Find the between graphs of y=x^2+2 and y=1-x between x=0 and x=1. Can you guys please check if my equation is right, I will type it below.

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Find the between graphs of y=x^2+2 and y=1-x between x=0 and x=1. Can you guys please check if my equation is right, I will type it below.

Mathematics
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Find what?
Area ?
Find the area oops

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Other answers:

Makes sense
[(1-x)-(x^2+2)]
limits 0 to 1
backwards
parabola is above the line, not below it
ooopsss I think my drawing is wrong
which line?
|dw:1438743560886:dw| i thought this is like this
not if it is \(y=x^2+2\)
upper strip - lower strip right?
http://www.wolframalpha.com/input/?i=x^2%2B2%2C1-x+domain+0..1
You had y=x^2+2, the parabola is shifted up to 2 units Yes, i graphed mines a little off |dw:1438743695117:dw|
|dw:1438743765513:dw|
When you major the rectangle, the top graph is the parabola y=x^2+2 and the bottom of the rectangle is the line graph y=1-x
Soooo it is [(x^2+2)-(1-x)]
\[\int\limits_{a}^{b}(Y(x)-y(x))dx\]
The limits of the integral are x=0 and x=1 So a=0 and b=1 Y=x^2+2 and y=1-x So: \[\int\limits_{0}^{1} [(x^2+2)-(1-x)]dx\]
\[[x^2+x-1]\] then this?
then integrate?
|dw:1438743978587:dw|
|dw:1438744033419:dw|
I got 1/6
-1/6
You integrate\[\int\limits_{0}^{1}[x^2+x+1]dx\]
Did I got it right?
11/6!!
No because your integral is not correct due to not distributing appropriately.
Yes ! 11/6 :)
It is 11/6 right???
Yeah
Thank you for your help!! Hoping you can help me sometimes again. Thankyou!
It was good to work with you!
No problem

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