ChillOut
  • ChillOut
Fun question
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Flvs.net
  • Flvs.net
Hit me.
Flvs.net
  • Flvs.net
Oh lord
ChillOut
  • ChillOut
Oops! Wrong one!
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Flvs.net
  • Flvs.net
Oh lol
Flvs.net
  • Flvs.net
God dang, what grade is this?
ChillOut
  • ChillOut
College Calculus
Flvs.net
  • Flvs.net
I am only in 7th grade. sorry
idku
  • idku
sen t ?
ChillOut
  • ChillOut
It's in portuguese. It's the same as "sine".
idku
  • idku
\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\)
ChillOut
  • ChillOut
Yep, That's it.
idku
  • idku
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?
idku
  • idku
too many variables, lol...
ChillOut
  • ChillOut
The inner integral gives me a strange answer in wolfram. There should be some change of coordinates or trig substitution that I'm missing.
ChillOut
  • ChillOut
http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bu^4%29dudt
idku
  • idku
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.
idku
  • idku
sorry, not a fan of world's most complex and longest expressions.
ChillOut
  • ChillOut
Haha, as I said, I'm missing something on this question, it can't be that hard.
anonymous
  • anonymous
@Chillout remember me its Peaches I see dat u blocked y
anonymous
  • anonymous
i only skipped 2 grades srry i can't help tip I'm on the next chapter xD
ChillOut
  • ChillOut
Some mates of mine were saying that all that is needed is FTC.
ganeshie8
  • ganeshie8
\[\cos x\sqrt{1+\sin^4 x}\]
ChillOut
  • ChillOut
Is really that simple? Just plug in sin and integrate over dt?
ganeshie8
  • ganeshie8
thats because of fundamental theorem of calculus : integral is the inverse of derivative
ChillOut
  • ChillOut
Yeah that's what I said earlier.
ganeshie8
  • ganeshie8
\[\dfrac{d}{dx} \int\limits_a^{g(x)}f(t)~dt~~=~f(g(x))*g'(x)\]
ChillOut
  • ChillOut
I can solve calculus problems just fine, but yeah, forgetting that is like... unforgivable XD
ganeshie8
  • ganeshie8
\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt\\~\\~\\ =\dfrac{d}{dx} \int\limits_1^{\sin(x)} \sqrt{1+u^4}du\\~\\~\\ =\sqrt{1+\sin^4 x} * (\sin x)'\\~\\~\\ } \)
ChillOut
  • ChillOut
Yeah, that was just naive of me, but thanks for your help!
anonymous
  • anonymous
sill need help
IrishBoy123
  • IrishBoy123
Holy Cow

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