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Evaluate the given definite integral.$$\int_{-1}^{1}|3 t-2| d t$$

$$13 / 3$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

Campbell University

Baylor 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 definite inte…

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Evaluate the given definit…

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Evaluate the integral.…

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so there's a couple ways of doing this problem, and I want to show you what I think is the easiest. But there might be a better option out there, but the main thing is the function three T minus two. Because it's the absolute value. Must be always positive. So if you were to try and graph, um, you know, even if you plugged in value strategy like negative one, I'm just starting with the first bound. Um, so you have negative three. Minus two is negative five, but absolute value would shoot that up too. Positive. Five. Um, And then as that graph goes to zero, you're going to hit zero or sorry. Hit positive. 20 minus two is negative two and then absolute value. Now, when you get to one, Um, okay, well, three times one is three minus two is one. Uh, and what's happening is when you get to let me set this equal to zero for a second, um, that function equals zero. Add to over. So three t will equal to and then divide by three at two thirds. That function bounces off of the X axis and goes back up. So to speak kind of looks like a check mark. So if you understand the problem, then you can see that you have two triangles. One triangle has a width of Well, if I go from negative one to positive two thirds, that's why solve this right here. Um, then So if I add one to it, which would be three thirds, uh, this base is five thirds with a height. That's five. Whereas let's do this other one in blue. Over here, we have a base that's one third with a height of one. So I found the area of each triangle. Let's do the right one first would be one half times the base times the height. That's the area of a of a triangle, one based on tight. So we're looking at 25 6 and then let's do the blue 11 half times one third the base times height and we're looking at 16 there, and we can add those two areas together because they're above the X axis. Everywhere we're looking at 26 6 or if you divide top and bottom by two, we're looking at 13 3rd. You're correct. Answer

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