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Express the limit as a definite integral.

$ \displaystyle \lim_{n \to \infty} \sum^{n}_{i = 1} \frac{i^4}{n^5} $ [$ Hint: $ Consider $ f(x) = x^4 $.]

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00:52

Frank Lin

00:40

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Missouri State University

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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Shown below.

writing this limit here. This limited some as a definite and girl. The first thing we could notices. We could rearrange this expression here. So let me just rewrite that as I to the four and keep end to the fourth as well. What? And then you have another factor of one over end because that one of Rennes RDX term and then we could even rewrite the to the fourth to the fourth as all toe the fourth there. And now we're really ready to identify what this limit some is equal to as a definite and a girl. So the one over N is equivalent to the width of the trek tangle. So that is a D X turn. The reason why we know the limit is from zero upto one is the definition of Delta X. So Delta X is equal to the minus. I all over. And and in this case, if you just end up with one over end, that means be an A only differ by one. And furthermore, there's nothing in front, um, added to it. So a must equal zero, so be must equal one. So that's how we know that those or zero in one. And then lastly, we're just left with this expression here, Um, which is this multiplied by I so would be won over end times I or I over end, and that is our X value. So that would just be X to the fourth there. And therefore, this is the rewritten form of what we had up top.

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