Question

Fun question

Answer 1

Hit me.

Answer 2

Oh lord

Answer 3

Oops!
Wrong one!

Answer 4

Oh lol

Answer 5

God dang, what grade is this?

Answer 6

College Calculus

Answer 7

I am only in 7th grade. sorry

Answer 8

sen t ?

Answer 9

It's in portuguese. It's the same as "sine".

Answer 10

\(\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt}\)

Answer 11

Yep, That's it.

Answer 12

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?

Answer 13

too many variables, lol...

Answer 14

The inner integral gives me a strange answer in wolfram. There should be some change of coordinates or trig substitution that I'm missing.

Answer 15

http://www.wolframalpha.com/input/?i=int+sqrt%281%2Bu^4%29dudt

Answer 16

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.

Answer 17

sorry, not a fan of world's most complex and longest expressions.

Answer 18

Haha, as I said, I'm missing something on this question, it can't be that hard.

Answer 19

@Chillout remember me its Peaches I see dat u blocked y

Answer 20

i only skipped 2 grades srry i can't help tip I'm on the next chapter xD

Answer 21

Some mates of mine were saying that all that is needed is FTC.

Answer 22

\[\cos x\sqrt{1+\sin^4 x}\]

Answer 23

Is really that simple? Just plug in sin and integrate over dt?

Answer 24

thats because of fundamental theorem of calculus :
integral is the inverse of derivative

Answer 25

Yeah that's what I said earlier.

Answer 26

\[\dfrac{d}{dx} \int\limits_a^{g(x)}f(t)~dt~~=~f(g(x))*g'(x)\]

Answer 27

I can solve calculus problems just fine, but yeah, forgetting that is like... unforgivable XD

Answer 28

\(\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)'\\~\\~\\

}


\)

Answer 29

Yeah, that was just naive of me, but thanks for your help!

Answer 30

sill need help

Answer 31

Holy Cow