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# 2015 AP Calculus AB/BC 3b

Riemann sum to estimate distance.

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• so if it doesn't have absolute value bars, is it just the distance in that period of time?
(1 vote)
• If you take the integral of the velocity function instead of the integral of the absolute value of the velocity function, you will have the change in position during the time period (displacement) instead of the change in total distance traveled during the time period.
• The Riemann sum (i = 1,50) ∑(i/50)²1/50 on the closed interval [0,1] is an approximation for which of the following definite integrals?
The answer is (0,1) ∫ x² dx. But I'm not sure how to solve it. Can anybody explain?
(1 vote)
• Converting Reimann sum to integral and reverse, you need to understand the "components" in each
In the Riemann sum, the upper and lower limits tell you how many smaller rectangles it has. The Riemann sum is the sum of those areas, with the component length x widths. Since it is from 0-1, and there are 50 rectangles, each width is 1/50, so that make (i/50)² the length. But this length is just the function f(x) and the width is delta x.

The formula for delta x = (a + b)/n, where a is the left bound, and b is the right bound, and n number of rectangles. and the other formula is
x_i = a + i*delta x

So from this problem, we see that n=50, a=0, b=1, so delta dx= (0+1)/50 = 1/50
x_i = 0 + i*1/50 = i/50

(i = 1,50) ∑(i/50)²1/50 = (i = 1,50) ∑(x_i)² deltax
When you convert this into a definite integral, the left and right bound become your lower and upper limit of the integral, i.e. [0,1] the function is still the function without i, and deltax becomes dx
so we have ∫₀¹x² dx
(1 vote)
• At , I thought that we needed to multiply each value of v(t) by (b - a)/n, where b is the right endpoint of the rectangle, a is the left endpoint of the rectangle, and n is the number of intervals (it might also sometimes be written as dx or delta x). I see Mr. Khan multiplying each value of v(t) by (b - a), but I don't think I see 1/n (in this case, 1/4) being multiplied by the total distance in meters. Am I missing something or are you not supposed to multiply by 1/4 in this case?
(1 vote)