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Please discover the area bounded by the curves: vex4 x2 & y=x2...

Question

Please discover the area bounded by the curves: vex4 x2 & y=x2

Please discover the area bounded by the curves: vex4 x2 & y=x2



Answers

find the areas bounded by the indicated curves. $$y=x^{2}, y=2-x, y=1$$

This problem focuses on finding the area of regions bounded by the curves. So our first function given is F of X equals X squared. And then our second equation given is f of X equals X. Because we're not given to balance that we have to determine Are balanced by where the two graphs intersect. We can do this just by equaling the two equations equal to each other or setting the two equations equal each other, X squared equals X. To do this, we have to subtract X both sides and then factor out it's this way we do not forget a factor. So we can determine that to grass cross at X equals zero and X equals one. So are bound for the integral are going to be 0-1. The next step is just to determine where or which graph mhm Is on top of the other two set up the integral correctly. So um uh an X squared. So we could say this is one, this is one. So both are going to intersect these two points right here, we change the color of this. These two points right here, the parabola X squared is gonna go wait this continue up, wow Y equals X. It's gonna go in a straight line like this. We're gonna find this area right here. So we can tell by this craft that Y equals X is on top of the X squared equation. So can set up X minus X squared D. X. Now if we solve this equation out or there's been a role, Our final answer should be won over six. Yeah.

Okay, so, in order do you figure out the area between these two functions? So I'm going to start by setting the functions equal so that I could be able to see their the intercept. So with that, I'm gonna have X squared X squared equal to X plus two. So I'm gonna bring everything to the left hand side. So that will give me X squared minus X minus two is equal to zero. From here, I can factor to figure out my, um, X value. So here I'll have X minus two times X plus one. So this will give me X equal to an X equal to negative one. So this shows me that this is negative blanc, and this is to okay, so keeping that in mind. So now we want to find that aereo's. So we're gonna take the top most function and subtract, um, the bottom most function and then put that under the integral so area is going to be equal to the integral of the top most function, which is gonna be exposed to and then minus the bottom most function, which is X squared and then times, DX or X range from negative born to chew as we found out here. So then, from here, I'm gonna just rewrite my integral so a is going to be equal to the integral from negative 1 to 2 of negative X squared plus X plus two times d x From here, I'm just gonna now take the anti derivative of each term. So the anti derivative off negative X squared is gonna be negative X cubed over three and then the anti derivative of X is gonna be X squared over two. Then the anti door of do of two is just gonna be two X and then we're gonna evaluate this from negative 1 to 2. So first, we're gonna plug in to so we have negative two cubed over three plus two squared what were too plus two times two and even now subtracts and I were gonna put in negative blanc. So negative negative one huge over three plus negative one squared over two plus two times negative one. So and with that, we will get a result of 9/2 or nine Hafs, which is equal to 4.5, and that's gonna be area of the shaded region


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