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Problem

Evaluate the integral. $ \displaystyle \int \f…

04:43

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Problem 48 Hard Difficulty

Evaluate the integral.

$ \displaystyle \int_0^1 x \sqrt{2 - \sqrt{1 - x^2}}\ dx $


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 5

Strategy for Integration

Related Topics

Integration Techniques

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Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Watch More Solved Questions in Chapter 7

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75
Problem 76
Problem 77
Problem 78
Problem 79
Problem 80
Problem 81
Problem 82
Problem 83
Problem 84

Video Transcript

let's start off by doing the use substitution here. So here I'm basically taking you to be this radical, which is inside of the larger rentable. And then here use the chain rule to take the derivative. So here we get negative X over square root, one minus x square and then here we should go ahead and rewrite this. So this is negative X over you the eggs and let's go ahead and write. Multiply the negative. Let's take this equation here and just push this negative in the D you to the other side by multiplying So we get negative. You do You equals X T X and receiving the X here with the DX term. So also for our integral Remember, we have a definite numeral here, so we will have to change the limits of integration. So plug in X equals zero and X equals one into this X over here. And after doing that, we have our new limits for one and zero. And then for the inner girl. Well, this big right uncle right here that's just radical to minus you and what's left over. And blue. It's just this x, The X, Sir and we already have that over here. That's just negative. You and then I also need my to you. So this is our new integral. And then here we can go ahead and pull out the minus sign looks one zero. And then the next thing to do would be we could do another use up here. Let's take Vita. Be two minus you. Oops. Two minus you. That's just based on the radical here was inside for the negative DVD equals do you? So here I have a minus from here. But then we'll also get another minus from over here. It is a definite integral still. So I will have to change the limits. So plug in U equals one and zero into this equation. Oh, so plugging in one first we get to minus one is one and then plugging in zero for you We get to So then hear this radical. This is just square root of thie and then we're left over with you. So here you is just equal to two minus V. So we just have square ruin and hear that schist two minus v. So before we start integrating, let's first of all cancelled, this double minus that's not becomes a plus and louse. Just rewrite this integral so that it will be easier to use the power rule. So I'm doing here is rewriting the radical as Vito the one half. And then I just multiplied this radical turbo terms. So going on to the next page, I used the power rule. Now that's to be Now we have three halves and then we have to divide by three house. And then the next one becomes me too. The five half divided by five house plugging your end points wants it too. So here first, let's just multiply the constant out before we plug into too. So it's for over three and then we have two to the three house. And then here we have two or five times two of the five over two. So this is from plugging into now plug in one. Now we're basically finished here. The last thing we could do is just go ahead and trying to clean up our answer a little bit to make it look a little nicer. So here you can write. This is to square, or if you'd like to the four over, too. And then go ahead and combine these two exponents. That's two to the seven over too. And we still have the three in the bottom. Similarly, over here we have to to the one or flee want two to the two divided by two. So combining the two over two and five over two and then here combining these fractions, we'LL end up with fourteen over fifteen. Such is combining these here get a common denominator. Also here we should get a come denominator as well. So let's multiply top and bottom by five top and bottom here by three. So then we'LL have five of these up here, but then we're subtracting three of them. So that will leave us with two times through the seven halfs. Once again, I'll simplify the exponents so that it's two to the nine halfs And then we have minus fourteen from this and all divided by fifteen. What? And that's your final answer

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Top Calculus 2 / BC Educators
Heather Zimmers

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Baylor University

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University of Nottingham

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Boston College

Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
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