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Evaluate the given definite integral.$$\int_{0}^{1}(3 x-2)^{4} d x$$

$$11 / 5$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

Idaho State University

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 definite inte…

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Evaluate the integral.…

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03:39

Evaluate the given definit…

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01:49

so I would do this problem using U Substitution, the girl comes here to one of three X minus two to the fourth Power DX. But you may not have learned u substitution yet. So what I'm showing you might not make a whole lot of sense, but I'll explain it in a second. So you let u equal this piece and then the derivative of that would be, uh, three d x. So what? That means one third d U. Is equal to of g X. Um, so what I'm talking about if you're using u substitution is we're talking about, like you to the fourth power and then replacing D X with one third do you and makes finding the anti directive easier because you add one to your experiment. So you do the fifth, and then you divide by your new explanation 1/15. But, uh, if you haven't done new substitution, this might not make a whole lot of sense. So let me show you what, uh what? Your answer is for the anti derivative is a three X minus two to the fifth power. I'm using this this tool of you. Uh, so I'm plugging three X minus two. Back in for you and then 1 15, um, from 0 to 1. And, uh, what I'm showing you is that you can methodically figure out what the derivative is by trying different values. Like replacing this with you. Um, so you can say Okay, Well, the anti derivative of U to the fourth is adding one to that exponent divide by that new exponent. But then also, if you did the chain rule, you'd have to take the derivative of the inside. Well, instead of multiplying by three, we better divide by three. So if you don't believe that this is correct, take the derivative of this and it should equal what we started with. Um, so at this point, we can evaluate by plugging in one. And if you notice plugging in one here, you get three times one, uh, which is three minus two is one to the fifth. Power is still one and then minus. When you plug in zero, you get negative. Zero minus two is negative. Two and negative. Two to the fifth power. We'll give you negative 32. And if you have 1/15 plus because of minus and negative turns into positive, uh, so one plus 32 gives me 33 15th. You're correct. Answer

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