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Evaluate the given definite integral.(a) $\int_{2}^{3} 3 x d x$

(a) $15 / 2$(b) $-15 / 2$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

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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So as we look at this problem, we're evaluating the integral from 2 to 3 of three x dx. That's just for Port 80. And for part B, you were doing the same thing. Just we change the monitor two or three x dx. So, as we are doing the anti derivative, we add one to our experiments. So explain. Here's one. But then we need to divide by that new exponents of three halves from 2 to 3. Now, here's the thing is the anti drift. It is going to be the same. No matter what. The only difference on part B is we change the balance around. So how does that affect our answer? Well, as we plug in our balance, So when you put equals in here, girls Yeah. Yes. And why they didn't show up? Um mhm. You know, three squared is nine and then minus three has two square to be four. Um, now, quick math would say to factor out the three halves. And really, I could have done that from the get go, because I could have put this in parentheses and brought the three in front. But anyway, uh, so what? I'm left with is three halves times five and you can multiply that five into the numerator. So that first answer for part a three times five is 15 15. And that, too, is still in the denominator. So there's your answer to part a, um, the only thing that's different for Part B if you go back to it, is all that work is the same. Not even factor out the three halves first, um, two squares four three squared is nine. What we end up with is a negative size. So how does change in the bounds affect our answer for a letter B? Uh, it just makes it a negative 15 halfs. So what we're supposed to learn from this is that if we just flipped up our bounds, we negate the answer. So it goes from positive. 15 have two mega 15 minutes, and that is true

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