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Determine the area of the region bounded by the given curves.The region bounded by $f(x)=x \sqrt{4-x^{2}}$ and the $x$ -axis.

$$16 / 3$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Missouri State University

Oregon State University

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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Determine the area of the …

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Find the area of the regio…

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Okay, So the first part of this problem is we need to figure out the bounds between X and the square root of four minus X squared DX where they cross the X axis. So that would be setting that equation equal to zero as a times right there. Uh, so there's one bound at X equals zero, and then the other bound would be both. I square both sides, so I'm ignoring the sex. Now. Um, if I said that this piece equal to zero and I square both sides, I would have this equals zero. So, addicts where it over. Um, so four equals X squared. So we actually have two bounds positive and negative two. So what I'm going to do is just do one bound from 0 to 2 and then double it because the areas will be the same. This isn't even Yes, the need, um, function should be. Anyway, from there, though, I would use U substitution four minus x squared. So do you. Or equal negative two x. D s. So then, from there, I could rewrite this and going over to x d. U is equal to DX. Um so as I'm looking at doing u substitution. What I would probably do is leave that X alone. Rewrite that as, uh instead of the square root of you, I'm gonna write as you to the one half power. And again the square root is won't have power. And I'm replacing you with that and then replaced DX with a negative 1/2 X. I'm just splitting them up. Oops, does not look like to you. But that way you can actually see these exes eliminate. And I actually forgot to rewrite this to down here, but from there I think you'd see that those change. And if you plug in your bounds, you plug in zero for X four, minus zero would be four and then plugging to infer x four minus four because two squared is four would give me zero, so I would actually rewrite this problem. I know I'm kind of getting messy here, but we had a negative in front. What I can do is I can just switch those bounds around, and now I'm ready to actually do the operation. And that's adding one to your exponent, um and then multiply by the reciprocal of that new exponents from 0 to 4. And what's nice about this is this Is the square root cubed. So the square to four is two. Cubed is eight times two is 16 3rd, and when you plug in zero, the square 20 is still zero. Cubed is still zero times. Anything is still zero. So that's your final answer.

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