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Find the area of the region bounded by the parabola $ y = x^2 $, the tangent line to this parabola at $ (1, 1) $, and the x-axis.

$\frac{1}{12}$

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 1

Areas Between Curves

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

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find the region bounded by the parabola. Why equals X squared the tangent line to this parable? A at the 0.11 and the x axis. So So to get a feel for this region, let's graft these functions. So for our parabola using are exploitable. We see that the Vertex is at the origin and then we have the following points here and then we have that additional point that I popped in there. X equals 1/2 y is 1/4. So when X is 1/2 why is 1/4 right there? And so let's plot those points and plus the entire function. Okay. The other boundary for our region is the green y equals zero. So this is the X axis where you want to put that in there. And the third function that's going to bound our region is the tangent line to our red parable, A y equals x squared. And that's going to be specifically at the 0.11 So recall when we're trying to construct a line because this is the tangent line, that is why equals m x plus be so What we need here is the slope and the why value of our Why intercept? That's our be value. So recall that the slope of the tangent line is equal to the derivative with respect to X of our function. So D y d x So this is again slope is the change in y over the change in X so d y dx we want to calculate the derivative of our parabola So d d x of why is equal to the derivative with respect to X of X squared. Okay. And so we're just taking the derivative of our y equals X squared. So this gives us that the wind, the X, or the slope right this is equal to the slope is equal to the derivative of X squared, which is two X. But remember, we're considering the slope of this line at the point 11 and here are X value is one. So we're gonna plug one into there into our our slope of the tangent line. So that's two times one. Our slope of the tangent line at the 0.11 is equal to two. And again, we're constructing a line here. We're constructing a tangent line, and so we're going to plug that too back into what we're constructing for our M. So we get why equals two x plus be. And now we need to find that be value, which is the y value of our white intercept. So we go back yet again to our 0.11 and plug that in here. So why is one and X is one and we salt for be So this is one equals two plus b Subtract two from both sides and we get that b is equal to negative one. So we're gonna take this b, and we're gonna plug yet again into our our line linear function that we're trying to build. So we have that. Why equals two X minus one. And this is tthe e equation for our tangent line to our parabola X squared at the 0.11 So let's grab that guy. We can do a quick little X y table to look at some points. Um, how about zero negative 1 1/2 and that would give us zero and then one when x x one y iss one. So let's plug those points into here and plot them. So when x zero. Why is negative one? So we're like down here when X is 1/2 we're at zero. And when X is one, we're at one right there. So we've got this kind of line going on, and we can see then that the region were trying to calculate the area of all Do it in black. It's very small, this very small little kind of triangular type region between those three lines. So what's interesting about this region is that when we're considering now bounce right, we're going to do an integral and we want to consider our bounce. If we're doing, they're integral with respect to X, then we have some difficulty with our boundaries. Um, in particular also, um, because if you think of the integral when you're doing an integral with respect to X, we want to do it. The top function minus the bottom function with respect to X, right. But here we've got a top function right, which is our parabola. But the bottom function there are two functions beneath that parable A. There's the blue one and there's the green one. So that makes things somewhat difficult. We could certainly split this and dio a split in a barrel with respect to X two into girls and add them together, we could split them over X equals 1/2. Um, or we could change our perspective And look at this as integrating Oh, by along the y axis And here when we're going to integrate along the y axis. This is such that we would to ups not red. Um, we would d'oh the right function minus the left function. So if we look at this kind of from the UAE perspective will, certainly we have a right, um, function, and that is the blue line. And then certainly we have a left function, and that is our red parabola. And so, yes, we could usually do this in one integral. If we're doing, um, this from why perspective? And then we have nice boundaries for our Why perspective as well. The bottom boundary is, um, the X axis and the top boundary would be this point of intersection right here, where the blue line and the red problem both share that point. So this is nice. So let's take this perspective since that works out nicely. So what we need to do then is, um let's give these guys some names. So let's call the, um let's call the blue function will call that g of X just just for some more clarification. And then let's call our our parabola. Let's call it F of X And then just for fun, let's call the green guy. Um h of X. So all these guys have some names, we can distinguish them. And so what we want here, then, is, um, um we want to convert g of X in, um, the Y perspective. So we want not g of X, but g of why So what would this be? So first of all, um, G of X is equal Thio Well, that's our That's our tangent line to the problems. That's two x minus one. And if we want g of why from that, then all we have to do is solve, um right again, this guy's why. So we have to solve for X. Okay, So, in solving for X, we would at one and divide by two. So we have that. Why plus one divided by two is g of why so this is the function that we're going to use this g of why, instead of g of X, that will be our, um, right function. And then we have to convert our left function, which is our red parabola f of x two f of y so f of X, which is X squared again. Remember, this is why and we're solving for X. So we're gonna take the square root of both sides. So we have f of why is equal to thesis square root of what? And I'm using the positive square root because we're on the positive, Um, positive. Why? Values and positive X values. So this is when when we when we feed, um, positive. Why values into this radical? We come out with positive exercise, so we won't want to be bigger than zero okay, or equal to That's fine as well. All right, so then we need to make sure that we've got our bounds in places. Well, sort of boundary and our bounds are going to be, um we want our bottom bound. Well, that'll be the smaller one. So that'll be when why is equal to zero. That's our X axis. That's the smaller one. So why is equal to zero and then we're going to go up to when? Why is equal to one right here. This is than the why value of that point of intersection Why is equal to one All right. So are right. Function is RG of why so it'll be g of why minus our f of Why do you why and then our bounds air from zero too once. All right, so let's do the calculation. So we have the integral still not red sticks that okay, we have the integral from 0 to 1 of our g of why? Which? I'm going to split that fraction. And I'm gonna call this 1/2. Why? I like to split things up. Just tow. See the pieces that I'm dealing with, plus 1/2 minus my f of y, which is the square root of why or wind to the 1/2. And then that's all. With respect to why the why and so let's calculate that integral and we have 1/4 because there's a two on the bottom two times two is four. Why squared plus 1/2. Why minus 2/3. Why? To the three halves and that's evaluated from 0 to 1. Now we're gonna plug in our upper bound subtracted value that's resulting after plugging in the lower bound. So we have 1/4 plus 1/2 minus 2/3. And when we plug in zero into each of these terms, they all go to zero so minus zero and then getting a common denominator, that common denominator being 12 we have three over 12 plus six over 12 minus for over 12. Well, this gives us up not for excuse me. Four times two is eight. This is an eight over 12. So we have that three plus six is nine minus eight. That's 1/12 and 1/12 is the area of that small little region between these three curves that are given.

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