Question

Find the area of the region between the parabola y= 1-x^2 and the line y=1-x.

Answer 1

Are you in multivariable or single variable calc? :P

Answer 2

i think single variable, whats the difference between the 2?

Answer 3

In multivariable you use multiple integrals :P Its okay. Using single variable is just as easy.
So, the first thing you want to find is the intersection of your two curves. So set them equal and you have:
1-x^2=1-x
or:
x^2-x=0
x(x-1)=0
So your limits of integration will be from 0 to 1 (the intersection points of the curves).

Answer 4

wouldnt it be x^2+x=0 ???

Answer 5

The next thing you need to do is see which curve is on top. Just plug in any value between 0 and 1. Try (1/2). You get that the parabola is above the line right? So that means you want to do top curve minus bottom curve. So your integral becomes:
\[\int\limits_{0}^{1}1-x^2-(1-x)dx=\int\limits_{0}^{1}-x^2+xdx\]
If you have 1-x^2=1-x
Add the x^2 you have:
1=1-x+x^2
subtract the 1
0=x^2-x

Answer 6

Does all that make sense? (sorry to send it in such big chunks)

Answer 7

why do we have to plug in any value? do we need too?
oh, and isn't it x^2+x=0 ???

Answer 8

If you have a good idea of what the graphs look like, then you don't need to. Its nice to double check, because if you say the line is above the parabola you'll get a negative area :P At the bottom of the last post I worked that last bit out :P It should be x^2-x=0 I believe D:

Answer 9

malevolence has it. answer is \[\frac{1}{6}\] yes?

Answer 10

Yes :)

Answer 11

whew!

Answer 12

oh okay, yea i barely saw the last part, sorry :)thats the area, 1/6?

Answer 13

Yes :) And that's the right graph. So now you can see it has to be x(x-1)=0 or else you would have an intersection at -1 and that isn't on the graph :P

Answer 14

ok, and how do u know that it would be x^2-x=0 and not x^2+x=0...sorry to keep bugging u with thishaha

Answer 15

Just work out the algebra when you set the curves equal
1-x^2=1-x
1=1-x+x^2
1-1=-x+x^2
0=x^2-x=x(x-1)

Answer 16

not to butt in, but here is a nicer picture than the one you sent. you can clearly see that the parabola is above the line, and also what the area between them is, and where they intersect
http://www.wolframalpha.com/input/?i=y%3D+1-x^2%2C+y%3D1-x.

Answer 17

I was gonna link that earlier but I was like mehhhhhhhhh :P

Answer 18

oh, i just did it in paint :D but thanks :) i need help with a similar problem but i dont have the question with me, can u guys help me tomorrow?

Answer 19

I'll be here I'm sure :P

Answer 20

k, thanks :)

Answer 21

i have the question i need help on:
Find the area of the curve between y=3-x^2 and y=2x.