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Evaluate the integral.

$ \displaystyle \int^{\sqrt{3}}_{1/\sqrt{3}} \frac{8}{1 + x^2} \,dx $

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$\frac{4 \pi}{3}$

00:47

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 3

The Fundamental Theorem of Calculus

Integration

Campbell University

Baylor 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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Evaluate the integrals.

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Evaluate the indefinite in…

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Evaluate the definite inte…

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Evaluate the integral usin…

Alright, so here is another integral to solve. And notice the limits kinda looks scary but there's things we can do and it's not so bad. First. What I'm gonna do is if I pull the eight out of the integral. Hopefully then you recognize this as a known derivative And this is a known derivative of Inverse. 10. So I can write then my solution as eight times in verse 10 of acts. And then I have to go from one over root well one over root three to route three. So when I substitute in then I get eight in verse 10 of route three um minus eight in verse 10 Of One Over Root three. And the trick with Tan is that Tan is really made up of sine over cosine. So in general if I have Tan of theta, it's sine of theta over cosine of theta. So you have to thank ratios. Uh just known like values. So for example if I have 10 of pi over six, That's equal to 1/2 over root three over to notice the twos cancel and I get one over root three. So that's where this one came from. So this whole value is just pi over six And if I flip it meaning I have signed that root 3/2 and co sign that is a half then um you can see how I get through three. So this part Is Pi over three. So this gives me 8/3 pi -86 pi. And that subtracts two B 8, six pi Or 4/3 pi. So that is our solution to our integral. Hopefully it helped to have an amazing day.

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