solve-the-given-problems-by-integration-using-eq-2715-show-that-int-bu-d-ufracbuln-bcb0-b-neq-1-2

Question

Solve the given problems by integration.
Using Eq. $(27.15),$ show that $\int b^{u} d u=\frac{b^{u}}{\ln b}+C(b>0, b 
eq 1)$.

Jaycee Glover · Accepted Answer

This is Problem 49 chapter 6.6. The problem is asking us to get parts A and B to be a multiple of this expression. So all we have locked on the inside is D W over W and then evaluate the expression at the given a M B values. So for part A, we have this expression here on its own. It doesn&#x27;t look like the multiple of D W over w w want. So we need to make it that way by using in this case, w substitution, which is basically you substitution this using a different variable. So we will make w equal to the more complex term which is one plus x squared. Taking the derivative of both sides gives us two x times. DX will solve her DX so we can plug it into our equation. And we have dw over to X. We rewrite this, we have x over w multiplied by D W over to X, the exes can cancel and the two comes outfront. We have 1/2 multiplied by the definite integral of D w over w, which is exactly what the problem asked for. Now we can take the anti derivative of this, which is Ellen. So w define terms of 10 You may think you can just plug one and zero into the six Russian and get your answer. Unfortunately, you can&#x27;t. The reason is because one and zero were for when the expression was in terms of X. But now we have in terms O W So really, you should write this as one equals x zero equals X. So we need to get these boundaries in terms of W we already have expression relating W and ex just w plus X squared. So to find these new boundaries for W, we plug in these values. So we have w plus one squared business, too, and W Siegel to one plus zero, which is one. So now we can rewrite this as Ellen W. Instead, we have two and one Now, since these air in terms of W, we can plug them into the equation. We have this 1/2 out front. We have Ellen, too. Maya&#x27;s Ln of one. Allen of one is equal to zero. So well, we have left here is Ellen of to Myers 1/2 and this is our answer or part, eh? So for part B and asking for us the same thing. It wants us to get in terms of DW over W and then to evaluate the expression. So right now we have a sign of axe over coastline of X. We need to make this simpler so you can choose either the terms I would just choose co sign. We&#x27;ll make W Eagle to co sign you would find actually, if you chose sign that the expression when they wouldn&#x27;t cancel with each other, they would be on the denominator. So the derivative of co sign is sign Negative sign times DX you saw for D. X so we could plug it into our expression. We have d w over negus sign of X when we rewrite our expression zero and then pi over four. Yeah. Sign of X over W multiplied I d w over Nega Sign of X, The science cancel And this negative which is really negative. One comes outfront We have this negative sign. Yeah, definite integral of 02 pi over four and we d w over w which is exactly what the question wants us to have. Now. We can take the integral of its We have Ellen W Tau pi over four zero, but these are in terms of X, similar to part A. So we need to plug these numbers in and get these boundaries in terms of w So we have w is equal to co sign of X. So we have plugging empire before. So co sign of pi over four is equal to route too over too. And then w eagles co sign zero Siegel toe one. We have l and the W the new boundaries Air now high over too and one. So we can rewrite this as Ellen. A cry over too over too. Mice Ellen of one and Ellen of 10 What we have left is Ellen of route to Over two. There actually is a way to simplify this even more. And we&#x27;re actually gonna get the same answer that we got in the previous model. It was Elena to over two. What show you over here? So if you remember, when you have a route to over two ain&#x27;t time. You have a a square root in the denominator area. You want to move it to the numerator because it doesn&#x27;t just doesn&#x27;t look nice to have a square root in the denominator. Actually, if you revert this back toe how it used to be before it was simple, fire made it look nicer. Actually used to be like this, and then you would take, you know, multiply square root to to the top and bottom in. This gives you this answer. You actually could rewrite this, cause these air their equaled. These are equal to each other. So you could rewrite this as Ellen of one over square root of two, which is the same as negative, Ellen. Two to the negative 1/2 power. And we&#x27;re using natural log rules. You can bring that exponents outfront negative. 1/2 times. This negative one will make it a pause of 1/2. Not too stays on the inside. So this is how you get the same answers, eh? Neck

Michael Jacobsen · Answer

in this video, we&#x27;re going to be calculating the indefinite integral of x times beat of the X DX. First notice that the base be is unspecified, that we still have certain requirements for being a the X To be defined be has to be positive And in order for this problem to be interesting be should not be equal toe one since one of the power of X is just one now, to get started on this problem, we can solve it by using the method of integration by parts. Let&#x27;s set you to be equal to X so that do you must then be DX next. All that&#x27;s left for Devi is B to the power of X times DX So we need the anti derivative of BX. Well, we know that that derivative with respect to X of B to the power of X is itself times the natural log of be. So if we&#x27;re taking the anti derivative we Formby to the X divided by the natural log of B, Now we&#x27;re ready to begin that integration by parts solution first, take u times V so that this will be equal to all together. X times B to the power of X divide by the natural log of B, then subtract a particular anti derivative formed by taking V times to you. In this expression, one over the natural log of B is concerned constant Sophie like we can place it here just to make it less of a distraction in the next indefinite integral, which will be Peter the X DX. On the next step, we see that this is equal to first copy X Times view the X over the natural log of B minus. Now the anti derivative of B to the power of X. We&#x27;ve in fact, already done once. It&#x27;s this expression here, since we already have natural rugby and a denominator will be race of power to so we have B to the X divide by the natural log of B squared. Then we throw in an arbitrary constant to get the most general anti derivative. And if we prefer, this can be cleaned up a bit by, say, factoring out B to the power vex. That would leave behind X divide by the natural log of B minus one. Divide by the natural log of B squared, plus a constant

Ramzi Deek · Answer

So we got integral of X Ln of X squared here. And we&#x27;re being asked to use, um, substitution as you is x squared to solve. So part a Let&#x27;s take u equals X squared Do you over dx his two x So in this case, do you over to X Is DX so integral? No, start plugging in so x times Atlanta of X square So that&#x27;s Ln of you times do you over two x so x in X cancel here factor half out So half Ellen you do you. So, Ellen, you has a standard. Uh, it&#x27;s a definite integral. So half into you Ln you minus you. So let&#x27;s just plug in you for X squared. So that&#x27;s half into X squared Ln X squared minus X squared Can add a constant See here. All right, so that&#x27;s for substitution. Let&#x27;s do integration by parts, So x Ellen X squared DX. So here, I&#x27;m gonna take you as island X squared. Remember, Ln of X squared is also we can also put that as to Ellen X. So do you Is two over X v is X devious X. So V is X squared over to so we know to integrate by parts into go a few Devi is UV minus inter Go of v do you? So that&#x27;s oh two Ln x or just keep it as ah Ln X squared here Times X squared over two minus integral off v do you? So that&#x27;s X squared over two times two over x to and to cancel x the next cancel here. So we&#x27;ll end up with integral of X. So our answer here is X squared. Ellen X squared over to minus X squared over two plus c. So what we have, we have the exact same integral and they don&#x27;t different anyway, if you, uh, distribute the half here, it&#x27;ll match what we have down here. So parts, he asked us Verify that your answers to part A and B are consistent. Well, see ah, just say a so results of a and B prove that answer is consistent

Robert Daugherty · Answer

section seven dot to Problem number 54. So in this case, they ask us to look at the anti derivative of X natural log of X squared DX they wanted to use to method to your first method. Let&#x27;s just say let&#x27;s do a substitution. Let&#x27;s let you equals X squared And then the second method we&#x27;re going to use is going to be integration by parts. So if you is equal to X squared than d&#x27;you is going to be two x dx. So I can write this original problem X natural log X squared DX is equal to 1/2 the integral of two x log of X squared DX. So this is just of the form 1/2 and in the anti derivative of natural log of you. Do you? So this is 1/2 you natural law give you minus one plus a constant of integration. So my final answer here is 1/2 X squared Natural log X squared minus one plus a constant of integration. Now, if I were to use integration by parts, I can let you equal the natural log of X squared. Then d&#x27;you is going to be one over X squared times the derivative of X squared. So this is simply going to be two over X. The X and then Devi is going to be X dx, which tells Mavy is going to be X squared over two. So in integration by parts, this just becomes U V minus the integral VD you. So that&#x27;s going to be 1/2 X squared natural log of X squared minus the integral of the D you so x squared over two times two over x the X So this is just the integral of X t X. So this is 1/2 X squared natural log of X squared minus X squared over two, plus a constant of integration. And I can factor out the 1/2 here X squared and this is just the natural log of X squared minus one plus a constant. So what you see is I got the same answer both times using two different routes to the to finding the derivative here

Stephen Hobbs · Answer

in order to prove that this integral here ends up going. B squared minus a squared over two will find Delta X, which is our upper limit. Minus are lower limit, all divided by N. The idea is we&#x27;re going to rewrite these as limit of the some and then simplify from there. So we&#x27;ve got the limit as an approaches infinity, and then we have the some from I There is one thing up to and multiplied by the width of each of our rectangles, which is B minus a and then all over and that multiplied by the next value, which is X actually. And we start with the lower limit a and then we add the F term of Delta X, which is B minus a all over end times I and this is what we want to simplify here, all of that. So the next thing we could do is go ahead and distribute each of these. And so we&#x27;ll go ahead and write that as a times be, I must say all over and and then we have plus, by the way, it&#x27;s still the limit in the some. So these two just dropped down I&#x27;m just leaving them off for a second. Um okay, so we&#x27;ve distributed those, then we&#x27;ll go ahead and distribute there, so that will end up killing us. Be minus a squared. Over this time we have m squared and let&#x27;s go ahead and apply the some toe Both of these. And of course, it&#x27;s from I was one toe end so that the sum if we completed on the first one, will just give us another factor of n which will actually end up canceling their and the some on the next one. We could replace this. I hear where m and plus one. And you could look up how that substitution works there. It&#x27;s just one of a given some nations for just I from has won the end. Okay. And then we&#x27;re going to rewrite this whole thing up here, so let&#x27;s do that. Maybe green. So what do we have? We&#x27;ve got Let&#x27;s distribute a here and here. So we&#x27;ve got, um, a B minus a sward. Then we have plus Okay, if we kind of mush this all together here, we&#x27;re going to have, we&#x27;ll rewrite it. We&#x27;re gonna have be minus a all of that weird. Then distributing the end will have and squared and then plus a term with n so that first one would be in squared over the bottom. We have a two and then end square and then finally plus, then we basically have that same thing with, um, end there. So be minus a all squared times and over to and squared. Now, in the very front, we still have the limit as I am approaches infinity. And it&#x27;s at this point that we&#x27;re just gonna go ahead and apply that rule. So when the higher power is in the denominator, all of that goes to zero. These ends here divide out. So really now we don&#x27;t need to worry about that limit. We could just go ahead and substitute, so that helped us simplify it quite a bit. So let&#x27;s write what we have at this point. So we&#x27;ve got a B minus, a squared still, let&#x27;s go ahead and expand this out. This will end up killing us plus B squared and then minus two a B and then plus a squared. Except all of these are divided by two. Still, because of that fraction, So it&#x27;s right that And in the end, let&#x27;s see what happens here. So the twos cancel that A B cancels with a negative A B does go away a negative, a squared with a plus. A squared over two ends up giving us a negative. When we combine these two, that basically ends up giving us a negative a squared over to and then notice. We already have a B squared over two, which just drops down. And so, in the end, that is the exact same thing as this B squared minus a squared Over two course, we could just go one step further. In fact, with one half out, one half and then we&#x27;re left with so her empathy, it is there one half B squared, minus X squared. So one of the questions that I had when I was over the years of kind of understanding integration is where does that b minus a come from? And this is ah, good example of how that appears there upon evaluating a and a girl such as this. So I hope that was helpful. There

Problem

Solve the given problems by integration. Find the…

Problem 37 Easy Difficulty

Solve the given problems by integration.
Using Eq. $(27.15),$ show that $\int b^{u} d u=\frac{b^{u}}{\ln b}+C(b>0, b \neq 1)$.

Answer

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Catherine R.

Anna Marie V.

Kayleah T.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus