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Evaluate the definite integral.

$ \displaystyle \int^a_0 x\sqrt{a^2 - x^2} \, dx $

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$\int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x=\frac{1}{3} a^{3}$

01:38

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 5

The Substitution Rule

Integration

Missouri State University

Oregon State University

Baylor University

University of Michigan - Ann Arbor

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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we know that Are you? Convene what's under the square root, which is a scored minus X squared which means r D X is now negative 1/2 do you over acts which means our bounds are now going to be from a squared to zero because we know a is a squared minus a score which is zero and be a zero squared plus a scored, which is simply just gonna give us a squared. Let's let us use the power rule, which means we increase the expert by one and we divide by the new exponents. Now we can simply plug in our bounds. This gives us 1/3 a cute

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