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Consider the graph defined by $f(x)=a x^{n},$ with $a>0$ and $n \neq-1 .$ When $x=b,$ call the $y$ -coordinate $h,$ where $b>0 .$ Show that the area bounded by the graph and the $x$ -axis between $x=0$ and $x=b$ is $A=\frac{1}{n+1} b h$.

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

University of Nottingham

Idaho State 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.

02:06

Sketch the graph, and find…

02:04

02:59

Consider the parabola $f(x…

01:57

01:39

01:32

Okay, so there's a lot of tidbits in this problem where they we have a X to the end power. We're finding the area. They tell you that a a is positive. Um, So the reason why that's important is so that you recognize that we are going to be above the X axis now and can be a negative number, so you might have a graph that looks like that. Um, I yeah, I think you want to stay away from n being a negative number. That's why they said and can't be negative one. But, um, anyway, the what the premise, though, is that when x s b you get a y coordinate, that is H is what they call it, which means that if you were to set this into an equation, what they're trying to get at is that a plugging be in for X to the power will be equal to H. And I'm going to use that fact as I do this problem. So we're doing the integral and they tell you that bounce as well, from zero to be of this function D x. So if you do the anti derivative, then you add one to that exponents and then you divide by that new experience. Oops. Um, actually, what I'll do is multiply by the reciprocal. So I'm going to rewrite that as one over and minus one. I'm gonna leave that alone, but I'm gonna plug in B and for X to the n minus one power. And I'm also gonna plug in zero in for X. But if you plug in 00 as long as N is not, um yes, I need to be careful that if n is zero, which, uh, anyway, I'm going to assume that they meant that that's going to be 00 to any power would be zero. And subtracting zero is not going to change the value of it. So what I would do is actually rewrite this problem as one over and minus one times a And instead of writing this out, I'll write it as B to the end. Power over be The reason why I like to write it that way is then I can establish that a times B to the end. This is an internal your mother, that bees and economic I hope you realize that that's equal to B to the N minus one. If you follow your rules of exponents you divided with the same base you subtract your experience. So what I just circled in blue is what I had earlier is equal to H. So what I have is I have that one over. N minus one. I have this one over B, but then that's a B to the power is equal to H as we established earlier. Um oh, rats. I realized I made a mistake. I subtracted. What I was supposed to do is add one to my experience and divided by the new experiment. Here we go. So that's plus one. That's plus one that's plus one. So this should have been times B. This should have been times be, uh, remember, if you're multiplying with the same base, you add the excellence. So that was a true statement as long as I did it correctly to begin with. And everything I said was correct. Except the final answer that they want you to write is I have in the right spot. Just I had the wrong answer that B is in front of the age because multiplication is communicative. and there we have it. But really, all of the work from the get go is necessary for this problem, okay?

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