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Solve the given problems by integration.Find the circumference of the circle $x^{2}+y^{2}=r^{2}$ by using the formula for arc length: $s=\int_{a}^{b} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x$.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 6

Inverse Trigonometric Forms

Integrals

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Suppose $C$ is the circle …

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$\int_{-1}^{1} \frac{4}{1+…

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Evaluate the line integral…

00:36

in this question were given this circle X squared plus y squared, goes to ask where we want to find its circumference by using this as equals two. This integration of dysfunction over here now we will be integrating from X equals two minus R. Two R. But if you do that there's only half the circumference over here. So to get the other half over here we need to multiply by two. Yeah, okay now we need to replace the D. V. D X over here. So differentiating the circle we respect to X on both sides. The first term when we differentiate we get two x second term. When we differentiate we'll get to y dy dx and our square is a constantly differentiate and zero so dy dx mhm is minus X over Y. Now we can support the wine. Why from this equation here is actually square roots of our square minus x square. I know there's a plus minus but I will be using the positive one. So over here are we putting in minus x over square root of our square minus x square. Now let's take the one side. Just this just this part over here. So I won't be using up too much space. You can see that minus X. When I square, I'll just get X square mm and at the bottom square or r squared minus x squared, I'll just when I square it, I'll just get our square minus X squared, not common base, I'll get our square minus x square. Classic square, you can see the X squares can sort out. So I'm left with our square at the top. But when you square really, it's just our and at the bottom is our square minus x square square roots. So let me replace that, it will be our over square root of our square minus x square the X. Now my arm is a constant so I can bring it out. So let me write the S. Yeah over here. Let me right here as one because I will bring up my are over here. Mhm. Great. Now this actually fill the form of the sign was integral. You can see the sign was in the growth. Mhm. Now my A is my are and my you is my ex. So let's integrate that by fitting into the sign in verse form. So it will be signing verse X over our. Yeah something in the upper and lower limit. You get signed in this are over our minus sign in this minus are over our. So a sign in verse one minus I invest minus one, evaluating you get power to minus minus power to. So the final answer will be two pi r.

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