Find Integral
https://i.gyazo.com/879bd5e4e91be768627fbf07ebba4897.png

- anonymous

Find Integral
https://i.gyazo.com/879bd5e4e91be768627fbf07ebba4897.png

- chestercat

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- Jhannybean

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

- Jhannybean

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

- misty1212

HI!!

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## More answers

- misty1212

replace the \(t\) in the integrand by \(4x\)

- misty1212

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

- misty1212

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

- 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

Sorry for the late reply. So the answer would be C?

- SolomonZelman

yes, correct

- anonymous

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

- SolomonZelman

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

- anonymous

https://gyazo.com/a6de7e8243555fc7379828fcf09d1fd9

- SolomonZelman

Oh, then you don't need splitting

- SolomonZelman

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

- 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

Did I make sense just now?

- anonymous

Yeah, I pretty much got it. Solving atm

- SolomonZelman

atm ?

- SolomonZelman

atmosphere?

- anonymous

At the moment

- anonymous

lol

- 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

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

- SolomonZelman

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

- SolomonZelman

Anyway, what did you get for your integral ?

- anonymous

I got 3 for the final answer.

- SolomonZelman

yes, that is right

- anonymous

Thanks. Wish I could give you another medal.

- SolomonZelman

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

- anonymous

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

- SolomonZelman

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

- anonymous

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

- anonymous

Just plug in 0-2?

- SolomonZelman

I refreshed the page, I have to do it again

- anonymous

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

- SolomonZelman

Actually not D.

- SolomonZelman

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

- anonymous

Ohhhhh

- 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

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

- SolomonZelman

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

- anonymous

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

- anonymous

Should be C?

- SolomonZelman

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

- SolomonZelman

yes, it should be C.

- SolomonZelman

Good job!

- anonymous

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

- anonymous

Should be C for first and A for second?

- SolomonZelman

yes, correct for both.

- SolomonZelman

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

- 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

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

You added what for absolute value/

- SolomonZelman

?

- 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

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

- SolomonZelman

I mean because - - = +

- SolomonZelman

I have to go right now. Be well:)

- anonymous

Bye. Thanks for the help.

- SolomonZelman

You are welcome!

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