area: A = f(x) dx:

`\text{Midpoint Sum:}\quad A=(1)/(2)\bigg((41)/(16)+(65)/(16)+(97)/(16)+(137)/(16)+(185)/(16)+(241)/(6)\bigg)\\\\\\.\qquad \qquad \qquad \qquad =(1)/(2)\bigg((766)/(16)\bigg)\\\\\\.\qquad \qquad \qquad \qquad =(383)/(16)\\\\\\.\qquad \qquad \qquad \qquad =23.9375`