Question
Ve 44} L J+ISvGive Dx Dumenc (Eull> 9 Ttouna R3-? 11
Ve 44} L J +ISv Give Dx Dumenc (Eull> 9 Ttouna R3-? 1 1
Answers
$19-44$ Evaluate the integral.
$$ \int_{1}^{2} \frac{v^{3}+3 v^{6}}{v^{4}} d v $$
Here we have the integral starting at zero. Going toe one of one plus one. Half you to before minus 2/5. You to the nine, do you? Let's go ahead and find the integral using. By undoing power rules, we have a U plus 1/10. You to the fifth, minus to over 50. You to the term quick from 01 Really? Here we can evaluate it one and also evaluated zero. And when we simplify, we get 53 over 50.
We have the integral from 1 to 9 x minus one, divided by skirt X dx. Now let's go ahead and write this out in terms of powers of X. So we fixed to the one half minus X to the negative one half DX. Now let's go ahead and undo powerful two thirds X to the three hats minus two X to the one half. Yeah. Now let's go ahead and evaluate at nine and let's go ahead and evaluate at one. Hmm. He This works out to be two thirds 27 minus two times three. Here we have two thirds minus two. So this is 12 minus and negative, 4/6 just equal to this or 38/3.
We have the integral of one over square it three to square it three of AIDS divided by one plus X squared DX. Now remember that the anti derivative of 1/1 plus X squared is 10. Adverse. Yeah. Now let's go ahead and evaluate tan inverse of these values remembering our were special triangles. So 10 inverse of square with three is 60 degrees or hi over three 10 inverse of one over a square three is going to be 30 degrees. That's, however, six. Okay, we have four pi or three.
So for this integral, what we're gonna do is we're just gonna split it up into three different integral Z. And then use this rule, I have written up in the top right here, which is kind of like the power role for derivatives, but instead it's a power role for integral. And so instead of the powerful for derivatives where u minus one to the power, we actually add one to the power and then we divide by that um power. And so we're gonna use this once we go ahead and split this integral up into the integral of this first term, just just one. So this is really just the integral of dx and then plus two times the integral from 1 to 3 of x. I'm just taking out the constant here. Another property of integral and then lastly minus four times the integral of X cubed times dx. And so we can see here that each of these integral is just going to be an integral of an X term. Aside from this first one, which is just the integral of dx which we know is just gonna be equal to X or go ahead and plug that in. And then for the second integral and is equal to one here. So we're going to have X. D n plus one or X squared divided by n plus one or divided by two. So this is going to be X squared divided by two and then this last one and is equal to three here we have x cubed. So we're going to have X to the three plus one, Power X the fourth power divided by four. And we're looking from 1-3, so this is equal to X plus X squared and then minus X. The fourth from 1 to 3. Um When X is equal to three we get three plus nine minus 3 to the fourth is 81, so we're going to have three plus nine minus 81 and then minus when X is equal to one, we get one plus one minus one. And so that's going to be 2 -1 which is just one. And so this is equal to 12 -81 and then -1 again, so -82. And so this would be equal to negative 70.