Let fxx3 and compute the riemann sum of f over the interval

Solution

For n = 2, the subintervals are (0,.5) and (.5,1). The length of each subinterval is .5. Using the midpoints of the subintervals as the representative points, the Riemann Sum is .5(.25^3) + .5(.75^3) = .21875 or 7/32 For n = 5, the subintervals are (0,.2), (.2,.4), (.4,.6), (.6,.8), and (.8,1). The length of each subinterval is .2. Using the midpoints of the subintervals as the representative points, the Riemann sum is .2(.1^3) + .2(.3^3) + .2(.5^3) + .2(.7^3) + .2(.9^3) = .245 or 49/200 For n = 10, the subintervals are (0,.1), (.1,.2), (.2,.3), (.3,.4), (.4,.5), (.5,.6), (.6,.7), (.7,.8), (.8,.9), and (.9,1). The length of each subinterval is .1. Using the midpoints of the subintervals as the representative points, the Riemann sum is .1(.05^3) + .1(.15^3) + .1(.25^3) + .1(.35^3) + .1(.45^3) + .1(.55^3) + .1(.65^3) + .1(.75^3) + .1(.85^3) + .1(.95^3) = .24875 or 199/800. Based on the results, the higher n goes, the Riemann Sum gets closer to .25 or 1/4. So that would be a good guess as the area of the region under the graph of f on the interval 0,1. (In fact, it\'s exactly .25 :D)