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

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

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

Campbell University

Oregon State University

Baylor University

Lectures

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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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Evaluate the definite inte…

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Evaluate the given integra…

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Evaluate the given definit…

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So we're looking at the interval from 1 to 2 of this function. Eat the negative to t d. T. Now, if I were doing this, I would not use u substitution. Um, just because I would think about how the derivative of E to the T is equal to either the tea. But if you go backwards, you have to consider the chain rule. You know, if I had I don't know what constant right here. So the derivative of and e the C t would be equal to fix that. Like I wrote, it would be e to the c t times the derivative of C t, which we see. Um, so if you're working backwards, it's almost like you have to divide by that piece. Um, So let me just give you a guess as to what the correct answer is it be divided by or multiply by negative one half, and you can verify that this is the correct anti derivative. Because if I did the directive of this, I need to get this to be the same. Well, if you did that, you'd have either the negative to t times negative two. And now we cancel out with this negative one half. So it is correct. And then from here, I can just plug in my bounce. Now, I would also just leave that negative one half factored out. So I'll leave as negative one half and then plug in eating the negative for power because you're plugging into. And then he minus e to the negative second power. Because when I plug in 1 82 times, one is negative two. Now, as I'm looking at the answer, what they actually do is they sort of distribute this in, but instead of writing it So instead of writing that times one half, they divided everything by two. But then you're dividing this negative in there. So it makes a negative e to the negative fourth over here, and this turns into a positive e to the negative second power. And hopefully I've explained sufficiently for you

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