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Determine the area of the indicated region.Region bounded by $f(x)=4 \mathrm{e}^{-2 x}$ and the $x$ -axis, between $x=-1$ and 1.

$$2\left(e^{2}-e^{-2}\right)$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State 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.

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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 …

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Find the area of the regio…

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So we have this area between the curve and the X axis. Um, so when you see a problem like this, you should automatically think Okay, it's going to be the integral of the function or eat the negative two x dx and just pay attention to the bounds. They tell you it's between X equals negative one and X equals positive one. And I like to actually write in from least to greatest. And you should. So, in the back of my mind, I'm just thinking how the derivatives of I'll just write E to some constant is equal to, uh that the chain role, which is that constant, would go in front of me to the c X. So I'm trying to work backwards, uh, and do the anti director, which is what I'm ready to do. I see that the exponent is the same no matter what, but I'm thinking about if I'm working backwards, I better take the concept that we have and divide by the concept that's in the experiment. So I'm looking at No, you just rewrite it for you. Negative. Two e to the negative two x from negative 1 to 1 and that should make sense, because if I take this derivative should equal the same thing is up here where the drift of Evita, the name to express itself and that in times the drift of the NATO's, which is native to times listening to his positive for So we're good, I'm just ready to Maybe what I would do is factor out that negative too. So it makes some math a little bit easier for me. I just have to plug in one in there. Now we've become e to the negative two power minus plug in negative one now. Well, negative. Two times negative One makes it positive, too. And, uh, I mean, I could leave my answer like this, but it looks like what they did is, uh if you factor out a negative, that would change that native to to be a positive, and then this term would turn into a positive. So we write it first, and then this term turns into a negative. So we right that second. But at this point, it's just preference. The boat that circled in green are correct

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