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Find the values of $ p $ for which the integral converges and evaluate the integral for those values of $ p $.

$ \displaystyle \int_0^1 \frac{1}{x^p}\ dx $

$\int_{0}^{1} \frac{1}{x^{p}} d x$ converges when $p < 1$ and the value of the integral is $\frac{1}{1-p}$

04:36

Wen Z.

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 8

Improper Integrals

Integration Techniques

Missouri State University

Campbell University

Idaho State University

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Hello. Welcome to this lesson in this lesson. Well, look at the value of feed that makes the whole integral converge then would also evaluate the integral. Now, if the value of P it's less than one dummies, that p is negative. Oh, zero. So that's a That's a very high core belief. She dad, the whole of that X, the puppy goes up. Okay. Yeah. So that would make the integral converge. Let's look at So you are looking at P. That is less than one. So, for example, if we have, if you have a P, we have a p of negative mhm one. Then it means that we have And we the whole thing here becomes X to the power, Okay, on the other side. Or if you haven't peer to be negative too, then we have quote us extra panic, add to the whole thing here, reserves to us about two. Okay, So prefer api. That is less than one. Okay, if we also have an API that is zero, it means that the whole thing here is one. Okay, So, as you can see, the whole integral convert, F p s less than one. Okay, that is a path. So let's look at the integral again. Yeah. So write it. Us this. Whoa, the X So glad. Well, in evaluation, we I'd want to Negative b. And this becomes okay. Yeah, the same. You know, at the same power. Okay. From 0 to 1. So here we can write it very well as one minus p. Then all over one, minus p. Again from 0 to 1. So we put in one fast. So one minus one all over one minus. Yeah. Uh, the access will change. So we have a woman. Okay. The excess school changes. Oh. Okay. So one in place of the X. The second path, we have zero in place of the X. Yeah, little. Yeah. So this gives us right one minus p as a power to the one to the pile of this. The whole thing still becomes one. Okay, then you have one minus p. The whole of this pod is zero. So the integral from 0 to 1, then x right one over X p. Yes, it is a call to 1 1/1, minus p. All right. So if PS one let's it's only fine And if P is anything that is less than one, for example zero. You see that? The whole of this becomes one. Okay, so this is the end of the lesson. Thank short time.

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