anonymous
  • anonymous
find the definite integral y = x(1-x) from -2 to 2
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you can multiply the x into the parenthese and obtain:|dw:1324593176181:dw|, then its just routine
anonymous
  • anonymous
integrate and evaluate at the limits
anonymous
  • anonymous
could you please further elaborate thankx

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myininaya
  • myininaya
\[\int\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C\] \[\int\limits_{a}^{b}f(x) dx=F(x)|_a^b=F(b)-F(a)\]
anonymous
  • anonymous
man, i caught up with something else, or else i would have explained it to you. But the one above me, did a fairly good job
anonymous
  • anonymous
thank you @ lagrangeson678 ^.^
anonymous
  • anonymous
myininaya what did you get as your answer?
myininaya
  • myininaya
i found the antiderivative now all you have to do is: (just plug in upper limit) then write minus (then plug in lower limit)
anonymous
  • anonymous
oh because i got a negative answer
anonymous
  • anonymous
and according to the solution booklet it is a positive answer i.e 17/3 square units
anonymous
  • anonymous
|dw:1324594394344:dw|
anonymous
  • anonymous
|dw:1324594437592:dw|
anonymous
  • anonymous
yes and i got -16/3 square units
anonymous
  • anonymous
Chek out the graph of this fucntion:http://www.wolframalpha.com/input/?i=integral+y+%3D+x%281-x%29+%5B-2%2C2%5D
anonymous
  • anonymous
Do you notice anything?
anonymous
  • anonymous
yes the area is below the x axis, but how come in the solution book to this question, their answer is 17/3?
anonymous
  • anonymous
OH I SEE!!!!!!!!!! THANKS HEAPS
anonymous
  • anonymous
but why were you saying 17/3
anonymous
  • anonymous
oh thats what the solution book says
anonymous
  • anonymous
The answer is negative
myininaya
  • myininaya
\[\int\limits\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C \] \[\int\limits_{-2}^{2}(x-x^2) dx=[\frac{x^2}{2}-\frac{x^3}{3}]^2_{-2}=[\frac{2^2}{2}-\frac{2^3}{3}]-[\frac{(-2)^2}{2}-\frac{(-2)^3}{3}]\] \[=[2-\frac{8}{3}]-[2+\frac{8}{3}]=2-\frac{8}{3}-2-\frac{8}{3}=2-2-\frac{8}{3}-\frac{8}{3}=0-\frac{16}{3}\]
anonymous
  • anonymous
THANK YOU @ MYININAYA

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