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Without using a spreadsheet, determine the approximate area of the region bounded by the given curve, the $x$ -axis and between the given lines, using the indicated number $n$ of rectangles.$f(x)=-2 x+3,$ between $x=0$ and $x=1, n=4$ using the (a) left endpoint, (b) right endpoint, (c) midpoint.

(a) 2.25(b) 1.75$(c) 2$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 4

Approximation of Areas

Integrals

University of Nottingham

Idaho State University

Boston College

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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Consider the region below …

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$$\begin{array}{l}{\text {…

so in this problem were given the formula F of X is equal to X plus two. In this case, we're working on the interval from zero 24 Okay, so let's kick and let's take a look at our interval. Here we have 0123 and four. Right. So 123 and four Notice. What's nice about this particular interval is 1234 equally spaced intervals. Okay, that have a with of one. Okay, So in order to make this problem work easier and be able to work through pretty efficiently, let's make a table off the off. All the values of all the output values at zero at one and two and three and at four. And will pick which ones we need based on the problem we're working with. So I'm gonna create a little table here, and I'm going to have 0123 and four is my input values. Zero will give us choose an output value. Three is an output value. 45 and six is outfit values. Now we want to work with the area using the left hand side. Well, I'm using the left side rectangles What's nice is the width of the rectangles will always be one. So it's one times and then I'm going to use just the outputs of the left hand side of each of the intervals. So in this case, I'm going to use the output values at 012 and three in this case, out the adding two plus three plus four plus five. In this case, I'm going to have five, 10 14 and 14 times one gives us 14 to the area. Using the left hand model gives us 14 using the right hand model again. My with is still one each time. So now I have to find the output value on the right side of the interval. So at 123 and four, So the output values and that those points would be three plus four plus five plus six, which is gonna give us seven, 12 18. Now, if we average those values 14 plus 18 divided by two will give us 16. Okay, let's check out the midpoint values well. My with is still one, but my height will happen at each of the middle of the values, which is what happens at between zero and one. Well, what happens between zero and one is halfway between two and three, which is 2.5 halfway between three and 43 and four is 3.5. Halfway between four and five is four point 5.5 way between five and six is 5.5. Adding up 2.5 plus 3.5 plus 4.5 plus 5.5. Multiplying by one gives us 16. Khazad. Check it out. We have the same answer as the average.

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