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Auextior 10baineAntotIF the following is reduction fommula J (In x)" dx (In x)" (In x)"(where n-7) then (a 5)*3...

Question

Auextior 10baineAntotIF the following is reduction fommula J (In x)" dx (In x)" (In x)"(where n-7) then (a 5)*3

Auextior 10 baine Antot IF the following is reduction fommula J (In x)" dx (In x)" (In x)" (where n-7) then (a 5)*3



Answers

Applying reduction formulas Use the reduction formulas in Exercises $44-47$ to evaluate the following integrals. $$\int x^{2} \cos 5 x d x$$

Good question About 10. We're gonna find the value of this definite integral using fundamental serum off calculus. It stays that if we can find entered derivative, then we can place the upper end of elements. So let's rewrite this. And in two parts, the first part is 10 times X rays to about 90 X in the second part is from minus two toe Okay, three X rays to the ball. Five dx 10 comes outside. X rays to the power 90 x integrated within the limits minus two and two. He had three comes outside. We have X rays to the power five d X again within the limits minus two and two because 10 and three are constants. Now we can find the inter derivative here. The integrative derivative would be X rays to the part 10 or 10 the limits off minus two and two. And here it would be X rays to the past six or six within the limits of minus two into again, the 10 gets cancer. If you place the parliament, we have to reach to the port and minus lower limiters minus two times X rays to the portend plus three times to a six. So we have one or two outside. If you place the A parliament, we have to raise to the past six minus. If you place the lower limit, we have minus two, uh, raised to the power six. So to raise the port and minus minus off and even powers plus so these two terms will cancel out. And likewise thes two terms will cancel out. So we had just left with zero, which is the final lungs.

Section 72 Problem 48 were asked to use the reduction formula that was proven in a couple of exercises earlier. Ah, here's the reduction formula. X to the end e to the A X DX is equal to, um one over a x to the n e to the a X um minus in over a the integral x to the n minus one e to the a x t X. So in order to use this reduction formula, I'm going to use it at first with in equal to and a equal three. So that gives me the integral of X squared you to the three eggs is going to be so one over a That's 1/3 and then x squared e to the three x minus the integral, and this is gonna be in over a That's 2/3 X to the N minus one. So that's gonna be X E to the three x dx. Now I do the same formula, but now in is equal to one in a is equal to three. So I just repeat the same process. And now what I end up with is 1/3 x squared e to the three X minus 2/3. And now you have, what, one over a So you've got 1/3 extra, the one e to the three X and then minus in over a That's 1/3. And then you've got X to the zero, which is one. So either the three x dx And now I can evaluate the integral and be done with everything. So this gives me 1/3 x squared e to the three x minus two nights X e to the three X and then plus to nineths and then the integral of eat of the three X as he to the three x I was 1/3 plus a constant of integration. Yeah, and so what? This gives me ISS 1/3 x squared e to the three x minus two nights X e to the three x plus 2 27th you to the three x plus a constant of integration. You can write this in another form. You see that I could factor out Let's just see 1 27th e to the three x and then that would leave me with nine x squared minus six x plus two plus a constant of integration

Question about 41. We gotta find the value of this, uh, this definite articles using the fundamental to Europe off calculus. Uh, eso for that when you first need to finance center derivative. So let's first of all, breakers in two parts the first part of three x dx and the second part is minus 2098 days to the pole tree X Uh, this can be related industries taken outside. Uh, this will be X dx within the limits of minus 2 to 0 on here, nine is taken outside. It is a constant it is to the party X dx within the limits off minus 2 to 0. Let's find the inter derivative of it is to the poor three x separately. So we substitute three Xs p The SMEs start three d excess. DT. This means start the excess d t o three. So the inter derivative comes out as each is to the poverty DT over three s. Oh, this comes out of one of the three is taken outside integral, obvious to the poverty. Easy it is to the party where t is replaced by three X so they're integral over here becomes three times this will be extra square were to within the limits of minus 2 to 0. Here the integral, which we just found as 1/3 times he raised to the power three X again within the limits off minus 2 to 0. Uh, so this comes out as three over to this will be zero minus minus two square would be four minus 39 or three is three. You place the parliament. This is it is to the ball zero. And if you place the lower limit, this is it is to the power minus six. So this comes orders. Two times two is four. That the minus sign and three times two is six. So we have minus six year minus three times. This is one minus CDs to the poor, minus six thistles minus six minus three plus three times CDs to the Paul minus six. On the final answer comes orders. Three times it is to the Paul minus six minus nine. So this is the final answer

Question about 37. We need to find the value of this definite integral using the fundamental to remove calculus one, uh, eso for that. We first got to find the inter derivative and enter derivative of D X, where X is pretty straightforward. There's Ln X. We placed the limits from to Putin. The upper limit is Ellen 10 minus the lower limit is five on. We have a property of log that log a minus log. These log away will be. So we have Ellen 10 over to which is nothing but Ellen five, which is the final answer.


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