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Evaluate the given definite integral.$$\text { (a) } \int_{1}^{2} \frac{1}{x} d x$$$$\text { (b) } \int_{5}^{10} \frac{1}{x} d x$$$$\text { (c) } \int_{6}^{12} \frac{1}{x} d x$$

(a) $\ln 2$(b) $\ln 2$(c) $\ln 2$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

University of Nottingham

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 Integral

This is kind of a film problem, and you'll see why when you get to the end as we're looking at the interval from 1 to 2 of one of our X DX, uh, same thing would let her be. We're doing the inner role. It's the same function. But now it's 5 to 10 one of our X DX, and then same thing with C is just different. About 6 to 12 of one over X DX. Well, no matter what you do, the integral is still the same for each of these functions and that integral is natural law, the anti derivative of one of our X's natural law events. Now I'm putting an absolute value just because we can only allowed positives. But it actually doesn't matter here, because we only do with positive numbers. So that goes from 1 to 2 same thing with this one natural log of X from 5 to 10 and same thing with like a C. We're looking at natural log effects from 6 to 12. But when we go to evaluate this when we plug in your bounce, what we get is natural log of two minus natural log of one. Well, what I would do is, uh, because all I want all of these to be the same is when you're subtracting with logs, you can put them together using division so two divided by one. Would you just reduce the natural log of two? That's your correct answer for a So to try to save you some time on the next one. We're looking at natural law of 10 minus natural. Log of five. Well, again, we have subtraction in here. So we're looking at natural log of dividing those two pieces and 10 divided by five is to. So I think you can tell the pattern going on here is that this last answer is the same thing as natural log of 12 divided by six or natural log of two. So they're all the same Answer.

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