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

$$\frac{e^{2}-e^{-2}}{6}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Missouri State University

Campbell University

University of Nottingham

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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01:03

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02:04

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04:43

02:31

Evaluate the integral.

02:23

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01:37

03:30

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s so we're looking at the interval from negative one toe one of this function X squared e ast times the native to execute d X. So if we use u substitution, which is what I like to do, I just let you equal the exponents in this problem for a couple of reason is I know the derivative of either of you is either of you. But then this derivative is very close to canceling out that x squared. And I have ah, kind of a foolproof way of figuring that out. Um, at least in my experience, my students have to solve her DX and divide that native six x squared over. And so then you can visually see that this new interval I have my senses leave x squared alone. Replace e guess I'm leaving the yield on a swell, but replaced the native to x cubed with you and then replaced DX with the quantity of negative 1/6 x squared. Do you? The reason why I do that is now my students can actually see the X squared is divide out, cancel each other out. So now we're all in terms of you accept my bounds, which I need to fix by plugging in a negative one. Cute, negative one cubed is still negative. One time save two is positive too. If I call him one, Cute is positive. One time Save two is negative two. Now what I would probably do. I hope this does not confuse you as I would factor that Negative 16 in front. But I also take out the negative. And that's okay. As long as you switch these to bounce around so negative to goes to the bottom and then to goes to the top. Either the you do. Hopefully that makes sense. I am following all the rules of math. I'm not not making anything. So the anti derivative of either the EU is e to the U and then your balance air now from negative to to to. And so now when you plug in your bounds, uh, notice I because I factor out that 16 I'm just leaving in front and you subtract when you plug in your bounce. Uh, this answer should be good enough. But as I'm looking at, the answer to what they did is seven multiplied by 16 they divided by six, which is equivalent divided by six of the same thing is multiplied by 16 So I'm gonna circle this answer just because it matches. But there's no reason why this is better than anything else.

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