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Estimate the area bounded by $f(x)=e^{x},$ and the $x$ -axis for $0 \leq x \leq 2$ using $n=100$ rectangles and (a) the left endpoint in each subinterval, (b) the right endpoint in each subinterval and (c) the midpoint of each subinterval.

(a) 6.3253785(b) 6.4531596(c) 6.3889496

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 4

Approximation of Areas

Integrals

Campbell University

Harvey Mudd College

Baylor University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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Approximate the area under…

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$$\begin{array}{l}{\text {…

So this problem we're working with the function f of X is equal to eat of the X plus one and we're looking at the interval from negative to two. Which means we're trying to find the area under the curve from negative to to to and based on the directions were wanna use Onley four rectangles which tells us that if I were to split up negative to negative 1012 If I were split up the distance between negative two and two into four equal pieces, we'd end up with equally spacing of one unit between each of the Internet, which is great. So what I want to do here is I want to be able to work with the rectangles now, in order, work with the rectangles. If I find the values at each of those beginning and ending of the intervals, then it will help us to be able to be able to find out exactly what the area would approximate two. Okay, so if I evaluate the function at negative two at negative one at zero at one and at two, we have values that would help us to come up with the information. So evaluating this at negative two gives us 1.1 35 Evaluating this at negative one gives us 1.3 68 evaluating and zero gives me too. Evaluating at one will give us 3.718 and evaluating it too. Well, give us 8.389 Now. The first thing we want to do is we want to evaluate an approximation using rectangles from the left hand side. Well, what's nice here is the rectangles will always have a with or a base of one unit. And if I want the height, I'm gonna use the values at the left hand side of each of the rectangles at negative to negative 10 and one. So it's gonna be 1.135 and 1.368 and to and 3.718 adding those up 1.135 plus 1.3 68 plus two plus 3.718 and multiplying those values by one. We get an approximate area of 8.2 to 1 to calculate the right hand side, the rectangles using the right hand side again my base will always be one. But now I'm going to focus in on the heights that air achieved on the right hand side of the function, which would be 1.36 AIDS plus two plus 3.718 an 8.389 that will give us an approximation of 1.368 plus two plus 3.718 plus 8.389 which would give us an area of approximately 15.475 If we average those values, which would be 8.2 to 1 plus 15.475 divided by two, we would have an average area of 11.848 Well, let's calculate what the value would be at the midpoint. So at the midpoint again, my base is gonna always be want. But now I want the height halfway in between each of those values. So that would be 1.135 plus 1.368 divided by two. That would give us the halfway point between those two. 1.368 plus two, divided by two. That's the halfway point between those two two and 3.718 Give us the halfway point between those two and finally 3.71 AIDS at 8.389 divided by to give you the halfway point between those two. So that's 1.135 plus one point 368 Divided by two Gives us one point 2515 1.368 plus two Divided by two Gives US 1.684 two plus 3.13 point 718 Divided by two gives US 2.859 and 3.718 plus 8.389 Divided by two gives US 6.535 So 6.535 plus 2.859 plus 1.684 was 1.2 515 Gives us an approximation of 11.848 and noticed that these two will be the same

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