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Sketch the area represented by the given definite integral.Integral in Exercise 7

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.

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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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Sketch the area represente…

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Sketch the region whose ar…

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Express the area of the re…

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Evaluate the definite inte…

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Sketch the area correspond…

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Evaluating a Definite Inte…

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So in this problem, they're not asking you to evaluate. They just want you to understand what's happening in the problem. Uh, and what's happening is first of all, the T is represented D T. That's your independent variable, sensibly X axis. It's the T axis, and we can keep This is the y axis. It could. It could actually be anything, but I'm going to write it down as if it's y equals y equals the square to t. So that's the first thing is to understand the square function. Well, when t is zero, you get the answer for y zero. When t is one, we get a Y value of one, 234 When t is four, the square to four is too, you know, and we can plot all these points. The next special one is actually nine. So 56789 square to nine is three. But we could actually graph where you can use a graphing calculator to verify that. That's true. So what the integral does is it finds the area under the curve smashed with this T axis. In this case, it's a T axis. Now where do you start, and that is the bounds that is established from this number and this number. So those r T equals four m t equals nine. Those are your vertical lines, their vertical because it's the T axis instead of the X axis. So we're starting at T equals four all the way over two t equals nine. Send that over. Um, And so what we're doing is we're finding the area under the curve from there. It's just like an infinite number of tiny rectangles, uh, that are very, very skinny. So if you look at this, it makes sense that the answer is bigger than it looks like. It's five by two. So bigger than 10. But then smaller than 15. If we worked it out, you could actually see exactly what it is. Um, yeah, but this is all you have to do. This graph

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