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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.$$\lim _{n \rightarrow \infty} \frac{4}{n} \sum_{k=1}^{n}\left(2 x_{k}^{3}-3 x_{k}^{2}+1\right),[1,5]$$

$$\int_{1}^{5}\left(2 x^{3}-3 x^{2}+1\right) d x=192$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

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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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for this problem we are asked to represent the given some is a definite integral and then to evaluate the integral to begin we take the limit as N approaches infinity of four over N times the sum from cables one up to end of two Xk cubed minus three X k squared plus one and we're going from one up to five. Now just by looking at our some, we should be able to see that we can write this as our function being to X cubed minus three X squared plus one and we can see that we are going from one up to five. The distance between the endpoints is four, which is what gives us that coefficient of four out front. So that means that we can directly represent this as an integral as or this an integral form whether as the integral from one up to five of two X cubed minus three X squared plus one D X. Now the anti derivative of that we can find simply by integrating each term separately. So we'll get two times X power 4/4 minus three times X cubed over three plus X. Now we're evaluating from one up to five where I'll note that we can do some simplification that two times X power for over four comes X power 4/2 that X cubed or three X cubed over three becomes just execute. So now we just need to plug in our endpoints. So we'll have five to the power of 45 to the power of 4/2 minus five cubed. No division there plus five minus once that are of 4/2. So it's just going to be 1/2 minus one plus one. So that's going to be minus one half and simplifying that down, evaluating everything out. The final result should be 192.

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