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find the definite integral y = x(1-x) from -2 to 2

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you can multiply the x into the parenthese and obtain:|dw:1324593176181:dw|, then its just routine
integrate and evaluate at the limits
could you please further elaborate thankx

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Other answers:

\[\int\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C\] \[\int\limits_{a}^{b}f(x) dx=F(x)|_a^b=F(b)-F(a)\]
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
thank you @ lagrangeson678 ^.^
myininaya what did you get as your answer?
i found the antiderivative now all you have to do is: (just plug in upper limit) then write minus (then plug in lower limit)
oh because i got a negative answer
and according to the solution booklet it is a positive answer i.e 17/3 square units
yes and i got -16/3 square units
Chek out the graph of this fucntion:
Do you notice anything?
yes the area is below the x axis, but how come in the solution book to this question, their answer is 17/3?
but why were you saying 17/3
oh thats what the solution book says
The answer is negative
\[\int\limits\limits_{}^{}(x-x^2) dx=\frac{x^{1+1}}{1+1}-\frac{x^{2+1}}{2+1}+C \] \[\int\limits_{-2}^{2}(x-x^2) dx=[\frac{x^2}{2}-\frac{x^3}{3}]^2_{-2}=[\frac{2^2}{2}-\frac{2^3}{3}]-[\frac{(-2)^2}{2}-\frac{(-2)^3}{3}]\] \[=[2-\frac{8}{3}]-[2+\frac{8}{3}]=2-\frac{8}{3}-2-\frac{8}{3}=2-2-\frac{8}{3}-\frac{8}{3}=0-\frac{16}{3}\]

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