Question
7 Use an area argument to compute [,1g x2 dx
7 Use an area argument to compute [,1g x2 dx
Answers
Use Theorem I to compute the shaded area in Exercise 7.
In discussion, we are required to establish a definite integral that can evaluate the area bounded by the region FX recalls to access square and Gs physicals too X cube. So let's see how to solve this question. First of all, let's draw the graph for these two equations to identify the region. So this is a graph for the equations Fx and GX. And now let's find the area of this region. From here. We can observe that FX is greater than equals two G X. Therefore we can conclude that X Lies between zero and 1 hands. The expression to calculate area. Can brightness integration 0 to 1 fx minus G X. D X. No society with all the values. So we get Area is because 20-1 access square minus X cube D X. So this is a vile answer for this problem. I hope you understood revolution. Thank you.
My problem 91 and a girl from 0 to 7 x times a cube root of x plus one d x Want to find the area of the region set you is You go to X plus one. Do you is to go to D. X So the areas the integral from 18 of U minus one times you to the third Do you in separate these two and integrate them individually to get 3/7 you There are 7/3 minus 3/4. You two are 4/3 integrated from 18 and that comes out to 1209 over 28.
In discussion. We are required to establish a definite integral that can evaluate the area of the region bounded by fx recalls to X -1 to the power three and gxe articles to x minus one. So let's see how to solve this question. First of all, let's draw the graph for these two equations and the graph is shown below this is the graph for these two equations. This is G X and this article is okay ethics from this graph we can observe that Okay, GX is less than equals two. Fx four X Lies between zero and 1 and G X is greater than it calls to. Fx food X lies between one and two. Since fx and gx both are symmetric about the 0.1, comma zero. So the area of the region can be written as To into integration 0- one. Okay, fx minus G x D x. No substitute all the values. So we get Go into integration 0-1 X -1 to the power three x minus one D x. So this will be calls to Integration will do one X -1 into access square minus two works D X. And the expression to calculate the area for the second region can grittiness To into integration 1 to 2 G x minus fx D x. Substitute all the values. So we get area is the calls to To into integration 1 to 2 In two x -1 minus X -1 to the power three D x. And this will be calls to To into integration 1 to 2 X minus one into to works minus X, esquire D X. So finally we will have area is the calls to minus two. Integration 1 to 2 X -1 into x squared minus two weeks. D x. Cool. These two are the final answers for this problem. I hope you under for the solution. Thank you.
Okay this problem says first graph that function and estimate the area underneath there. So let's have a look at this. Um Each X value is one each y. Is one. So let's count the whole boxes. 1st 123456789 Whole boxes I think those to make 10. Um Let's see um 11:12 13:14:15 16 17. I'm going to guess about 18 is the answer here. All right so first it says break it up into two inner girls so I'll do that. Okay and the first one I'm gonna let oh I forgot the dx I'll let you before minus X squared. And then D. U. S minus two X. Dx. So I need a minus two in there minus one half out in the front So get -1 half. Um If you is too I get zero and if you as minus to get zero so I get 0 to 0 square to you. D you which is zero. All right now over here I'm gonna have to this is a plus. Right here else we're going to get a negative answer this one. I'm going to have to just figure out by what's it a picture of. So it's not this anymore. I'm just talking about this part. So if you let y equal four minus X squared and you square both sides you get y squared equals four minus X squared. So X squared plus y squared equals four. This is a circle center 00 radius too. But remember it was only the positive square root. So it's the top half of the circle From -2 to 2. Okay, so this will give us plus three times the area of a circle which is pipe half a circle. Well, three times the area of half a circle, half by two square. So for two, two pi times 36 pi zero plus six by six pi is the answer. Let's see. That's a 3.14 times 64 to 8, 18.84. And what did I guess? 18? Okay, that's that's not a bad guess. Amy.