Worth 99 points!Use the midpoint rule with n=4 to approximate the area of the region bounded by y=x^3 and y=x. Must use the Midpoint Rule! Please help, thank you.
Accepted Solution
A:
The objective is to find total bounded area. Intersections occur at x=-1, x=0, and x=1. Since we need 4 subintervals, we split the line segment from -1 to 1 into four line segment.
That would be [tex]\Delta x = \dfrac{b-a}{n} = \dfrac{1 - \text-1}{4} = \dfrac12[/tex]
So subintervals would be [-1, -0.5] [-0.5, 0] [0, 0.5] [0.5, 1]
Since segment was split evenly, because of rotate symmetry, sub-area for [-1,0] and [0,1] would be same so we can just focus on interval [0,1] with n=2 and then multiple the result by 2.
Now we are talking about midpoint rules, the height of rectangle for [0,0.5], [0.5,1] would be [tex]f(\frac{0+0.5}2)=f(0.25), f(\frac{0.5+1}2)=f(0.75)[/tex] respectively. Width of rectangle would be [tex]\Delta x = 0.5[/tex].
Hence for [tex]x[/tex], that would be [tex]f(0.25)\Delta x + f(0.75)\Delta x = 0.25\cdot0.5 + 0.75\cdot0.5 = 0.5[/tex]
And [tex]x^3[/tex] would be [tex](0.25)^3\cdot0.5+(0.75)^3\cdot0.5 = 0.21875[/tex]
So the approximate area is 0.28125.
Now double that to account for area from [-1,0] and our answer is 0.5625.