anonymous
  • anonymous
Find Integral https://i.gyazo.com/879bd5e4e91be768627fbf07ebba4897.png
Calculus1
chestercat
  • chestercat
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Jhannybean
  • Jhannybean
\[\sf \frac{d}{dx}\int_1^{4x} \sqrt{t^2+1}dt = \frac{d}{dx}\left[F(4x)-F(1)\right]\]
Jhannybean
  • Jhannybean
Remember that \(\sf (F(t))' = f(t)\) and \(\sf f(t) =\sqrt{t^2+1}\) So what would \(\sf (F(x))'\) = ?
misty1212
  • misty1212
HI!!

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misty1212
  • misty1212
replace the \(t\) in the integrand by \(4x\)
misty1212
  • misty1212
then, via the chain rule, multiply the whole thing by 4
misty1212
  • misty1212
\[\sqrt{\left(\text{ put 4\(x\) here}\right)^2+1}\times 4\]
SolomonZelman
  • 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: \(\large\color{blue}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right]=F'(g(x))\cdot g'(x) }\)
anonymous
  • anonymous
Sorry for the late reply. So the answer would be C?
SolomonZelman
  • SolomonZelman
yes, correct
anonymous
  • anonymous
@SolomonZelman How would I solve an integral for an absolute value?
SolomonZelman
  • SolomonZelman
You would split the function for x>0 and x<0
anonymous
  • anonymous
https://gyazo.com/a6de7e8243555fc7379828fcf09d1fd9
SolomonZelman
  • SolomonZelman
Oh, then you don't need splitting
SolomonZelman
  • SolomonZelman
The function from 0 to -3 is never positive. So you can just write \(\displaystyle \int_{-3}^{0}-(x+2)dx \)
SolomonZelman
  • 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.
SolomonZelman
  • SolomonZelman
Did I make sense just now?
anonymous
  • anonymous
Yeah, I pretty much got it. Solving atm
SolomonZelman
  • SolomonZelman
atm ?
SolomonZelman
  • SolomonZelman
atmosphere?
anonymous
  • anonymous
At the moment
anonymous
  • anonymous
lol
SolomonZelman
  • 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.
anonymous
  • anonymous
Oh well, lol meant laugh out loud, just in case you didn't know.
SolomonZelman
  • SolomonZelman
I know "lol" "jk" and some others perhaps, but I think this is pretty much it.
SolomonZelman
  • SolomonZelman
Anyway, what did you get for your integral ?
anonymous
  • anonymous
I got 3 for the final answer.
SolomonZelman
  • SolomonZelman
yes, that is right
anonymous
  • anonymous
Thanks. Wish I could give you another medal.
SolomonZelman
  • SolomonZelman
I don't really care about medals, I got over 12K:)
anonymous
  • anonymous
LOL, then you wouldn't mind helping me with another? :p
SolomonZelman
  • SolomonZelman
I guess not, if I will be able to:)
anonymous
  • anonymous
This one should be easy, just can't remember what to do. https://gyazo.com/eb75236776b0079f4699d2e338fe6ec5
anonymous
  • anonymous
Just plug in 0-2?
SolomonZelman
  • SolomonZelman
I refreshed the page, I have to do it again
anonymous
  • anonymous
Ok, well anyways, if I recall properly the answer should be D.
SolomonZelman
  • SolomonZelman
Actually not D.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle e^0=1 }\)
anonymous
  • anonymous
Ohhhhh
SolomonZelman
  • SolomonZelman
Proof: \(\large\color{black}{ \displaystyle a^0=a^{b-b}=\frac{a^b}{a^b}=\frac{\cancel{a^b}}{\cancel{a^b}}=1 }\)
anonymous
  • anonymous
By the way would I plug in 0,1, and 2 or just 0 and 2?
SolomonZelman
  • SolomonZelman
Just plug in 2, and then 0 F(2)-F(0)
anonymous
  • anonymous
This is what I did. \[\frac{ 1 }{ 4}e ^{4x}|_{0}^{2}\]
anonymous
  • anonymous
Should be C?
SolomonZelman
  • SolomonZelman
yes, correct, that is what you had to do:)
SolomonZelman
  • SolomonZelman
yes, it should be C.
SolomonZelman
  • SolomonZelman
Good job!
anonymous
  • anonymous
One last one please. Actually two but one I already solved. https://gyazo.com/f8561038cb4b1113268276fbaf824c2e https://gyazo.com/e7aea1e2a2b3ca9e81319fcbe8b70cb1
anonymous
  • anonymous
Should be C for first and A for second?
SolomonZelman
  • SolomonZelman
yes, correct for both.
SolomonZelman
  • SolomonZelman
But, you have to still know that as \(x\rightarrow0\), then \(\ln(x)\rightarrow \infty\).
anonymous
  • 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.
SolomonZelman
  • 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, \(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[~~\log_{10}\left(10^{-n}\right) \right] = \lim_{n\rightarrow \infty} \left[-n\log_{10}(10)\right]=\lim_{n\rightarrow \infty} \left[-n\right]=-\infty}\)
SolomonZelman
  • SolomonZelman
You added what for absolute value/
SolomonZelman
  • SolomonZelman
?
anonymous
  • 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. :/
SolomonZelman
  • SolomonZelman
yes, because F[0]+F[-3]
SolomonZelman
  • SolomonZelman
I mean because - - = +
SolomonZelman
  • SolomonZelman
I have to go right now. Be well:)
anonymous
  • anonymous
Bye. Thanks for the help.
SolomonZelman
  • SolomonZelman
You are welcome!

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