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Represent the given sum (which represents a partition over the indicated interval), by a definite integral, and then evaluate the integral to determine the value of the sum.$$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} e^{2+\frac{k}{n}},[2,3]$$

$$\int_{2}^{3} e^{x} d x=e^{3}-e^{2}$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Campbell University

University of Nottingham

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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Represent the given sum (w…

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Definite Integral as the l…

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First recognize the given …

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Evaluate each limit by int…

for this problem, we are asked to represent the given some as a definite integral into evaluate where some is the limit as N approaches infinity of one over N. Of the some from K equals one up to n of each. The power of two plus K over N over the interval from 2 to 3. Now this um our term that we're going to be summing up over each car of two K or two plus K over and can recognize that when we start at K equals one, um, as we take that limit as N approaches infinity, our K equals one term is essentially going to be just eat the power of to. So that indicates to us that we'd be starting off the initial values eat starve to and as we take K up to N, then we'd be approaching E to the power of three. And if we look at exactly what's being summed up well, it's E to the power of something which indicates to us that our function is going to be E to the power of X. And we are going from 2 to 3. So our integral form is going to be the integral from 2 to 3 of each. The power of X. Dx And the anti derivative of E to the power of X is itself. So we get each power of x evaluated from 2-3. So our final result is going to be eat power three minus eat power of two.

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