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Sketch the region enclosed by the given curves and find its area.

$ y = \sinh x $ , $ y = e^{-x} $ , $ x = 0 $ , $ x = 2 $

$A=0.59$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

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We want to find the area enclosed by these four equations. So first, let's do a quick sketch to see what they with these equations look like. The 1st 1 is Why's the hyperbolic sign of X? Or some people say Cinch X? Um, the second equation is e to the negative X. Are these air just sketches? So they're not 100% accurate. Let's get the 3rd 1 For now, let's actually show the 4th 1 X equals two, which is somewhere there. And if we were to pluck from the 3rd 1 X equals zero Hey, we wouldn't have an enclosed region. So the way that this question is posed, there's actually a mistake. This should say why equal zero, which would give us this region from this line over here. And so we get this enclosed region here. So that's the area that we want to determine. I'm so of course we're gonna be integrating. So we're going to integrate from X equals zero two X equals two of whatever our top of function is on the region minus our bottom function. However, we do have this point over here where the top function changes so we want to figure out first. What is the X value here? X equals question Mark. So the way that we solve this is to equate the two equations that cross at this point the 1st 1 beings sin checks and the 2nd 1 being e to the minus X. So let's just equate those himself for X meat to the miners X. So we recall that cinch X is e to the X minus e to the negative X and that over too on we create that eat a negative X and then we just solve for X so we can multiply both sides by to e to the X And in here we have e to the minus X and each of the ex canceling to give us one. So we have a two on the right side. We have two's canceling here and we distribute e to the X. So we get e to the two x minus one. So we bring this one over, we have a three and then we take the natural log a rhythm that's an X equals natural logarithms of three. So X is equal to ah half of London of three or bringing this half inside. We can write this as long of screwed three. So that's the point of intersection over here. So X is lunch of route three. So to integrate over, to integrate, um, to find the area of this region, we're going to writers to inter girls first going from X equals zero X equals one over three, and the second Integral is going to go from X equals a lot of room 32 X equals two. So let's write that up. Uh, area is equal to integral. We said X equals zero to London of route three. Um, and what we have to do is take the in the top function, minus the bottom function. Over here, the top function is cinch X, and the bottom function is X equals or y equals zero s. So we just subtract those cinch x minus zero. I I won't write the minus zero d X plus integral. We're going from lawn of Route three two X equals two. And so we have our top function minus their bottom function. Here, the top function is eating negative X, and the bottom function once again is y equals zero So we integrates e to the negative X minus zero. This is D X and then so we recall that the anti derivative of cinches, cautious or hyperbolic, co sign Kash X. We're going from zero to lawn of route three plus Ah, the integral or the anti derivative of Eton. Negative X is negative E to the minus X And here we're going from loan over three up to two. And so the second piece should be easy enough. But the first piece, we have to recall that caution X equals E to the X plus e to the minus X over two by definition, and then so plugging everything in here we have a route three plus one over route three all over, too. We subtract the one No, we add, um, we plug in X equals two so negative e to the minus two or one of re squared, Uh, and then we subtract. But then we have a minus on them. Another minus. So that becomes plus so e to the negative. Long over three will work out to be won over three. And after simplifying everything, we get screwed three minus one minus one over East Squared, which works out to be approximately 0.59

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