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Use Riemann sums and the limit to find the area.Find the area of the region bounded by $f(x)=x^{3}$ and the $x$ -axis, between $x=0$ and $x=1$.

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Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 5

Sigma Notation and Areas

Integrals

Missouri State University

Baylor University

University of Nottingham

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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Use the method of Riemann …

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Determine the area of the …

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01:29

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Use Riemann sums and a lim…

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right. So this is a interesting problem with area. A lot of people associate area as always, being positive. Um, so if you're in that category of saying the area is always positive that you want to make sure that you do the absolute value of the problem. Um And when you see the X axis is telling you to do the integral of the function that they give you two over x dx and they give you the bounds from X equals negative 32 X equals negative one. Uh, now I like to go in order from least to greatest. But again, if you're considering area to be always positive and you can absolute value that answer, So if you get a negative, which I believe you will, um you just have to think Oh, okay. Well, area needs to be a positive value. So the derivative of natural log of X is equal to one over X. So for working backwards here, it makes sense that the anti derivative or the integral of this problem is to natural log of X. Now, notice that I'm putting an absolute value here because we can only log positive numbers If you're confused, why, it's two times we'll just think of it this way is you can move a constant in front and becomes two times one over X. And then you can apply this This rule, I guess, from negative 32 negative one. So, as we're looking at that problem, um, we're gonna plug in to natural your upper bound and being one minus two natural log. And again, I'm plugging in negative one. But the absolute value of next one is possible. And so I guess I did that already. Natural log of three. Now, this is where we're going to get into that debate on our answer because and I expect my students, you know, the natural log of one is zero and zero times. Anything is zero minus. That's going to give you a negative answer, and that's where we go back to. An area needs to be positive. So it's an absolute value. All of this, and you're left with just the positive answer, which would be to natural log of three, because two times 00 minus this is that answer. Um, but it has to be positive, so we're left with that

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