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Evaluate the given integral and check your answer.$$\int \sqrt{2 x} d x$$

$$\frac{2 \sqrt{2}}{3} x^{3 / 2}+c$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 1

Anti differentiation - Integration

Integrals

Missouri State University

University of Nottingham

Idaho 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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Evaluate the given integra…

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Evaluate the integral.…

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Find the Integral

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Evaluate the indefinite in…

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answer please

I'm not trying to over explain this problem, but I feel like some students need a step by step process, so that's why I'm going to show you. And sorry if this video is ridiculously long. But what I like to point out to students is that that's the same thing. Is the square root of two Times Square defects like we can split up a radical? It's two different products. The reason why I would do that is now I realized that the square to to is just a rational. I'm sorry. It's just a constant. I would rewrite X to the one half power, and that's a rational exponents. So from there I could do the integral by adding one to the exponents and then multiply by the reciprocal of that new exponents. Eso We're ready for the final answer because if I add 1 to 1 half, I get three halves and then I need to multiply this constant this route to by the reciprocal of three have. So that flips it, Um, and it's true that the numerator eyes being multiplied, you know, the two goes into the numerator multiplied by this constant. Don't forget about plus C on DWhite. Do we need a plus? C is when we do the derivative of this to check our work, we shouldn't get this answer on. That's why we need a constant is because we don't know if if there was a constant because the drift of of a constant zero we don't write down plus serum. But it is true that when I do, the derivative of the three has brought in front what can solve the three and the to your left with route to X to the one half power once you subtract one from that next moment. So that's to check your work that my work in green is correct and it iss

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