jackkim4
Find the interval [a,b] for which the value of the integral b
s(2+xx^2)dx
a
is maximum



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jackkim4
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help me :'(

UnkleRhaukus
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is this your question ?\[\int\limits_a^b(2+xx^2)\text dx\]

jackkim4
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yes!!! sry first time here....

jackkim4
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is the answer [1, 2] or [1,1/2] ??

helder_edwin
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as @UnkleRhaukus pointed out, you have maximize
\[ \large F(a,b)=\int_a^b(2+xx^2)\,dx \]

helder_edwin
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which one gives u the greatest area?

helder_edwin
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*value, sorry.

jackkim4
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doesnt [1, 1/2] wait no... it's [1, 2]

UnkleRhaukus
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what did you get when you integrated

jackkim4
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27/6 ???

helder_edwin
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which interval?

jackkim4
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[1,2]

UnkleRhaukus
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\[\int\limits_a^b(2x^0+x^1x^2)\text dx=\frac{2x^1}1+\frac{x^2}2\frac{x^3}{3}\Big_a^b\]\[=2(ba)+\frac{(ba)^2}2\frac{(ba)^3}{3}\]
right?

jackkim4
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yes!

UnkleRhaukus
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actually i think i made a mistake

jackkim4
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it's separated right

jackkim4
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[2b+b^2/2b^3/3][2a+2a^2/2a^3/3]

UnkleRhaukus
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\[\int\limits_a^b(2x^0+x^1x^2)\text dx=\frac{2x^1}1+\frac{x^2}2\frac{x^3}{3}\Big_a^b\]\[\qquad\qquad=\left({2b}+\frac{b^2}2\frac{b^3}3\right)\left({2a}+\frac{a^2}2\frac{a^3}3\right)\]yeahs

UnkleRhaukus
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so when is this maximum

jackkim4
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at a= 1 b= 2 ! right?

sirm3d
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Maximize F(b) by derivative, minimize F(a) by derivative.

UnkleRhaukus
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\[=b\left({2}+\frac{b}2\frac{b^2}3\right)+a\left({2}\frac{a}2+\frac{a^2}3\right)\]

UnkleRhaukus
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\[=6b\left(12+3b2b^2\right)+6a\left(2a^23a12\right)\]

sirm3d
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\[\large F \prime (b)=2+bb^2=0\]\[\large \text{max at } b=2\]\[\large F \prime (a)=2+aa^2=0\]\[\large \text{max at } a=1\]
interval: [1, 2]

sirm3d
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wow, my answer is among the choices given.

jackkim4
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omg lol

sirm3d
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typo mistake. it should be min at a = 1

jackkim4
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yayyyyyy tytytyty! :')