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

$$\frac{e^{2}-5 e^{-2}}{4}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 9

Two Integration Techniques

Integrals

Missouri State University

Campbell University

Oregon State University

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 given integra…

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Evaluate the definite inte…

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evaluate the definite inte…

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Evaluate the integral.…

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evaluate Integral

Our task is to find the integral from 1 to 2 of E to the two x over e to the two x minus one d x. So we're using new substitution, and I was strategically let u equal the denominator. And the reason why that's to benefit is because then the derivative of that is using the changing role to e to the two x dx. Now, what I always do is because I've had the most success doing this is dividing that one over to E to the two X over Um, do you? So then when I go to rewrite the interval, I just leave that to the two x alone right there. I rewrite the denominators you because they're equal to each other on then I rewrite DX as one over to E to the two x time to do you. And that way my students can actually see the Either the two exes will cancel. You also need to change your bounds. Just you know, you have to plug in for in terms of X so e to the two times one power. That's just he squared minus one and then you have to plug into a swell two times two is four. So we're looking at E to the fourth minus one. So as far as the anti derivative of this goes, you should know that you have one half natural log of the four. That kind of looks like you natural log of you from these two values. He squared minus one into the fourth minus one. And from here is just a little bit of algebra manipulation. I don't know. Uh, maybe even factor out that one half will mess with that at the end. Natural log of E to the fourth minus one minus natural of plugging in your lower round. Oh, and because we're subtracting two different logs, we can combine them as a single log using division eight of the fourth minus one over. He squared minus one. Oh, my goodness. I think some students might stop right there. I don't know how many of you would recognize this, but you can actually do a difference of squares in the numerator either. The fourth minus one is the same thing. All right. This in blue as e squared times e squared, dizzy to the fourth one times. Negative one gives you negative one. And if you really distribute this out, you'll get the outside and inside terms of canceled. So by doing that, you know, rewrite this blue by doing that. And factoring that out is e squared plus one e squared minus one. You can actually cancel out. I don't know how many of you would think about doing this. E squares minus one. So the best answer as one half natural log of just e squared plus one. Which is an interesting answer, sure.

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