Find the interval [a,b] for which the value of the integral b s(2+x-x^2)dx a is maximum

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Find the interval [a,b] for which the value of the integral b s(2+x-x^2)dx a is maximum

Mathematics
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help me :'(
is this your question ?\[\int\limits_a^b(2+x-x^2)\text dx\]
yes!!! sry first time here....

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is the answer [-1, 2] or [-1,1/2] ??
as @UnkleRhaukus pointed out, you have maximize \[ \large F(a,b)=\int_a^b(2+x-x^2)\,dx \]
which one gives u the greatest area?
*value, sorry.
doesnt [-1, 1/2] wait no... it's [-1, 2]
what did you get when you integrated
27/6 ???
which interval?
[-1,2]
\[\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b\]\[=2(b-a)+\frac{(b-a)^2}2-\frac{(b-a)^3}{3}\] right?
yes!
actually i think i made a mistake
it's separated right
[2b+b^2/2-b^3/3]-[2a+2a^2/2-a^3/3]
\[\int\limits_a^b(2x^0+x^1-x^2)\text dx=\frac{2x^1}1+\frac{x^2}2-\frac{x^3}{3}\Big|_a^b\]\[\qquad\qquad=\left({2b}+\frac{b^2}2-\frac{b^3}3\right)-\left({2a}+\frac{a^2}2-\frac{a^3}3\right)\]yeahs
so when is this maximum
at a= -1 b= 2 ! right?
Maximize F(b) by derivative, minimize F(a) by derivative.
\[=b\left({2}+\frac{b}2-\frac{b^2}3\right)+a\left(-{2}-\frac{a}2+\frac{a^2}3\right)\]
\[=6b\left(12+3b-2b^2\right)+6a\left(2a^2-3a-12\right)\]
\[\large F \prime (b)=2+b-b^2=0\]\[\large \text{max at } b=2\]\[\large F \prime (a)=2+a-a^2=0\]\[\large \text{max at } a=-1\] interval: [-1, 2]
wow, my answer is among the choices given.
omg lol
typo mistake. it should be min at a = -1
yayyyyyy tytytyty! :')

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