You can approximate the area under the curve by taking the upper sum and the lower sum and then getting the limit of each as n approaches infinity.

to get the upper sum..

f(Mi)* delta X

If the interval over which we are integrating is [a,b], then delta X = (b - a)/n, where n is the number of partitions

Mi = 2i/n for upper limit

and 2(i -1)/n for lower limit

f(x) = x^2, n = 4, interval is [0, 2]

Mi= 2i/4 = i/2

f(Mi) = (i/2)^2

= i^2/4

delta X = 1/2

Use special summation formulas for sum of f(Mi)* delta (x)

The area will be 8/3

The easier way is just to take the integral of x^2 from 0 to 2

(1/3)x^3 evaluated at 2 = 8/3

(1/3)x^3 evaluated at 0 = 0

8/3 - 0 = 8/3

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