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

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 3

The Exponential Form

Integrals

Missouri State University

Campbell University

Harvey Mudd College

Lectures

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02:25

Verify the integration for…

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(a) Make the indicated $u$…

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'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'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'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'll make W Eagle to co sign you would find actually, if you chose sign that the expression when they wouldn'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'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't time. You have a a square root in the denominator area. You want to move it to the numerator because it doesn't just doesn'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'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

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