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Evaluate the integral.

$ \displaystyle \int^{\sqrt{3}/2}_{0} \frac{dr}{\sqrt{1 - r^2}} $

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00:49

Frank Lin

00:52

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Nathan C.

November 26, 2021

what about the d, I am confused about that part?

Oregon State University

Harvey Mudd College

Baylor University

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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this problem, I would just say to my students to memorize this rule. Otherwise there's a lot of math to work out um to really understand this would be a very long video. So I'm a keep this simple and I'm going to tell you ah that you could google or use your resources to look up the derivative of inverse sine. That's it's equal to this one over the square root of 1 -1. square. Uh So what I'm getting at is if we're going backwards then this anti drift will be inverse sine of our because that's the independent variable D. R. From 0 to Route 3/2. So as you plug in your upper bound and your lower bound, it's just saying like, okay, hey um if we're in the first quadrant we want the where the y coordinate on the unit circle is route 3/2. And that's at pi over three radiance. That's this angle. If if you remember your unit circle. And then the same thing is when you plug in zero Wins The Y Coordinate Equal to zero Radiance. Well the white cord and zero at zero radiance. I said that wrong. So when you plug in zero you actually do get zero. So your answers pirate 3 0, which is simply pi over three. Now I'm just double checking before I move on. Yes, that is the correct answer.

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