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Consider the following functions_ sin cos 2X { sxss Sketch the region bounded by the graphs of the equations_Find the area of the region: (Round your answer to thre...

Question

Consider the following functions_ sin cos 2X { sxss Sketch the region bounded by the graphs of the equations_Find the area of the region: (Round your answer to three decimal places.)

Consider the following functions_ sin cos 2X { sxss Sketch the region bounded by the graphs of the equations_ Find the area of the region: (Round your answer to three decimal places.)



Answers

Find the area of the region bounded by the graphs of the given functions. $$ y=2 \cos x, y=-\cos x, \text { on }[-\pi / 2, \pi / 2] $$

Suppose you want to find the area of the region bounded by why coz Carson square decks and why call sign X. Cosign X over the interval -5 or two. To buy over four. Now looking at this graph, the area of the region we're looking at is this area. And to find the value of the area we choose this trip that would represent it. And when we received at the area of the region this is equal to The integral from -5 or two. To buy over four of the height of the strip. That will be the top function minus bottom function. So we have co sign squared X minus sign X. Cosign X times the with which is the X. Now note that go sine squared X. We can write this as one plus. Co sign up to X over two. And cynics school sign exodus is just 1/2 of sign of two x. And so we right The integral from -5 or two two pi over four of one plus co sign of two x over two minus. We have sign of two x. over two D. X. And so from here we get The integral from negative pi over two To buy over four of one half plus 1/2 co sign of two x minus one half sine of two X. D. X. And so integrating we have one half X plus we have one half times sine of two x over two minus. You have one half negative co sign of two x over two. This evaluated from Negative by over 2, 2 by over four. And so from here we have 1/2 x sign of two x over four plus Co sign of two x over four. There is evaluated from -5 or 2 to buy over four. Now in Texas by over four we have 1/2 times by over four plus we have signs of two times by over four. That's pi over two over four plus. Co sign of two times by reform. That's by over two over four. And then minus when X is negative five or two, that's one half Times -5 or two plus. You have sign of two times negative five or 2. That's just negative by All over four. And then plus co sign of negative by all over. For now, sign of by over two is just one Co Sandpiper to zero Sign of negative by zero and co sign of negative by is negative one. And so simplifying this, we have pi over it Plus 1/4 Plus that's by over four Plus 1/4. We have three pi over eight plus one half. And so this is the area of the region.

The region that is bounded by this function any X axis between X. Zero and pi over two. So here is a graph of our function sine of X to the 4th power times cosign effects. We're interested in this region right here. When exposed from 0 to I overtook. You can see that um our function stays positive between zero and pi over two. So we're looking for the area underneath uh the blue curve and above the X axis between zero and pi over two. So since our function is positive on this interval, the area will equal the definite integral of this function between zero and I were to so area Will equal the integral from 0 to Pi over two of our function. Sine of X to the fourth power. This is just another way to write uh sign to the fourth power backs. All this really means is that sine of x rays to the fourth power. Uh And then times cosine X. So the area underneath the graph of our function uh and above the X axis between X zero and high over two is equal to the definite integral of our function between zero and pi over two definite integral will equal the area as long as the function is positive on the unit. So we have to do. Now the catholic area is evaluate this definite integral. Now it looks complicated but we're going to use u substitution, You equal sine of X. So that means d'you dx the derivative of sine of X. Co Synnex. And of course that means that D. You multiply both sides by D. X. D. You is equal to cosign X. D. X. So we see this is you. So we really have you to the 4th power and co sign next D. That's that's R D. U. So the integral we'll go back to the zero and pi over two in a minute. But the integral the integral Of U. to the 4th. D. You equals U. to the 5th over five. So this being that U. Is equal to sine of X. This is equal to sine of X To the 5th power. Of course that could have been written as sign you know to the 5th power. X over five. So The integral of U. to the 4th. You to the 4th d. You which of course is the integral of sine of X. To the fourth power times. Cosine. X dx is equal to this right here. So that means this integral anti derivative of sine of X. To the fourth power times. Cosine Xd Xd anti derivative of that function. Okay, the integral of this function was sine of X. To the fifth. Power over five. So this definite integral equals the anti derivative which is sine of X to the fifth. Power over five Evaluated between zero and Pi over two. No, when exes pie over to sign up by over two is one. So we have one to the fifth which is 1/5. We need to subtract from that. This expression evaluated when X0 will sign up 000-50, divided by 50. So our area comes out to equal 1/5 so the area underneath the graft dysfunction and above the X. Axis between X zero and excess pi over two Comes out equal to 1/5.

Number 94 areas you goto integral from zero to pi sine X plus co sign Thio X dx. This is equal to negative co sign X plus sign to X over to evaluated from zero to pi that gives us one plus one just a equals two.

Hello. So here the area of the region between what we have Y. Is equal to three to the coastline of back center backs. We have wild 0 to 0 X. Is equal to zero and execute a pie. That's gonna be the the area here is the integral from zero to pi of three to the coastline of X. Times sine of back to the X. So to do this integral we're going to use a U. Substitution and let um you be equal to just co sign of acts. We have co sign of X be equal to you. And then we get that negative sine of X. The X. Is equal to D. You. So we get that when X. Is equal to zero U. Is equal to one and when X is equal to pi U. Is equal to negative one. So therefore we then get here while we get negative the integral going from one negative one of just three to the U. D. You. Okay so we're valuing this integral. This is going to give us um Well we could flip the basically instead of having negative integral this could be positively integrated. Instead of going from 12 negative one. Now we're going from negative 1 to 1 of three to the U. Do you. Okay so we evaluate that integral and we get just three to the U. Over the natural log um of three. Um And then we evaluate that from 12 negative one. So so evaluated from one Over negative 1-1. So that's going to give us three over the natural log of 3-1 over three times the national of three, which is going to give us eight Over three times the natural log of three.


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