Areas between graphs of 2 functions

- anonymous

Areas between graphs of 2 functions

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- anonymous

y=x^2, x+y=2

- anonymous

Alright, I'll lead you through the steps, how about that? :)

- anonymous

1) Sketch a graph of both functions

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- anonymous

sounds fantastic! thank you! lol

- anonymous

np ^_^

- anonymous

graph was given lol (way easier)

- anonymous

lol, okay!

- anonymous

2) find the intersection points when y1= y2

- anonymous

set equal to zero?

- anonymous

yep, find the x's :)

- anonymous

got em...-2 and 1

- anonymous

alright, good :)

- anonymous

now, we must find out which function is bigger :) in other words, which graph is on the top of the other?

- anonymous

I think y= x^2 is on top of the other one, right? :)

- anonymous

yeah i remember that from class but it sort of looks like an upside down "V" so neither is on top. Do i use the right one because its a graph of y not x?

- anonymous

exactly! you can see the intersection in the graph right? which curve is on the top?

- anonymous

i think for y he said right-left and for x top-bottom

- anonymous

let me sketch the graph lol just to make sure :)

- anonymous

yep i see the intersection lol thanks. i would upload the pic but that would take longer than a sketch

- anonymous

hold on :)

- anonymous

okay got it, look at your sketch, color the region bounded (where the intersection lies/ closed region lol)

- anonymous

which line is on the top? :)

- anonymous

in my sketch, I see y = 2-x on top :)

- anonymous

##### 1 Attachment

- anonymous

hmm, I got a different sketch , I've sovled for y for the second equation (y=2-x)

- anonymous

try to sketch the graph using y = x^2 and y = 2-x :)

- anonymous

A plot related to this problem is attached.

##### 1 Attachment

- anonymous

yeah, that's what I got robto ^_^, birdsell are you following?

- anonymous

its the same graph its just the given one is a snippet

- anonymous

LOL, then draw your own to make your life easier :) , look at robto's sketch , which line is on the top ? ^_^

- anonymous

intersection point of the 2 lines in Q1 is at 1,1

- anonymous

are you sure? I got x = -2 and x = 1

- anonymous

so did i :-X but the picture shows it at 1,1

- anonymous

birdsell, dear :) , let's take it step by step
1) sketch both graphs, on your own , ignore the given :)

- anonymous

you have y =x^2 and y = 2-x

- anonymous

sketch them

- anonymous

you'll have to get the same sketch as robto's ^_^

- anonymous

im so confused :( most of them are clear that one is on top of the other

- anonymous

calm down, did you draw the graph?

- anonymous

and im attempting to use my calculator to graph the points but cannnt

- anonymous

don't!
you can try out values, alright for example, sketch y = x^2 first, take values for x and find y, then plot them :) same with the other equation, do that

- anonymous

after that, you'll get a graph similar to robto's

- anonymous

try :)

- anonymous

The second plot version uses the same formulas except the vertical and horizontal scaling is the same.

##### 1 Attachment

- anonymous

so x^2 is on top?

- anonymous

nope, look at the sketch again :)

- anonymous

The curve passing through (0,0) is x^2.

- anonymous

look at the closed region,=)

- anonymous

lmfao im dumb :)... got it

- anonymous

no you're not, you panicked :D, alright so which line is on the top?

- anonymous

y=2-x

- anonymous

excellent, so this wil be you g(x) :) and the other curve is f(x) okay?

- anonymous

got it

- anonymous

I actually think I got it! lol wanna help with another?

- anonymous

now to find the Area, the general formula is :\[A =\int\limits_{a}^{b}[g(x) - f(x)] dx\]
where g(x) is the curve that's in the top and f(x) in the bottom, replace the equations and solve :
\[A = \int\limits_{-2}^{1} (2-x) -(x^2) dx\]
\[A = \int\limits_{-2}^{1}( 2-x-x^2)dx\]
\[A = [2x-x^2/2 + x^3/3]\] from -2 to 1 so you'll get :
\[A = [2(1)-(1)^2/2-(1)^3/3] - [2(-2) -(-2)^2/2 - (-2)^3/3]\]
compute it and you'll get A lol :)

- anonymous

sure , I'll be glad to help ^_^

- anonymous

hanks! new thread or current?

- anonymous

thanks**

- anonymous

lol whatever you want :), np ^_^

- anonymous

not sure if you are familiar with slicing?

- anonymous

yes I am , it has to do with finding the volume right =)?

- anonymous

yep! :):)

- anonymous

I had a midterm on it last sunday lol!

- anonymous

alright, which method does it want you to use? :)

- anonymous

ughhh college... calc is killin me... up until now i have been great

- anonymous

lol, calculus II right? practice, work hard, you'll perfect everything. Work hard now to relax in the end, believe me I sucked in this part =-= but now it's just SIMPLE :)

- anonymous

Sorry to intterrupt, your question ask area of between x^2 , y+x=2 and x axis according to photo but the solution answers area of two function.I am confused

- anonymous

you're finding the area of the 2 functions between the 2 curves Hakan, to do that you have to find the enclosed region :)

- anonymous

actually calc 1. I go to a tough school and this is calc 1 carriculum

- anonymous

serious? I'm taking this in calculus II in college lol

- anonymous

Ill start a new thread sstarcia

- anonymous

well, I took the area b/w the curves in calculus I but the volume in calculus II

- anonymous

alright :)

- anonymous

ahhh it isnt working lol

- anonymous

how come?

- anonymous

anyway... The solid lies between planes perpendicular to the y axis at y=0 and y=2 the cross-sections perpendicular to the y-axis are circular discs with diameters running from the y-axis to the parabola \[x=\sqrt{5y^2}\]

- anonymous

But the photo of question doesn't ask that as i see.Anyway.Thanx :)

- anonymous

lol we re-sketched it, look at robto's sketch :) np :)

- anonymous

alright, to make your life easier bird, find y of this equation

- anonymous

\[y = x/\sqrt(5)\] I guess

- anonymous

i got y=x/5 lol

- anonymous

lol, take y^2 out and you'll get y, then divide by sqrt of 5 :)

- anonymous

nvm i see now lol gotta take Sqrt of both top and bottom

- anonymous

gotcha

- anonymous

okay :)

- anonymous

now:
1) sketch the given in a graph, y=0, y=2 and y=x/sqrt(5)

- anonymous

use your calculator for y = x/sqrt(5) you'll get something like a slight curve on the right, right?

- anonymous

it wont give me anything... when i type y=x/sqrt[5]

- anonymous

lol, forget your graphying calculator and do it by hand. Take values for x and calculate y using the calculator :)

- anonymous

graphing*

- anonymous

okay

- anonymous

got it

- anonymous

I think you'll get something like this , I think

##### 1 Attachment

- anonymous

did you get something similar to it?

- anonymous

yep

- anonymous

alright, now what did the question say, how does the curve rotate? about the y-axis, x-axis?

- anonymous

find vol of disc that runs perpendicular top y axis w diameter running from x to the parabola

- anonymous

so you'll have to use the disk method :). So that means it's rotating about the y-axis right?

- anonymous

so cut the curve perpendiculary to the y-axis ( draw a horizontal line in the bounded region I showed you )

- anonymous

ughh im so sorry but i have to go to class now. thank you for all of your help :)

- anonymous

lol, sure :) good luck, and try to finish it ^_^

- anonymous

thanks!

- anonymous

:)

- anonymous

:):) hiya ppl!!!!

- anonymous

need any more help @birdsell ???

- anonymous

lol, he's in class :)

- anonymous

hey sstarica, couldn't type ther in grp chat ..lol...its stuck again !

- anonymous

I don't like it either ways :) it's alright.

- anonymous

which grade are you? ^_^

- anonymous

im gonna be engineer too :)

- anonymous

which field? :)

- anonymous

software

- anonymous

nah, I'm in the hardware field =P

- anonymous

never mind..we can help each other:)

- anonymous

lol, yep. I'll deal with the hardware, you deal with the software, then we'll combine them together in the end

- anonymous

sure :):)

- anonymous

I find it rather weird in someway. I thought software = more girls work in that field and hardware = men, but now it's the opposite LOL

- anonymous

half of the guys I know are all wanting software and girls working in hardware.

- anonymous

lol..girls work on both like us:)

- anonymous

the majority of the boys in here are all hardware. It is very rare to see a girl working in the hardware field :)

- anonymous

they say it's a guy thing, but I disagree.

- anonymous

good...we can work on both :) right :)

- anonymous

yes ^_^, anyways, I wish you the best of luck :)

- anonymous

thank you !!! best of luck to u too...:) lets see wen we meet here again :)

- anonymous

np :) and thank you ^_^, good night ~

- anonymous

good night star..:):)

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