Question

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

Answer 1

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

Answer 2

Remember that \(\sf (F(t))' = f(t)\) and \(\sf f(t) =\sqrt{t^2+1}\)

So what would \(\sf (F(x))'\) = ?

Answer 3

HI!!

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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:

\(\large\color{blue}{ \displaystyle \frac{d}{dx}\left[\int_{\rm C}^{g(x)} F'(t)dt\right]=F'(g(x))\cdot g'(x) }\)

Answer 8

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

Answer 9

yes, correct

Answer 10

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

Answer 11

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

Answer 12

https://gyazo.com/a6de7e8243555fc7379828fcf09d1fd9

Answer 13

Oh, then you don't need splitting

Answer 14

The function from 0 to -3 is never positive.

So you can just write \(\displaystyle \int_{-3}^{0}-(x+2)dx \)

Answer 15

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.

Answer 16

Did I make sense just now?

Answer 17

Yeah, I pretty much got it. Solving atm

Answer 18

atm ?

Answer 19

atmosphere?

Answer 20

At the moment

Answer 21

lol

Answer 22

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.

Answer 23

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

Answer 24

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

Answer 25

Anyway, what did you get for your integral ?

Answer 26

I got 3 for the final answer.

Answer 27

yes, that is right

Answer 28

Thanks. Wish I could give you another medal.

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Just plug in 0-2?

Answer 34

I refreshed the page, I have to do it again

Answer 35

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

Answer 36

Actually not D.

Answer 37

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

Answer 38

Ohhhhh

Answer 39

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

Answer 40

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

Answer 41

Just plug in 2, and then 0

F(2)-F(0)

Answer 42

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

Answer 43

Should be C?

Answer 44

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

Answer 45

yes, it should be C.

Answer 46

Good job!

Answer 47

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

Answer 48

Should be C for first and A for second?

Answer 49

yes, correct for both.

Answer 50

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

Answer 51

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.

Answer 52

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,
\(\large\color{black}{ \displaystyle \lim_{n\rightarrow \infty}\left[~~\log_{10}\left(10^{-n}\right) \right] = \lim_{n\rightarrow \infty} \left[-n\log_{10}(10)\right]=\lim_{n\rightarrow \infty} \left[-n\right]=-\infty}\)

Answer 53

You added what for absolute value/

Answer 54

?

Answer 55

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. :/

Answer 56

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

Answer 57

I mean because - - = +

Answer 58

I have to go right now. Be well:)

Answer 59

Bye. Thanks for the help.

Answer 60

You are welcome!