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25 May, 13:40

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].

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  1. 25 May, 13:53
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    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
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