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# Midpoint sums

Approximating area under a curve using rectangle where the heights are the value of the function at the midpoint of each interval.

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• Which type is the most accurate. Is it the midpoint sum or the left or right Riemann sum?
• Midpoint is better than the other two as it somehow comes in btwn, so your approx is neither too low, nor too high, hope it helped :)
• At , , and , why do the rectangle heights change?
• The heights are changing as different approximation methods are being used. At midpoint formula is being used, then left Riemann sum approximation, and finally right Riemann sum approximation (respectively). The instructor is demonstrating how to solve with each different kind of approximation.
• In this video, how is Sal getting the height of each of his rectangles?
• Hi @win148! For the midpoint example, Sal is getting the height of the rectangles by using the midpoint. So for example, the midpoint between x=-1 and x=0 is x=-0.5, so the height of the rectangle whose endpoints are x=-1 and x=0 is f(-0.5) = 1.25.
• At , Sal defines the height of all three rectangles- the first two's height as 5/4 and the third ones as 13/4. What login is behind taking such heights to find the area by mid point method?
• Let one of these rectangles have its left endpoint at 𝑥 = 𝑎 and its right endpoint at 𝑥 = 𝑏, which means that its midpoint will be at 𝑥 = (𝑎 + 𝑏)∕2.

Within the interval [𝑎, 𝑏] it is much more common for a function 𝑓(𝑥) to be strictly increasing/decreasing rather than not, which means that its lowest and highest values within that interval will most likely be 𝑓(𝑎) and 𝑓(𝑏), while 𝑓((𝑎 + 𝑏)∕2) is somewhere in between.
Therefore, a midpoint sum is more or less guaranteed to be a better approximation of the area under the curve than a left- or righthand sum.
• for midpoint, do you multiply the heights of each rectangle by the width of each rectangle? or did he not do that because the question states all the rectangles have equal widths?
• I hope you guys have figured it out in the last 4 years but I'll post an answer for any others who have the same questions nowadays. Usually, you do have to multiply the height of the rectangles by the widths. When the widths are all equal, you can add up the heights and then multiply by the width once, since each number is being multiplied by the same width. (ex. ab + ac + ad = a(b + c + d)). In this video, however, the widths are all equal to 1, so he doesn't need to multiply.
(1 vote)
• The Fundamental Theorem of Algebra states that a polynomial will cross the x-axis n times for n equaling the degree of the highest degree term. However, x^2 + 1 does not cross the x-axis. Please explain!
(1 vote)
• This is the FTA: every non-zero, single-variable, degree n polynomial with complex coefficients has, counted with multiplicity, exactly n complex roots. There are n roots, but not all roots are necessarily real, so the polynomial may not cross the x axis. In fact, the roots of x^2 + 1 are i and -i.
• What is the summation notation for midpoint sums?
• Assuming a is the start, b is the end and t is the number of rectangles in between, Sum from x=0 to t-1 of (b-a)/t*f(a+(b-a)/(2t)+x(b-a)/t)

That may be a little complicated but if you break it up (b-a)/t is the length of each rectangle and the function is the height at a certain point, so we needed to get what each midpoint would be. Also, we go to t-1 instead of t because the first rectangle's midpoint is going to be x=0, the second midpoint is x=1, and we do this all the way down to the t-th midpoint being x=t-1. Hopefully that makes sense.

To find the midpoint we want to start at a, then go to the midpoint of the first rectangle, which is half the length so plus (b-a)/t divided by 2 which leads to (b-a)/(2t) and then finally we want to add another rectangle length to get to the next midpoint, and we want to add one midpoint length over and over again for as many rectangles there are and add them up. so x*(b-a)/t.

I really really hope that made sense. Let me know if it didn't and I can try a different way of explaining.
(1 vote)
• Couldn't you take the left sum, subtract the right sum, divide the difference by two, and then add that difference to the underestimation or subtract it from the overestimation to get the exact value?
(1 vote)
• Not nessesarily. In some instances what you suggested may work ,but there is no guarantee that the amount that one is more than is exactly twice the value that the underestimation is less than. You would get somewhere close to the exact value but it wouldn't be the exact value
• Is there a way to know how to divide the area under the curve in such a way that the Riemann approximation results in the least error?
(1 vote)
• To the best my knowledge no. The first step to solving this problem would be constructing the mathematical model.

So we end up the objective function minimise |riemann sum - integral f(x) dx|. Also need to define the domain of each variable.

To further make the problem more complicated there could be 100's of variable. So you end up with nonlinear optimisation with a large number of variables.

You may be interested in looking into operations research.
• -1^2+1=0. I am confused how 2 is
(1 vote)
• You are correct -1^2 + 1 =0, however that is not the calculation the video is performing.

The calculation perform is (-1)^2 + 1 which equal 2.

What is going on with -1^2 + 1 is the exponentiation is performed first. This leaves us with -1 + 1 = 0. This is because going by order of operations unary operator i.e. negative sign is applied after exponentiation

BENDMAS:
Bracket
Exponentiation
Negative sign
Divison/Multiplication # note equal hierarchy