Compute the lower riemann sum for the given function f(x)=x2 over the interval x∈[−1,1] with respect to the partition p=[−1,− 1 2 , 1 2 , 3 4 ,1].

21/64

Step-by-step explanation:

First, we need to note that the function f(x) = x² is increasing on (0, +∞), and it is decreasing on (-∞,0)

The first interval generated by the partition is [-1, -1/2], since f is decreasing for negative values, we have that f takes its minimum values at the right extreme of the interval, hence -1/2.

The second interval is [-1/2, 1/2]. Here f takes its minimum value at 0, because f(0) = 0, and f is positive otherwise.

Since f is increasing for positive values of x, then, on the remaining 2 intervals, f takes its minimum value at their respective left extremes, in other words, 1/2 and 3/4 respectively.

We obtain the lower Riemman sum by multiplying this values evaluated in f by the lenght of their respective intervals and summing the results, thus

LP(f) = f(-1/2) * ((-1/2) - (-1)) + f(0) * (1/2 - (-1/2)) + f(1/2)* (3/4 - 1/2) + f(3/4) * (1- 3/4)

= 1/4 * 1/2 + 0 * 1 + 1/4 * 1/4 + 9/16 * 1/4 = 1/8 + 0 + 1/16 + 9/64 = 21/64

As a result, the lower Riemann sum on the partition P is 21/64

answer:   14

step-by-step explanation:          

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step-by-step explanation: