anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zale101
  • Zale101
Find what?
Zale101
  • Zale101
Area ?
anonymous
  • anonymous
Find the area oops

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More answers

Zale101
  • Zale101
Makes sense
anonymous
  • anonymous
[(1-x)-(x^2+2)]
anonymous
  • anonymous
limits 0 to 1
anonymous
  • anonymous
backwards
anonymous
  • anonymous
parabola is above the line, not below it
anonymous
  • anonymous
ooopsss I think my drawing is wrong
Zale101
  • Zale101
which line?
anonymous
  • anonymous
|dw:1438743560886:dw| i thought this is like this
anonymous
  • anonymous
not if it is \(y=x^2+2\)
anonymous
  • anonymous
upper strip - lower strip right?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=x^2%2B2%2C1-x+domain+0..1
Zale101
  • Zale101
You had y=x^2+2, the parabola is shifted up to 2 units Yes, i graphed mines a little off |dw:1438743695117:dw|
Zale101
  • Zale101
|dw:1438743765513:dw|
Zale101
  • Zale101
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
anonymous
  • anonymous
Soooo it is [(x^2+2)-(1-x)]
Zale101
  • Zale101
\[\int\limits_{a}^{b}(Y(x)-y(x))dx\]
Zale101
  • Zale101
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: \[\int\limits_{0}^{1} [(x^2+2)-(1-x)]dx\]
anonymous
  • anonymous
\[[x^2+x-1]\] then this?
anonymous
  • anonymous
then integrate?
Zale101
  • Zale101
|dw:1438743978587:dw|
Zale101
  • Zale101
|dw:1438744033419:dw|
anonymous
  • anonymous
I got 1/6
anonymous
  • anonymous
-1/6
Zale101
  • Zale101
You integrate\[\int\limits_{0}^{1}[x^2+x+1]dx\]
anonymous
  • anonymous
Did I got it right?
anonymous
  • anonymous
11/6!!
Zale101
  • Zale101
No because your integral is not correct due to not distributing appropriately.
Zale101
  • Zale101
Yes ! 11/6 :)
anonymous
  • anonymous
It is 11/6 right???
Zale101
  • Zale101
Yeah
anonymous
  • anonymous
Thank you for your help!! Hoping you can help me sometimes again. Thankyou!
Zale101
  • Zale101
It was good to work with you!
Zale101
  • Zale101
No problem

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