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Evaluate the given integral.$$\int_{0}^{1} 2 x e^{-3 x} d x$$

$$\frac{2\left(1-4 e^{-3}\right)}{9}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 9

Two Integration Techniques

Integrals

Missouri State University

Baylor 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 integral.…

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Evaluate the integrals.

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04:54

Evaluate the integral.

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Here. I want to work out the integral between zero and 1 of two X. E. Power minus three X. Dx. And this one is a part question. So what I would do is I will take this as the U. And this has the T. V. By dx I'm working out from there. So it becomes the parts formula you two X. The integrate E power negative three X. I will get negative one third. The negative three X. Again think backwards from differentiation That will give me -3 times that function Or cancel out with -1 3rd. So minus the integral of negative one third e minus three X. Which is found Times differential of 2X. Which is to The X. and limits. I'll put over here like this 10 next step a bit simpler. So it's minus two thirds of X. E. To the minus three X. Plus Take out a factor of 2/3 integral of E. to the -3 x. The x Limits of one and 0. So we end up with minus two thirds X. E minus three X plus two thirds. If I integrate that Again negative 1 3rd the To the -3 x. And again 10. So minus two thirds X. E minus three X -2/9. He -3 X. With the limits of one and zero. Okay that's plugging the 1 1st So I get -2/3 E. to the -3 minus two nights. Eat -3 Plug in the zero and the first term goes away. So we have zero and we have minus just two nights. He power zero being one. So that becomes here, it's a minus 6, -2 nights, so negative eight over nine E. to the -3 plus two nights. And that's fine for the answer.

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