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ChillOut
 one year ago
Fun question
ChillOut
 one year ago
Fun question

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Flvs.net
 one year ago
Best ResponseYou've already chosen the best response.0God dang, what grade is this?

Flvs.net
 one year ago
Best ResponseYou've already chosen the best response.0I am only in 7th grade. sorry

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2It's in portuguese. It's the same as "sine".

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1lets work this part I suppose: \(\large\color{slate}{\displaystyle \left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\) u will get \(\large\color{slate}{ \int\limits_{1}^{\sin(t)}\sqrt{1+\sin^4t}dt}\) is that right, or am I off?

idku
 one year ago
Best ResponseYou've already chosen the best response.1too many variables, lol...

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2The inner integral gives me a strange answer in wolfram. There should be some change of coordinates or trig substitution that I'm missing.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bu^4%29dudt

idku
 one year ago
Best ResponseYou've already chosen the best response.1Jesus was crucified on this answer, it is not the Romans or Jews that killed him, but for real, this is where my good feelings towards math stops.

idku
 one year ago
Best ResponseYou've already chosen the best response.1sorry, not a fan of world's most complex and longest expressions.

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Haha, as I said, I'm missing something on this question, it can't be that hard.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Chillout remember me its Peaches I see dat u blocked y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i only skipped 2 grades srry i can't help tip I'm on the next chapter xD

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Some mates of mine were saying that all that is needed is FTC.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\cos x\sqrt{1+\sin^4 x}\]

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Is really that simple? Just plug in sin and integrate over dt?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2thats because of fundamental theorem of calculus : integral is the inverse of derivative

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yeah that's what I said earlier.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\dfrac{d}{dx} \int\limits_a^{g(x)}f(t)~dt~~=~f(g(x))*g'(x)\]

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2I can solve calculus problems just fine, but yeah, forgetting that is like... unforgivable XD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt\\~\\~\\ =\dfrac{d}{dx} \int\limits_1^{\sin(x)} \sqrt{1+u^4}du\\~\\~\\ =\sqrt{1+\sin^4 x} * (\sin x)'\\~\\~\\ } \)

ChillOut
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, that was just naive of me, but thanks for your help!
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