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Evaluate the integral_7/4 (sec(8) tan(8)) d82V 2...

Question

Evaluate the integral_7/4 (sec(8) tan(8)) d82V 2

Evaluate the integral_ 7/4 (sec(8) tan(8)) d8 2V 2



Answers

Evaluate the integral.

$ \displaystyle \int^4_0 2^s \,ds $

Okay, So, um and we look at this problem the integral from 227 of four DX. The answer makes perfect sense. If you look at the graph, you know you have a horizontal line at, like was for well supported. That's the equation. And your ex bounds are from two, 234567 So from 2 to 7 has within five. That's the birth of this rectangle in the height of four or five times four gives you an area of 20 so that should be your answer. Um, so let's actually work it out where you do the integral of this, which is the anti drift if you add one to your experiment. And if you need to think about this is actually zero power and makes sense, that's actually the first. You divide by any experiment from 2 to 7, and then you plug in your balance four times seven four times to plugging those balance in for X, and you subtract Well, now you're at 28 14, 7 minus 84 times to to give you exactly 20 the correct answer

Let's evaluate the definite integral From 1 to 2, we have a polynomial divided by a polynomial, and the denominator is already factored. So using what the textbook were called, case one here in the section they refer to this as case one. We have three distinct and linear factor. So the partial fraction decomposition should look something like over y be over y plus two see over y minus three. And then we could go ahead and multiply both sides by this denominator over here. So after multiplying four y squared minus seven Y minus 12 on the left, on the right, we have a no y plus two y minus three. And then for B, we have y and y minus three. And finally for C. Why? But why in a Y plus two. So let's just go ahead and rewrite this left hand or the right hand side. Just multiply things out. Y squared minus Y minus six. This is for a and then for me. Why squared minus three y and for C y squared plus two y. Now, one more thing to do here on this right hand side is to just rewrite it by factoring out first, we'll factor Ry Square. We'll have a plus B plus C one here, one here, one there. Yeah, Then we'll have. For after we found her to lie, we'll have a minus a a minus three B plus to see And then finally, the constant term. On the right hand side, it's just minus six. A. So on the left hand side, the term the constant in front of the White Square is a four on the right hand side, A plus B plus C that gives us an equation. Similarly, for the white term, we have a negative seven on the left, and this is what we have on the right. It gives us our second equation minus a minus three B plus to see equals minus seven. And then on the left hand side, the constant term was minus 12. Whereas on the left hand side it was minus six A. So we can go ahead. And those are three equations. We have a three by three system and A B and C. So, taking this last equation here we see that A s two. So there's already one of our values. And now let's just go ahead and plug this a value into the previous two equations. So we'll replace that A with the two and this over here that will become minus two. And then we can rewrite these equations. So this first one here that just becomes B plus C equals two. And for the other equation, add that negative two to the other side. We have minus three B plus two C equals negative. Seven plus two is minus five. So let's go on to the next page, running out of room here. But we will have this two by two system and B and C to solve, and then we'll have our coefficients. So we had two equations from the previous page. We had B Plus C is, too, and a minus three B plus to see is minus five. So let's just go ahead and solve this for B or C. So taking this first equation C equals two minus B and then plug this into the second equation. We'll have negative three b plus two, and then R C is two minus B, and the right hand side is negative. Five. So let's just go ahead and solve this for B, so we'll have minus three B and then minus two more B. That's minus five B and then we have a plus four. So maybe add the five to the left and then push the five beats on the other side. And then we C B is just 9/5 and then using this equation up here, let's solve proceed. So plugging B equals 9/5. In we have C equals two, which is 10/5 minus B, which is 9/5. That leaves us with 1/5. So now we've found our coefficient A, B and C. On the previous page, we found that a was to sell me. Just rewrite that here from page one. And then Now let's go on to the next page. Mhm. So let's rewrite the original integral 1 to 2. Yeah, why? Why? Plus two y minus three. This is what we intended to evaluate. We just did the partial fraction decomposition. We found a B and C. So now let's go ahead and plug those in. So a over Y becomes too over y be over Y plus two becomes 9/5 over y plus two. And then we found that C was 1/5 over Y minus three d Y Yeah, and this is a much easier integral for us. If this plus two in minus three or throwing you off, you go ahead. And do you use up here? U equals Y plus two or for the second one, U equals my mind. History. Yeah, let's go ahead and evaluate those integral the first one to natural log Absolute value y The second one That's a five in the denominator. 9/5 and that natural log y plus two plus 1/5. Natural log. Absolute value. Why my mystery and then our endpoints 1 to 2. Let's go ahead and plug in those endpoints. So plug in the two first we have natural log of four plus 1/5, and then we have two minus three, which is minus one. But then absolute value gives us a one so natural log of one. Now we plug in one for why we have 1/5 natural log of negative, too. But then we take absolute value. That's a two there. And now we cancel and simplify as much as we can. We know that natural log of one is zero. So these go away zero and zero. That's the whole the whole term there. And now we combine as much as we can, so we have to natural log two minus 1/5. Natural log, too. We can write. That is nine number five l N two. And then we have 9/5. Ln four minus 9/5 l N three. Mm. And here we can. That can be our final answer. It's a number, but we can go ahead and combine. So here, let's write this for is two squared. And then using the properties of the algorithm, we can pull out this two in front of the nine and then multiply. So we have 9/5 Ln two and then two times nine. So we have 18 there, over five, minus 9/5. That should have been Ellen. Mm. Ln of two minus 9/5 l n three. And then just combine those first two fractions and we have 27/5 natural log of two minus nine, Number five Ln of three. And there's a final answer

Mhm. Okay, so we have the integral, uh, D X over two x squared plus four x for seven. Yeah. So this is equal Trio the integral E X over two claims X squared, plus ah, two X What's happened? So that is nearly a square. So let's complete it by pulling out a four from the seven. So here's sorry. By pulling out to for so you get two X Plus two her expert plus two x plus one We brought out to into this or brought into this we have to bring out to from there. So you get plus five. Okay, so there is equal to the integral DX, too. Times x squared, plus one square loss for life. Okay. And so that is equal to integral. So if I pull out this one half so I'm calling on the factor of one half. This five becomes five over chill. All right, so now we have one half you enter role X squared plus ones. Where was five halves? All right. And so what does this equal to eso? This looks like in our attention, integral. And the rule for our attention is one of her a where this five house would be a times are contingent time if I perhaps would be a squared, um, times you there are tension of you over a so a and our case is equal to the square root of five over to. So this is equal to one over. That is just the square root of 2/5. We have our factor of one half that we need to bring in. Then we have to York Tianjin's ah, so again we're going to get one over a so speaker room here, one over is going to be Thio five and then you is going to be, uh, X plus one apologies. Serves an error here. Uh, this is X plus one squared. You see if I wrote that again? Yeah. So that should be X plus one squared small mistake, But beak on it. Easy to make mistakes. Okay? And so that's just about done. We need our constancy, But let's simplify this a little bit more. And so home we're going to get here is this is square to this is 1/2. So we're going to get this divine out in this meat. The screwed into. So we'll have one over the square root of 10 or contentions off square roots. Oh, see if we can make that cleaner square root to over five X plus one us, yeah.

Let's evaluate the following integral. We should go in and check to see if this denominator this quadratic turmoil factor before we do partial fraction to composition. So here we look at the discriminative B squared minus four a. C. In our case, that's just negative for squared, then minus four times one time six. Now this is a negative number. That means that the quadratic is irreducible. It will not factor so using with the book calls case for a partial fraction the composition should be of this following form. And then we have another linear factor. See, explicit e. And then this time we have the same factor. But we'LL square it all right again. That's case for Then let's go ahead and multiply both sides of this equation by the denominator on the left. When we do that and on the right, we have eggs plus B and then we have times the quadratic and then just see explicit E. We can go ahead and expand that right side as much as we can be ex players and then combining depending on the power of X, for example, we could pull up X Cube. We just have a So me. Rewrite this and then pull out a X squared B minus four, eh? Pullout of X. And we have six a minus four b plus e. And then the constant term leftover is six. B plus de. There's Prentice's around that. So now we look at the coefficient on the left than on the right on the left notice that there's no X Cube here. This must mean that a zero. So then we also have B minus for a that must equal one. So since a zero we get B is one, So these are two of our values. Now let's plug these in for Andy over here. And if we look at the left hand side, this should be equal to negative three. So we have negative. Three is six a minus for B plus C. I got it. And solve that for sea. We get C equals one. And finally, let's offer d sixty plus de. That's the constant term on the right that must equal seven the constants from on the left and then go ahead and solve that for D to get the equals one. So now we have our four values A, B, C and D. Let's go ahead and plug these in to the constants up here and then we'LL take the integral of the right hand side. Let's go to the next page so plugging in our values for a, B, C and D. So this is our inaugural. So let's go ahead And maybe let's just split this into two parts. Let's call this and be so let's look at a first. The first thing we should do for either of these in a girl's is complete the square. So let's look at that quadratic. We can go ahead and complete the square here and we'LL end up with Explain Is Too Squared plus two. So let's look at a party first we're DX up top X minus two squared and that I could write. This is square root of two square that will make my choice for the tricks of more obvious X minus two is rude to ten data. Therefore, the X squarer too. Seeking squared data data data now squat and plug these in root too Sequence where data defeat us over replacing the X Using this on the bottom, we have X minus two square. So that will be this thing over here Square. So that's two tangent square data and then root to square is also too. And instead of writing that too there, let me just put a one here and then I'll factor out that two in the front Now recall tan squared plus one is equal to C can square so we could cross those off. We have to go over to and then the sequence canceled. We just have integral d theta. That's just data. And we could go ahead and software data by using the tricks up. So this we can rewrite. This is tangent equals X minus two over square room, then soft for data by taking our captain on both sides. Thanks so plainly that in for data, let me not worry about the constancy because I still have to deal with this other and there will be ill added to see at the variant. And then we have X minus two radical, too. So that takes care of our first in girlie. Let's go ahead and start around the next integral part B. So for B and recall, we completed the square on the previous page. And then the denominator this time had a square on the outside. What? So this is our interval using the same tricks of that we just used. Maybe write that well, go ahead into the same tricks of his before. So we have radical, too Sequence where? Dictator? No, we should also be careful in the numerator up there, we have X plus one. So how do we get X plus one from this equation? Over here, you just add readable signs. So if you add three, the law becomes X plus one in the right hand side plus three. So we can replace X plus one with the right hand side over here. Route to tan three Santa plus three. Then we multiplied by the X, and that's over here. I want to sequence where? Data on the denominator On the previous page, we already saw that X minus two square plus radical two squared. Well, that's just using. But the protagonist identities we had to see can't square. But this time this is also square. So it looks like here because this denominator, this is for seeking to the fourth so we could cross off two of the sea cans, but not all of them. And we can also perhaps blood thiss distribute this route to toe both terms. So here, let me rewrite this as in a girl. So technically, I let me not write this. C can't because we already cancelled. I have a ten and then we have force. He can't swear we could cross off into another top and bottom over there and then for the other in the rule crossed off the sequence already. So we just have four and two more sequins on the bottom. All right, so this is just distributing this room, too, and then breaking this into two other girls. And now let's evaluate each of these separately. So here. Simple. Find these in roles. We have one half tan Taito co sign squared data for the first integral. And for the second in a rule, just rewriting that is co sign squared data data. I'm running out of room here. Let me go on to the next page. Well, simplifying from the previous equation, we have one half scientific coastline data That's the first integral. And then using the half angle identity for co sign, we can rewrite coast and square is one half. That's where this that one half is coming from one plus co sign Tuesday. Now for the first integral Over here. Just use the use up. So we have It's got a different color. One have sine squared data over to after you do the u substitution. And then we have three root too. Combine that foreign the tomb to get eight down there and then taking Mina Girl, we have data plus sign to date. Oh, over too. Now here, let's just rewrite this a little bit. Science weird over for three roots who over a data and then three Roots who and then here I'm also let me go ahead and news we write. This is to sign data co signed data using the double angle for sign. Those twos will cancel and after using that, I still haven't ate And then now signed data Cose, Aunt Ada. So now I could go ahead and go back and we write everything in terms of X, probably the triangle to do this. So we had X minus two equals room to tan data so we can go ahead and find the sides of this triangle used for the algorithm to find the hypothesis and you could simplify it to look like this. So let's go on and use this. We could have evaluate, sign and co sign And that's all we need here. And we've already found data on the previous page when we did partner So sign we'LL write those out and then we'LL go to the next page to write up the final answer your State X minus two over the radical and then co sign adjacent over the radical. So for X down there, plus six. So let's go ahead and write Our final answer First will add Walls will simplify. B will go ahead and replace this on the right hand side with terms of X. And then we'LL finally add and be together. So be it that just becomes plugging in Science Square. And then we had over four. That's where the force coming from. And then when we square, the radical goes away just rewriting data in terms of tan inverse Here we did that party and then we have three route to over a and then for co sign route, too. If you'd like you can go ahead and take this and read You use Long Division here to rewrite This is one fourth and then we'LL see here we would have after doing long division Oh, we could write that So the last step here. So just take our answers for part a part being Adam together so that I'll write that in here and this will be the last up. So for a we have room to over too ten members That was our data. So that's just for party there enough for part B. So that first term we could have simplified if if using long division and then we have the two over four. So I should say one half ex players for its plus six plus And then we have three routes to over eight. This is all from part B and then we also have three over four x minus two x squared, minus for eggs. Plus it's and then finally will go ahead and add that Sian So this is our final answer. But this can be cleaned up a little bit, so the last step will be just simplification and that will be our final answer. So combining the tannin vs, we can write this and then we also have to re X minus eight four and then X Square for explosives. Plus that constancy and there is that's our final answer.


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