## anonymous one year ago Find Integral https://i.gyazo.com/879bd5e4e91be768627fbf07ebba4897.png

1. Jhannybean

$\sf \frac{d}{dx}\int_1^{4x} \sqrt{t^2+1}dt = \frac{d}{dx}\left[F(4x)-F(1)\right]$

2. Jhannybean

Remember that $$\sf (F(t))' = f(t)$$ and $$\sf f(t) =\sqrt{t^2+1}$$ So what would $$\sf (F(x))'$$ = ?

3. misty1212

HI!!

4. misty1212

replace the $$t$$ in the integrand by $$4x$$

5. misty1212

then, via the chain rule, multiply the whole thing by 4

6. misty1212

$\sqrt{\left(\text{ put 4$$x$$ here}\right)^2+1}\times 4$

7. SolomonZelman

Anytime, in general: $$\large\color{black}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right] }$$ $$\large\color{black}{ \displaystyle \frac{d}{dx}\left[F(t)~{\Huge |}^{g(x)}_{\rm C}~\right] }$$ $$\large\color{black}{ \displaystyle \frac{d}{dx}\left[~F(g(x))-F({\rm C})~\right] }$$ $$\large\color{black}{ \displaystyle F'(g(x))\cdot g'(x)-0=F'(g(x))\cdot g'(x) }$$ ------------------------------------------------- Conclusion: $$\large\color{blue}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right]=F'(g(x))\cdot g'(x) }$$

8. anonymous

9. SolomonZelman

yes, correct

10. anonymous

@SolomonZelman How would I solve an integral for an absolute value?

11. SolomonZelman

You would split the function for x>0 and x<0

12. anonymous
13. SolomonZelman

Oh, then you don't need splitting

14. SolomonZelman

The function from 0 to -3 is never positive. So you can just write $$\displaystyle \int_{-3}^{0}-(x+2)dx$$

15. SolomonZelman

because absolute value function on the interval (0,3) is negative (at least not positive). So, it is just a line y=-(x+2) at this interval.

16. SolomonZelman

Did I make sense just now?

17. anonymous

Yeah, I pretty much got it. Solving atm

18. SolomonZelman

atm ?

19. SolomonZelman

atmosphere?

20. anonymous

At the moment

21. anonymous

lol

22. SolomonZelman

No seriously, I am terrible at text language. I have hopefully managed to learn proper English, but text language is not for me. I guess it is too informal.

23. anonymous

Oh well, lol meant laugh out loud, just in case you didn't know.

24. SolomonZelman

I know "lol" "jk" and some others perhaps, but I think this is pretty much it.

25. SolomonZelman

Anyway, what did you get for your integral ?

26. anonymous

I got 3 for the final answer.

27. SolomonZelman

yes, that is right

28. anonymous

Thanks. Wish I could give you another medal.

29. SolomonZelman

I don't really care about medals, I got over 12K:)

30. anonymous

LOL, then you wouldn't mind helping me with another? :p

31. SolomonZelman

I guess not, if I will be able to:)

32. anonymous

This one should be easy, just can't remember what to do. https://gyazo.com/eb75236776b0079f4699d2e338fe6ec5

33. anonymous

Just plug in 0-2?

34. SolomonZelman

I refreshed the page, I have to do it again

35. anonymous

Ok, well anyways, if I recall properly the answer should be D.

36. SolomonZelman

Actually not D.

37. SolomonZelman

$$\large\color{black}{ \displaystyle e^0=1 }$$

38. anonymous

Ohhhhh

39. SolomonZelman

Proof: $$\large\color{black}{ \displaystyle a^0=a^{b-b}=\frac{a^b}{a^b}=\frac{\cancel{a^b}}{\cancel{a^b}}=1 }$$

40. anonymous

By the way would I plug in 0,1, and 2 or just 0 and 2?

41. SolomonZelman

Just plug in 2, and then 0 F(2)-F(0)

42. anonymous

This is what I did. $\frac{ 1 }{ 4}e ^{4x}|_{0}^{2}$

43. anonymous

Should be C?

44. SolomonZelman

yes, correct, that is what you had to do:)

45. SolomonZelman

yes, it should be C.

46. SolomonZelman

Good job!

47. anonymous

One last one please. Actually two but one I already solved. https://gyazo.com/f8561038cb4b1113268276fbaf824c2e https://gyazo.com/e7aea1e2a2b3ca9e81319fcbe8b70cb1

48. anonymous

Should be C for first and A for second?

49. SolomonZelman

yes, correct for both.

50. SolomonZelman

But, you have to still know that as $$x\rightarrow0$$, then $$\ln(x)\rightarrow \infty$$.

51. anonymous

Appreciate the help. Just so you know I got the absolute value one wrong, but I see where my mistake is. I added instead of subtracted.

52. SolomonZelman

just in case... and this you can easily demonstrate by defining zero as: $$\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[10^{-n} \right] }$$ And then, $$\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[~~\log_{10}\left(10^{-n}\right) \right] = \lim_{n\rightarrow \infty} \left[-n\log_{10}(10)\right]=\lim_{n\rightarrow \infty} \left[-n\right]=-\infty}$$

53. SolomonZelman

You added what for absolute value/

54. SolomonZelman

?

55. anonymous

instead of F[0]-F[-3] I put F[0]+F[-3] which...........now that I look at it I did do right no? That would give me 3. :/

56. SolomonZelman

yes, because F[0]+F[-3]

57. SolomonZelman

I mean because - - = +

58. SolomonZelman

I have to go right now. Be well:)

59. anonymous

Bye. Thanks for the help.

60. SolomonZelman

You are welcome!