The rectangles in the graph below ilustrate a right endpoint Riemann sum for f(z

ID: 3198833 • Letter: T

Question

The rectangles in the graph below ilustrate a right endpoint Riemann sum for f(z)2z on the interval [4,8 The value of this right endpoint Riemann sum is 23.0833 the area of the region enclosed by y-f(x). the x-axis, and the vertical linesx- 4 andx-8 and it is an an underestimate of tn Right endpoint Rlemann sum for y +2z on |4, 8] Using left and right Riemann sums based on the diagrams above, we definitively conclude that 6 Hint: For the last integral, you should consistently choose either to underestimate or overestimate the area. This may require that you use the left Riemann sum for some x-intervals and the right Riemann sum for other x-intervals

Explanation / Answer

Given f(x)=(-x^2/6)+2x on the interval (4,8)

To find left and right endpoints riemann sum we must substitute the given interval in the given f(x)

taking interval from x=4, f(x) = (-4^2)/6 + 2(4) = 32/3 = 10.66

x=4.5 f(x) = (-4.5^2)/6 + 2(4.5) = 12.375

x=5 f(x) =(-5^2)/6 + 2(5) = 14.16

x=5.5 f(x) = (-5.5^2)/6 + 2(5.5) = 16.04

x= 6 f(x) = (-6^2)/6 + 2(6) = 18

x=6.5 f(x) = (-6.5^2)/6 + 2(6.5) = 20.04

x= 7 f(x) = (-7^2)/6 + 2(7) = 22.17

x=7.5 f(x) =(-7.5^2)/6 + 2(7.5) =23.375

x=8 f(x) =(-8^2)/6 + 2(8) =26.67

1.. the first question is to find integral from 4 to 6

taking the above f(x) from 4 to 6

the left endpoints are given by 4,4.5 and 5 then f(x) are given by 10.66,12.375,14.16