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Question

Find the exact area under the given curves between the indicated values of $x$. The functions are the same as those for which approximate areas were found in Exercises $5-14$.
$y=4 x-x^{2},$ between $x=1$ and $x=4$

Alex Johnson · Accepted Answer

okay. And this problem? We&#x27;re trying to find the area contained by the four curves shown in the graph The Blue Line X equals negative to the green line. X equals one. The red curve Y equals two X squared plus five and the X axis where y equals zero. We&#x27;re gonna set up in a girl, uh, shows this area. The integral will be from negative 2 to 1. And the function that were integrate is two x squared plus five with a lower boundary of zero. The anti derivative of this integral is 2/3 x to the third plus five x. And we&#x27;re gonna evaluate this from negative 2 to 1. Can we first plug in the upper limit, which is one. And we will then subtract that from the same anti derivative evaluated that negative too. Okay, In the first set, we&#x27;re gonna get 17 over three subtracted by and in the second set. Cool. Get negative. 46 3rd This gives us 63 over three, which we can simplify to 21. So the area contained by those four turn those four curves is 21

Alex Johnson · Answer

this problem. We&#x27;re gonna find the area between the proble and read Michael&#x27;s X squared in the line. Michael 70. Next, it&#x27;s this little yellow sliver in between. The intersection of these two curves is at zero and 17. So are integral are 0 to 17 of the upper function. 17 x minus. The lower function X squared are anti derivative. This 17 over to X squared minus 1/3 x cubed. When we evaluate from 0 to 17 this gives us the fraction 4913 over six.

Alex Johnson · Answer

this problem. We&#x27;re gonna find the area between the four curves pictured in the graph above. And our first step is that we&#x27;re gonna try to figure out the intersection point right here between this blue line and the red curve. So we&#x27;ll set the two equations equal to each other. We have X minus one over two, equal to one over X plus one, and through cross multiplication, get X plus one at quantity times the quantity X minus one equal to two. And then we can distribute and set X squared equals three. And so what we get is positive or negative, Right? Three. And just for clarification. This red graft does have another portion across the vertical Assam toad on the side here. And so the blue graft will cross that red one twice. Ah, but what we&#x27;re what we&#x27;re finding here is just the area that&#x27;s bordered by these vertical green lines. Okay, so we have ah, point of intersection, which is right here at the square root of three. And so we&#x27;re gonna split up the first region and find the area between the I read in the Blue crafts. Our in a girl will go from zero to the square of three. And our upper function will be the red one, which is one over X plus one minus the lower function, which is X minus one squared, and it might help us tow simplify this slightly before we integrate. So we have one over X plus one, and then we can split this up into X over to minus one hand. Excuse me. Plus 1/2. Okay. So our anti derivative the first would be the natural log of the absolute value of X plus one minus 1/4 X squared, plus 1/2 X. And we&#x27;re going to evaluate from zero to square up three. And this is going to give us decimal approximate of 1.12 107 Okay, Now, for the second region, which is ah, here between skirt of three and four are integral will be square, with three as a lower boundary to four, and then our upper function will be the X minus. One over to I&#x27;m gonna read it is X over two minus 1/2 and then we&#x27;re gonna subtract the lower function. Just one over X plus one. Okay. Are anti derivatives will be 1/4 X squared, minus 1/2 X minus the natural log of X plus one and we&#x27;ll evaluate it from Skirt of three. Four in this decimal approximation is one point 51164 If we were to combine these two areas that we just found, we get a total of 2.633

Ernest Castorena · Answer

for this. They want us to keep the area between the curves EADS and the negative Rex. And why&#x27;s he was X squared on the interval. One is less than equal taxes last night, equal to four and well, experience definitely above each of the negative X and then interval because it is the a T rex on Lian decreases from the height of one starting at the origin. So let&#x27;s see one department. We have X squared minus e to the negative X yet and now lead scene. I think the anti derivatives X cubed over three plus ease the negative x applied at foreign one. So we got for cube touches 64/3 plus e to the negative part subject 1/3 plus e to the negative one. So we end up with Let&#x27;s see 63/3. Well, that one&#x27;s just Let&#x27;s see anyone. They were gonna plus e to the negative poor and subtract e to the negative one. All right,

Taylor Shimono · Answer

So for this problem in order to find the area of the region between our two curves. We&#x27;re going to want to begin by graphing are two curves. Um And so we can do this um We can go ahead and start by graphing our problem. Uh So this first one, y is equal to x squared minus four X minus five. Um So with a little bit of algebra we know that our vertex is at X is equal to negative B over two. A. So that will be 4/2, which is X is equal to two. And we plug this in, we find that this is going to correspond to a y value of -9. So we&#x27;ll go ahead and plot this point right here And then we also are going to be able to find that the Y intercept is going to be a -5. Um And then also we know that the point A 67 is going to be on our graph. Um so we&#x27;ll go ahead and say that&#x27;s right about up here 67. Um And so when we connect these we are going to find that our graph should look about like this of our problems. And then we can go ahead and do the same thing For our second equation, which is why is equal to two X -5. So this is just in slope intercept form. Um so we know that it has a slope of two In AY intercept of -5. Um So when we end up graphing these, we can see that they are going to intersect right here, X is equal to zero And also at the point where X is equal to six. And so essentially um we can see that this blue the second one is on the top so that means that we&#x27;re just going to take the integral of our x. region which is from 0 to 6 of our top function which is two x minus five. The blue one minus our bottom function which is our parabola. So that&#x27;ll be minus x squared minus four X minus five the X. Um And so our next step is going to be distributing the negative in front of our second function right there. Um so this is going to be the integral from 0 to 6 of two X -5 minus x squared plus four x plus five dx. And so we can see these fives are going to cancel out. So we&#x27;re just left with the integral from 0 to 6 of negative x squared plus six x dx. And we already know how to integrate these using the power rule. So we know that the in the role of negative expert is going to be negative 1/3 x cubed Evaluated from 0 to 6 plus. The integral of six X is going to just be three x squared Evaluated from 0 to 6. And when we solve by playing in our sixes and Zeros we&#x27;re going to find that the area between our two curves is equal to 30.

Problem

Find the exact area under the given curves betwee…

Problem 19 Easy Difficulty

Find the exact area under the given curves between the indicated values of $x$. The functions are the same as those for which approximate areas were found in Exercises $5-14$.
$y=4 x-x^{2},$ between $x=1$ and $x=4$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Heather Z.

Samuel H.

Joseph L.

Precalculus Review - Intro

Functions on the Real Line - Overview

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Heather Z.

Samuel H.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus