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Evaluate the given definite integral.$$\int_{0}^{1} e^{t} d t$$

$$e-1$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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 definite inte…

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Evaluate the given definit…

So we're looking at the integral from 0 to 1 of e to the t d t double check, and I got that right. Um and the whole premise of this is that you recognize that the derivative of E to the T is e to the T. So it makes sense that if we're working backwards, the anti derivative of E to the T is also each of the T. V. So with that in mind, it makes perfect sense that you know what I have in greens derivative is this. So I'm doing this correctly, and then you have your balance from 0 to 1. There's not a whole lot else to do because all you can do is plug in your balance sheet of the first and e to the zero plug in your bounds. And don't forget to subtract. Now, I would also ask my students to simplify that in the sense that anything to the first power is itself. So it's just e 2.718 and e to the zero power is equal to one. Yeah, and just leave your answer like that. But as I kind of mentioned that if you know E is equal at 2.718 You know, if you have that memorized, it makes sense that 2.718 minus one would be 1.718 And I don't think there's no evidence of me using a calculator there. I just had it memorized.

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