Find the area pf the region enclosed by y = 3 - x^2
and y = -x + 1 between x = 0 and x = 2

Question

Find the area pf the region enclosed by y = 3 - x^2 
and y = -x + 1 between x = 0 and x = 2

solomonzelman · Answer

You need to make a sketch of the region frst.

solomonzelman · Answer

and as soon as I said that stoped responding.

solomonzelman · Answer

|dw:1460065297579:dw|

solomonzelman · Answer

it looks rouphly like this.
The first function is a parabola reflected across the x-axis, and then shiften 3 units up. The second function is just a line y=-x shifted one unit up. And this is between x=0 and x=2.

amenah8 · Answer

so I am looking for the anti-derivative of (3 - x^2) - (-x + 1)?

solomonzelman · Answer

yes, but not just an antideri vative. you need the limits of integration.

amenah8 · Answer

0 and 2?

solomonzelman · Answer

Yup,

$\color{#000000}{    \displaystyle   \int_0^2 [(3-x^2)-(-x+1)]~dx           }$

solomonzelman · Answer

So all you need to do is to integrate and plug in the limits.

amenah8 · Answer

so just to be clear, you graphed the two equations and found their intercept to get 0 and 2?

solomonzelman · Answer

their x and y intercept?

edisonvalentimlopes · Answer

To help you, if necessary, the graphics is:

solomonzelman · Answer

you have 4 equations.

amenah8 · Answer

sorry! i am confused on how you got 0 and 2

anonymous · Answer

Refer to the attachment.

solomonzelman · Answer

I am not looking for the intercepts, nor am I looking for anything else but the region and the integral that would represent the area.

At first I know that your region is bounded by x=0 and x=2 (given), so let's draw this out.|dw:1460065772100:dw|

Then, I know that they are bound below and above by $y=-x+1$ and $y=3-x^2$, so let's draw this as well.|dw:1460065875219:dw|