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Determine the area of the indicated region.Region bounded by $f(x)=\frac{2}{\sqrt[3]{x}}$ and the $x$ -axis, between $x=1$ and $x=8$.

9

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

Lectures

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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.

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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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Determine the area of the …

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Find the area of the regio…

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So when we're finding the area between X's goes one and X equals eight, I'm just going into order. By the way, they're just telling you to find the interval from 1 to 8 of the function that they give you, which is to over the cube root of X dx. Now, what I would do is have my students rewrite. This problem is still the interval, but the two uh, you can actually move that, too in front. But I'm finally met you right there and rewrite the rational exponents to the one third power for the Cuba. But it's in the denominator as well. So negative one third power still DX. So hopefully all of this makes sense because now we're ready to do the anti drip where you add one, uh, just a reminder that adding one is the same thing as adding three thirds so you can see that it's two thirds as your new experiment, and then you have to multiply by the reciprocal of your new excellent, which is three halves. This, too, is still from the problem, but it will cancel out with the two that's from the experiments from 1 to 8 now what you can do is actually factor out. The three do this. That's a property of the integral and just think about the when you plug in your bounds. It means the cube root of eight squared well. The cube root of eight is two. Squared is four, and then same thing with plugging in one. Well, one day, any power? Just one. So it's really easy. Uh, but it is the cube root of one square, which is still one, and you subtract those values and now it's pretty easy. I think you can do this in your head. That four minus one is three times at three, gives you nine, and that would be a correct answer.

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