ChillOut one year ago Fun question

1. Flvs.net

Hit me.

2. Flvs.net

Oh lord

3. ChillOut

Oops! Wrong one!

4. Flvs.net

Oh lol

5. Flvs.net

God dang, what grade is this?

6. ChillOut

College Calculus

7. Flvs.net

I am only in 7th grade. sorry

8. idku

sen t ?

9. ChillOut

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

10. idku

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

11. ChillOut

Yep, That's it.

12. 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?

13. idku

too many variables, lol...

14. 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.

15. ChillOut
16. 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.

17. idku

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

18. ChillOut

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

19. anonymous

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

20. anonymous

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

21. ChillOut

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

22. ganeshie8

$\cos x\sqrt{1+\sin^4 x}$

23. ChillOut

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

24. ganeshie8

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

25. ChillOut

Yeah that's what I said earlier.

26. ganeshie8

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

27. ChillOut

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

28. ganeshie8

$$\large\color{slate}{\displaystyle \frac{d^2}{dx^2}\int\limits_{0}^{x}\left(\int\limits_{1}^{\sin(t)}\sqrt{1+u^4}du\right)dt\\~\\~\\ =\dfrac{d}{dx} \int\limits_1^{\sin(x)} \sqrt{1+u^4}du\\~\\~\\ =\sqrt{1+\sin^4 x} * (\sin x)'\\~\\~\\ }$$

29. ChillOut

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

30. anonymous

sill need help

31. IrishBoy123

Holy Cow