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Evaluate $ \displaystyle \int^2_{-2} (x + 3) \sqrt{4 - x^2} \,dx $ by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.

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06:08

Frank Lin

01:56

Amrita Bhasin

03:08

Linda Hand

02:00

Suman Saurav Thakur

Calculus 1 / AB

Chapter 5

Integrals

Section 5

The Substitution Rule

Integration

Campbell University

Oregon 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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Alright, here's a fun integral. We're going to integrate from -2 to 2 of X-plus three times square root of four minus x squared dx. And uh it looks kind of scary at first but we're gonna do a trick. We're actually going to distribute um the square root to both parts and write it as two separate integral. So this will be equal to uh The integral from -2 to 2 of X times square root of four minus X squared dx Plus the integral from -2 to 2 times three times the square root of four minus x squared dx. So I split up the integral after distributing the square root. Now we can solve each of these by different methods. This one will solve by use substitution and this one we're going to solve by geometry. So this is kind of cool. We have two different ways to solve the different parts. So let's focus on the one on the left first by U. Sub. We'll let you be the inside of the square root and then D. U D X is minus two X. So do you over minus two. X equals D. X. And then we'll be able to sub in. We have to remember when we do so that we have to change our limits. Okay? So when I plug in X is minus two, I get zero and when I plug in excess to I also get zero. So even though I could go ahead and plug in and get it in the form of use this is going to be an integral of zero because if the limits are the same, there's no no area you could find under the curve and that equals zero. So that's pretty cool. So what's left then is the one with geometry, Let's figure out what this is. I'm going to write it as three times the integral From -2 to 2. The square root of four minus x squared dx. And what we can do is do a quick graph and um basically we have y equals square root of four minus x squared. If I square both sides I get y squared equals four minus x squared. If I add x squared to both sides, I get the familiar X squared plus y squared equals four. So that's a circle but we're looking at just the top path, top half of the circle, the minus have would have a minus side in front of the square room. So this then I'll make a note here. This is our graph of the second half. The only difference if we stretch it three times, but the area before we stretch it three times is gonna be one half of that of a circle. So let's go ahead and do it down here will change color. So basically we're going to get equal three times pi one half because it's a half circle. The radius um is um just to because you can see that uh when I have the form of a circle, this is like two squared, the radius is the two, so I will get three times pi over two times two squared because my radius is too. And this cleans up to be six pi. So it turns out that this whole integral just equal six pies. So pretty cool. Lots of like geometry and this one interesting stuff. So anyway, hopefully that helped to have a wonderful day.

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