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Evaluate the given definite integral.$$\int_{-1}^{2}\left(3 x^{2}-2 x+4\right) d x$$

$$11 / 2$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Oregon State University

Idaho State University

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 given definit…

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Evaluate the definite inte…

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Evaluate the integral.…

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all right. This problem is doing the integral want to of this function. To execute minus three X squared plus four X minus one the X and basically all we need to do is the anti directive of each of these where you add one to your exponent. Divide by your new exponent to divide by four reduces to one half, um, adding one to your experiment, but three divide by three is one. Um, add one to your experiment four divided by two is two and then minus one X. I don't know if you need to see that one there, and we're going from oops from one to choose. So now you have to plug in your bounds. And first I'm plugging, too. And for all these values and what I see versus one of the choose to cancel. So I'm looking at two cubed minus two cube. Those two things are the same. So little cancel two squared is four times two is eight minus two. And now I need to plug in, um, my lower bound, which is one. What's nice about plugging in one is one too many. Power is just one, so I just have to look at the coefficients of all books. And as I'm looking at that, what do I see is I see negative one plus two minus one that will all cancel each other out. So I have six minus one half. So 12 halves is the same as six. Minus one half is 11 halves is your correct answer. We're just throwing some equals here so you can see my step by step process.

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