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Determine the area of the region bounded by the given curves.The region bounded by $f(x)=x e^{-x^{2}}$ and the $x$ -axis, between $x=-1$ and 1.

$$1-e^{-1}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Missouri State University

Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

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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Determine the area of the …

05:30

Find the area of the follo…

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Find the area of the regio…

08:21

Okay, so there's several words in this problem. But if you found in the area under the curve of X E to the negative, uh, X squared, it's basically saying doing the integral and what you'll notice is the vertical lines that they give you Mexico's native on to expose positive one are your bounce. So you're still doing U substitution. I like to use the exponent in the exponential because I know the anti directive of the heat of the U is either the u. So then the derivative of that is named two x d. X. So then I would solve for DX by dividing that negative two x over. The reason why that helps me is now I can look at this problem as the integral I like to leave X there, and then I can replace, uh, the negative x squared with you because that's what the equals. And now I can replace D s with that one over negative two x d. U and my students will actually see the X's cancel, um, and then plug in your bounce like negative one when you square. It is positive one, but it's still negative. One. There and that one When you square it, uh, that's still negative. One, uh, means I wonder if I did something different. Well, if your bounds are the same, you should get an answer. Where the area is equal to zero just feels incorrect. Thank you. X e to the New York's work. I'm going to commit to that answer that since the bounds are now the same, um, that our answer is going to be zero.

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