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Rroblem 3 (part a) Priv e +Rat 0 Sr (h+3x) dx 7 R,d5 Por all n > 0 (Fpts ) X) (pat 62. By mAan$ 0 P +Re karul 6at a), Trave tRatSindxh . Ti Raldc for all h> 0...

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Rroblem 3 (part a) Priv e +Rat 0 Sr (h+3x) dx 7 R,d5 Por all n > 0 (Fpts ) X) (pat 62. By mAan$ 0 P +Re karul 6at a), Trave tRatSindxh . Ti Raldc for all h> 0 . (AmSr (X)

Rroblem 3 (part a) Priv e +Rat 0 Sr (h+3x) dx 7 R,d5 Por all n > 0 (Fpts ) X) (pat 62. By mAan$ 0 P +Re karul 6at a), Trave tRat Sin dx h . Ti Raldc for all h> 0 . (Am Sr (X)



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Given $\mathrm{f}(\mathrm{x})=\mathrm{e}^{2 \mathrm{x}}$ and $\mathrm{g}(\mathrm{x})=\sin ^{-1} \mathrm{x}$ Column I Column II (a) $\int_{1}^{0}\left(\mathrm{f} \circ \mathrm{f}^{-1}\right)(\mathrm{x}) \mathrm{f}^{-1}(\mathrm{x}) \mathrm{dx}$ (p) $\frac{\mathrm{f}\left(\frac{\pi}{2}\right)-\mathrm{f}(0)}{2}$ (b) $\int_{0}^{1}(\mathrm{f} \circ \mathrm{g})(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x}) \mathrm{dx}$ (q) $\frac{1}{4}[\mathrm{f}(-3 / 2)-\mathrm{f}(-1)]$ (c) $\int_{0}^{\pi / 2} \mathrm{f}(\mathrm{x}) \mathrm{g}^{-1}(\mathrm{x}) \mathrm{dx}$ (r) $\frac{\mathrm{f}(1)+\mathrm{f}(0)}{8}$ (d) $\int_{-1 / 2}^{-1} x f(x) d x=$ (s) $\frac{2 \mathrm{f}\left(\frac{\pi}{2}\right)+\mathrm{f}(0)}{5}$

Hello. My name is Butler. And in this video I'm sure how to find the iterated integral of this function here. So here for X -3 more applied by sign Y. This means that it's a function of X and Y. Because you see we have two variables here, the X and Y. We also have to integral so have the inner one inside the three brackets from 0 to 3. And we have dx we also have from zero to pi over three. We have a DY so the approach you wanna use? Yes firstly we integrate this in a part with respect to X and dx. And then lastly we integrate Dis occupied from zero two pi over three with respect to Y. Hence the DY. So answering this like I said firstly you're gonna start with the integral From 0 to 3. And what do we have inside? We have four x minus three and we have the same way and it is with respect to X. Now integrating this, like I said before the Dy means sorry, the dX means it is with respect to X. But we also have some wine here. So how do we treat this son? Y. Everything Which is not an X. Which is as a constant meaning sign, Y is a constant. So how do we treat constants? When we are integrating? We know that you can factor out the constant outside the integral sign. So that's what I did here. We're from Sarah 23. It was left inside four X -3. Great. Now progressing further, we rate our constant here. It is which is saying why And yeah, was the integral of four X. You know that it becomes four X to the power to over two and then before and the two council so that we have to expect. So you have to expect yet. And the integral of three. We know that it becomes three X. And then we have boundaries from Syria 23 great. Now progressing for that, you still have a constant again the same right? And plugging in these boundaries. So we're gonna have to and then we have or Upper boundary which is three. And then You will use three. Also, I won't write this zero because if you say two more purposes or to still zero and he also still zero. So don't try this every day. Simplifying this, you will have the same way. Yes. Before then. Yeah. So here we have three to the part two which is nine, Two times 9, -3 x three, which is nine. So we found that we find that end result becomes nine. So this sign wine applied by nine is the inner integral of what we have here on top. So what are we left with? We now need to find this out, integral with respect to why? So progress further. We need the integral from zero. Remember it was from up to by over three. And now what we have we have nine saying why. And remember now it is with respect to y now progressing with this. So the nine were treated as a constant like we did before. So we affected outside in the cell, We're all pi over three. Then we have same way inside here. Now was the integral of sine Y. You know that the integral of sine Y. It's minus course. So here way of minus. Of course. Why we have our boundaries zero two pi over three over three here. No progressing with this. Now what we need to do is to plug in our boundaries. So you have minus course. What's our first boundary? It's pi over three and then we're not minus. And then what so what are the boundary? That's fine as close zoo. Great. Now progressing with this year we have on nine again it's a constant. Yeah. So what's the value? Of course by over three we know that the value of Of course pay over three. It's just half. So we're gonna have a negative half year. Do you have a negative? And here was the value of course zero. Know that it's one. So we're gonna have a negative one. Great. Progressing favor here will see that it's here with the name and here we have minus half. He is negative and negative makes it a positive. So we have a positive one. And then now you can easily see that negative half plus one becomes half. So we're going to have nine over to. So this is the iterated integral of the function we hit above. So this was the function yet four X -3. Same Y, which again is a function of X&Y. Thank you for your time.

Yeah. In this problem we wish to evaluate the iterative integral. The integral from zero to pi over two integral 0 to 2. Absolutely the X and Y. Dx Dy This question is challenging understanding of integration in particular, is challenging our understanding of iterative intervals to solve. We must use single variable integration techniques. Step by step in step one we evaluate the integral that's integral zero to exceed the X and Y. The X. Instead to we played the result of step one into the integrated out of integral integral zero to pi over two. Dy and unsolved. So following these steps we first have are integral evaluating as integral zero to pi over to sign Y E to the X X minus X from zero to do I. This anti directors have found from integration by parts plugging inbound 0 to 2 gives integral zero to pi over to sign Y times X squared plus one. Dy this evaluate this anti rated E squared plus one times negative. Close N Y from zero to pi over two. Plugging in our bounds give solution E squared plus one times zero plus one or e square plus one

Hello. My name is Butler. And in this video I'll show out find iterated integral of this function here. So we have X cause Y. And here you see that we have an integral inside the square brackets which is from zero to pi over three. And then we have a DY which means the test with respect to Y. We also have the order part which is from 0-2 with the DX Michigan cities with respect to X. So how do you approach this firstly? We integrate this inner part and say the square brackets. And then lastly We integrate this outer part from 0 to 2 with respect to X. Because we have a Dx there. So answering the question that they said We're going to start with the 90 girl, this is from zero two Pi over three. And inside we have X. Course Y. Yes. Mm. Yeah of course. Why? And then it is with respect to Y and stick DY. So here you see that we don't only have why we also have X. Here but then we're integrating with us for two. Why? Because we have a dy so how do we treat this X. Here? This X. We're gonna treat it is constant because it's not a Why? So whatever we have in the year, it's not a why we treat it as a constant. So how do we deal with constance when you're integrating, you can figure them out outside the into your sign and then we are integration zero pi over three. And what's left inside? It's just cause why? Great. Now progressing with this, we have our course here. Sorry, our X. Which is a constant. What's the integral? Of course, Why we know that. The integral. Of course. Why is sign? Why? So we're going to have same way here And we have our boundaries from zero two Pi over three. Great. Now, what are we left with now? We need to plug in our boundaries. You still have a constant year, which is X. And replacing this. We know that here. We're going to have slang pi over three. Apologies. by over three minus putting the lower boundary which is signed to you, visit you here. Mhm. Great. Now how do you progress further history about constant X here? And what? Sign? Pay over 3? We know that sign pay over three brings us Square root of three over to. And what sign? 090 is just zero. So I won't write when a zero yet. So this means that this is the inner integral of what we have here. So what are we left with is now to find this out outside integral from 0-2 with respect to X. So we come here, we're now into greed from zero two, what are integrating skirt of three over to and we have our X here and it is with respect to X. Now, as we did before, our square root of 3/2 is a constant. So we can put it outside the integral sign and was left here. You're left with X. Great. Now progressing further was the integral of X. You know that the integral of X is just next to the port to over to. Then we have our boundaries from zero two. Great. Now progressing further, we are now we now need to plug in our boundaries. So here we have third of three. So here what you can do Our two is a constant. So we can just put it outside when you plug in our boundaries. So is to mean that I can have three over floor here and then with our daughter. And then now we're plugging our boundaries. So we're gonna plug too. So yeah we're gonna have to which is to to the part two which becomes four. If we plug in zero you're just gonna have zero. So I'm not gonna write you just gonna hold me right this to which we found that to do the power to became four. So you have four here. Now you can see that the force can cancel each other. And then we're left with On the Spirit of three. So this is the iterated integral. The function will give up up which was dysfunction here. She was X. Cause Y. So remember this is a function of X. And Y. Then iterated integral Square root of three. Thank you for your time.

Uh huh. In this problem we wish to evaluate the given entering an integral integral from 0 to 1 of the integral from zero to pi over two of x E Dfc and wide idea. This question is child you understand the integration particularly multi very integration. Remember that to evaluate able to integral. We use single variable integration techniques. Step by step and step one. We evaluate our inner integral here. That's integral zero to pi over two X. The X and Y. Dy. And step to plug in the results of step one into our outer integral here. That's integral 0 to 1 the X. Then we solve. So proceeding to step one are integral evaluates as integral 01 X E X negative press Y from 0.2 Dx. Plugging in our balance for inter integral gives are integral, simplified as integral 0 to 1 X E x times one dx. Thus the integration by parts are integral evaluates as E x times x minus one. From 0 to 1. This is either the one minus one minus 0 to 0 minus one, the first term canceled, and the solution is simply negative. Zero minus one equals positive one.


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