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Calculus1
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Hit me.
Oh lord
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Oh lol
God dang, what grade is this?
College Calculus
I am only in 7th grade. sorry
sen t ?
It's in portuguese. It's the same as "sine".
\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\)
Yep, That's it.
lets 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?
too many variables, lol...
The inner integral gives me a strange answer in wolfram. There should be some change of coordinates or trig substitution that I'm missing.
http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bu^4%29dudt
Jesus 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.
sorry, not a fan of world's most complex and longest expressions.
Haha, as I said, I'm missing something on this question, it can't be that hard.
@Chillout remember me its Peaches I see dat u blocked y
i only skipped 2 grades srry i can't help tip I'm on the next chapter xD
Some mates of mine were saying that all that is needed is FTC.
\[\cos x\sqrt{1+\sin^4 x}\]
Is really that simple? Just plug in sin and integrate over dt?
thats because of fundamental theorem of calculus : integral is the inverse of derivative
Yeah that's what I said earlier.
\[\dfrac{d}{dx} \int\limits_a^{g(x)}f(t)~dt~~=~f(g(x))*g'(x)\]
I can solve calculus problems just fine, but yeah, forgetting that is like... unforgivable XD
\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt\\~\\~\\ =\dfrac{d}{dx} \int\limits_1^{\sin(x)} \sqrt{1+u^4}du\\~\\~\\ =\sqrt{1+\sin^4 x} * (\sin x)'\\~\\~\\ } \)
Yeah, that was just naive of me, but thanks for your help!
sill need help
Holy Cow

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