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Determine the area of the indicated region.Region bounded by $f(x)=\frac{3+2 x}{x}$ and the $x$ -axis, between $x=2$ and $x=4$.

$$3 \ln 2+4$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Baylor University

University of Michigan - Ann Arbor

Idaho State University

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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this area problem just involves doing a trick. But first of all, make sure you understand that when you find the area between the curve and the X axis that you are doing the anti derivative of the function which they give you as three plus two x all over X and then the other piece that they give you is that it's bounded between the X values of two and four. Now it's important that you do go in order from 2 to 4 because we want to make sure that we get the correct sign for our answer. In this case, it should be a positive. Before doing this anti derivative, I would make my students recognize that because we're dividing by a mono mule excess amount, you can just divide each piece. So three divided by X will be my first piece. Go back to this made with black, and then you can divide this and the exit would cancel and then be elected to Most students are really good, um, at combining fractions together to get this. But then they struggled to go backwards, and they don't realize that this is equal to things that you've learned in the past. So why would I require my sense to do that? Is because now you can easily see that that anti derivative is three natural log of X. Now I'm gonna put in absolute value. You don't actually need it because we're only dealing with positives. And then the anti derivative of two is to Lex. You can double check. That's correct by, um, taking the derivative of this the drift of three natural out of access three over X and the directive to access to 2 to 4. You may have heard me say it. I don't know if it made sense to you, but we do ask the value because we can only log positives. And that's what I meant by plugging in. And we're going to be logging positives. Any look at three natural log of four plus eight and then minus plugging in your tubes in here. Three natural log of two. Um, plus four. Now you can combine like terms. Um, I'm trying to think of how I would combine these. For instance, I can get these logs to be the same number because four is equal to two squared and one of the law of logs is that you can move that, too, in front. So what I'm saying is this is actually equal to six match Well, because three times two, let's say this is how I would do the problem, not saying it's perfect, but it does work. And then I can combine like terms and notice I'm distributing in here. And so six minus three would give me three natural out of two and eight minus four. Well, give me four and this is a perfect answer. There's no need to simplify any further. This is good enough.

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