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Solve the given problems by integration.The St. Louis Gateway Arch has a shape that is given approximately by(measurements inm) $y=-19.46\left(e^{x / 38.92}+e^{-x / 38.92}\right)+230.9$ Find the area under the Arch by determining the area bounded by this curve and the $x$ -axis.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 3

The Exponential Form

Integrals

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Section 76 problem 41. We're dealing with problems here where we're using, um, Internet. Look up tables to find are integral. Ah, and we're also having to do some other things with intervals we've done before. Substitution is and in this case, our clients and want us to find the length of the curve, like what? You to the X from zero to the natural log of to. So we have a formula for the art length of a curve. Um, spell that correctly. So the art lent is going to be the integral from A to B square root one, plus the derivative squared the X. So the curve here is f of X equal. Either the ex and we know that it's derivative iss the same thing. So either the x DX. So what we need to do now is to figure out then what is going to be the art lent from zero to the natural log of two square root of one plus either the X square, that seat of the two x TX. Okay, Now, I may not be able to find this particular one on my integral look up table, so reach into your bag of tricks and say, Well, what if I were to let you, um equals e to the X Then d you is going to be e to the x d x which tells me that D eggs is going to be one over e to the X Do you or one over you? Do you now in changing this from the variable X to the variable You you know that when X is equal to zero to the zero is one so you is equal to one. When, um, look at the other limits of integration. Um, when excess equip the natural log on to e to the natural law, go to legal that too. So you is equal to two. So this integral that is expressed in terms of X can now be written in terms of the variable uses the interval from 1 to 2 and you're gonna have one plus u squared the square root of one plus you squared and then we know that d x is one over you d you So times one over you you. So now I'm looking for the interval from 1 to 2 square root of one plus u squared over you. Do you? And I'm hoping I can find that in my, uh, look up table. So if I go to my look up table, I do see that particular one right here. Okay, so these, um the integral of X squared plus C over X, where C is a positive quantity. So that's the one I'm gonna need to get to my solution here. So let's just go rewrite that particular interval. So what I've got from my look up table is thean, a girl of the square root of a X squared plus c over X d. X is equal to the square root of a X squared plus C plus the square root of C natural log the square root of X square plus C minus the square to see over X plus a constant of integration. So in my particular case, I now see that a the coefficient of the quadratic term is one, and the constant term is one. So that's gonna make this a little bit easier to substitute. So we're going to use that particular integral, and we're gonna use the formula here. So this is going to give me The square root of you squared plus one plus the square to see is just one so natural log you're gonna have the square root of you squared plus one minus one over you. And then all of this needs to be evaluated from 1 to 2. Calculus is over. It's arithmetic time and hoping that I don't make a mistake as I go through here. So now let's just substitute these values. So I'm going to substitute a to and a one into that result here. So if I substituted to, I get four bullets one. So this is the square root of five plus the natural log. And now what I get is the, um, square root of five minus one over to and then minus. When I substitute a one in here, I'm going to get the square root of two plus the natural log. And now you're going to have when you substitute a one, you'll get this squared of two minus one. That's positive over one. So that's just grand, right? That here and now, let's just go forward and see if we can simplify. So I've got the square root of five minus the square root of two, and then it looks like I've got Plus, if I try to bring together these Ah, logarithms. So I've got plus the, um so plus the natural log squared of five minutes, one minus the natural log of two and then minus the natural log of the square two to minus one. So if I bring all the log terms together, what I'm gonna end up with this is to see this is the square to five minus the square root of two. Um, plus the natural log squared of five minus 1/2 started to minus one. Now that if I use my calculator, that's approximately 1.22 which is the value for that particular are Clint. Now you're many different forms. This answer can come in. I think the book gave a slightly different form, and we can show how that turns out to be, Um I think in the book, Let's see the square root of five. My net square to to plus a natural log. Um, for some reason, they rationalize. So if I had squared of five minus one over to for two to minus one, if I multiply this by the skirt of five plus skirt of five plus one. Gratified plus one and square to two plus one. Escorted two plus one for some rationalization. Here. What you're going to see is that this particular term right here, that's what about it. This is just gonna be five minutes. One so normally it squared of five. It's great. A five minutes in squared two plus the natural log. So you're gonna have a four squared a two plus one divided by we're going to have a two and then these radical two terms. This comes two months, one that's just one scored a five plus one, and so that for over two is to And so what you end up here is the square to five, minus the square to two plus natural log, and you end up with, um, to root two plus two over one, plus Route five. So, again, when we got to this spot right here, this was an exact answer. Here was the approximation, and I think it was just stated in your book a little bit differently. But these are equivalent forms, um, of the of the same answer so again, the key was arc length formula, which led me to a integral that I could make a substitution and then find that showing up in an integral table substitute in a row table arithmetic for the definite integral to get to my final answer.

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