## Argonx16 Group Title Calculus! (See comments) one year ago one year ago

1. Argonx16 Group Title

$\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}$ Preferably with steps+rationale shown. Thanks!

2. Argonx16 Group Title

If the equation is too small to see: Find the derivative of the integral of squrt(1+t^2)dt from cos3x to tan 3x

3. Argonx16 Group Title

(Second fundamental therom)

4. mathg8 Group Title

check FTC 2

5. mathg8 Group Title

:)

6. wio Group Title

Suppose you have the following: $F(x) = \int_a^{g(x)}f(t)dt$

7. Argonx16 Group Title

So @mathg8 , so far in my class we have only solved problems with one of the limits of integration being an expression, and one would be a constant (ie 3), how would this differ.

8. wio Group Title

Let $$y = g(x)$$

9. kirbykirby Group Title

You would have to split the integral at some constant (say a)

10. wio Group Title

Then suppose : $H(y) = \int_a^yf(t)dt$Notice that $$H(y) = H(g(x)) = F(x)$$

11. wio Group Title

By the fundamental theorem: $\frac{d}{dy}H(y) = f(y) = f(g(x))$

12. kirbykirby Group Title

$\frac {d}{dx} \int\limits\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2}dt = \frac {d}{dx} \int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}dt+\frac {d}{dx} \int\limits\limits_{a}^{\tan 3x} \sqrt{1+t^2}dt$

13. wio Group Title

See where I'm going with this?

14. wio Group Title

@Argonx16

15. kirbykirby Group Title

and $\int\limits\limits_{\cos 3x}^{a} \sqrt{1+t^2}=-\int\limits\limits_{a}^{\cos3x} \sqrt{1+t^2}$

16. Argonx16 Group Title

@wio I'm reading through it (:

17. kirbykirby Group Title

If you know how to work with an expression other than just "x" on the upper integration bound, then you should have no problem computing it the way I presented it by splitting your integral into 2

18. wio Group Title

Basically we have: $\frac{d}{dy}F(x) = f(g(x))$

19. Argonx16 Group Title

@kirbykirby Ok, thanks!

20. Argonx16 Group Title

@wio here, let me put what you said in words to see if I get it

21. wio Group Title

@Argonx16 I'm not finished yet but sure.

22. Argonx16 Group Title

If you have f(x) = integral of f(t) from a to g(x), and y=g(x), and $H(y)=\int\limits_{a}^{y}f(t)dt$ Then we can see that H(y)=H(g(x)) because of the fact that y=g(x) (Not quite sure how H(y)=H(g(x)) though) @wio

23. wio Group Title

When we say $$y=g(x)$$ we're just saying they are interchangeable.

24. wio Group Title

The reason I did that was to make it so you can see that a function is sort of just like a variable.

25. wio Group Title

also there is a difference between $$F(x)$$ and $$f(x)$$

26. Argonx16 Group Title

Sorry, my reply was written wrong. Let me restate it.

27. Argonx16 Group Title

I can see that H(y) = H (g(x)) (y=g(x)), but I /don't/ see how H(y)=F(x) (or H(g(x)=F(x))

28. wio Group Title

It's the transitive property of equality. $H(y)=\int\limits_{a}^{y}f(t)dt = \int\limits_{a}^{g(x)}f(t)dt =F(x)$I'm not saying they are the same function though. I'm just saying they are equal.

29. wio Group Title

Look if you want, just ignore $$H(y)$$. If I said: $F(x)=\int\limits_{a}^{y}f(t)dt$Do you understand that by the fundamental theorem: $\frac{d}{dy}F(x) = f(y)$

30. Argonx16 Group Title

Ok, I see what you did to make F(x)=H(y)=H(g(x)) And, yes I understand the second fundamental theorem.

31. wio Group Title

Now the trick here is to multiply both sides by $$dy/dx$$ $\frac{d}{dy}F(x) \frac{dy}{dx} = f(y)\frac{dy}{dx}$

32. wio Group Title

By the chain rule: $\frac{d}{dy}F(x) \frac{dy}{dx} = \frac{d}{dx}F(x) = f(y)\frac{dy}{dx}$

33. wio Group Title

@Argonx16 this is sort of the tricky part, but do you get it?

34. wio Group Title

Another way to write this is: $\frac{dF(x)}{dy}\frac{dy}{dx} = \frac{dF(x)}{dx}$Or $\frac{dF}{dy}\frac{dy}{dx} = \frac{dF}{dx}$

35. Argonx16 Group Title

Yes I got the chain rule step (esp after you rewrote it in a different way).

36. wio Group Title

So we got:$\frac{d}{dx}F(x) = \frac{d}{dx}\int\limits_{a}^{y}f(t)dt = \frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt$And: $f(y)\frac{dy}{dx} = f(g(x))\frac{dg(x)}{dx} = f(g(x))g'(x)$

37. wio Group Title

Putting it all together: $\frac{d}{dx} \left[ \int\limits_{a}^{g(x)}f(t)dt \right] = f(g(x))g′(x)$

38. wio Group Title

$\frac {d}{dx} \int\limits_{\cos 3x}^{\tan 3x} \sqrt{1+t^2} = \frac {d}{dx} \int\limits_{z}^{\tan 3x} \sqrt{1+t^2} -\frac {d}{dx} \int\limits^{\cos 3x}_{a} \sqrt{1+t^2}$

39. wio Group Title

To expand on the earlier realization: $\frac{d}{dx} \left[ \ \int\limits_{h(x)}^{g(x)}f(t)dt \right] = f(g(x))g′(x) -f(h(x))h'(x)$

40. wio Group Title

@Argonx16 Does it all make sense now?

41. Argonx16 Group Title

Yes! @wio Thanks for spending so much time helping me on this! I know how much patience/time/effort it takes to help so in depth, so I am very appreciative of your help :D

42. wio Group Title

I've done this thing like 2 or 3 times and every time I it gets a bit easier to teach.

43. Argonx16 Group Title

Yeah. It definitely helps so much more. OS is also a good way to retain math concepts taught earlier :D