Question

help please
find the derivative of the function

Answer 1

\[g(x) = \int\limits_{-8}^{x} \frac{ 1 }{ t^2 +2 } dt \]

Answer 2

@Photon336 would we apply the FTC 1?

Answer 3

Yes, nice easy problem :d replace t with x.

Answer 4

okie so itll be
\[\frac{ 1 }{ x^2+2 }\]

Answer 5

I see a pattern here :S

that integral looks alot like this

\[\int\limits \frac{ 1 }{ 1+x^{2} } = \tan^{-1}x+C\]

Answer 6

They gave you an integral that's actually "do-able".
So yah, you could do it that way, applying FTC2 instead,
and then take derivative of your arctangent :)
just another option.

Answer 7

oh so it doenst matter it theres a x^2+2 ?

Answer 8

You mean instead of x^2+1?
Yes it does matter. It would change the arctangent slightly :)
It wouldn't give us arctan(x).

Because \(\large\rm x^2+2=2\left(\frac{1}{2}x^2+1\right)=2\left(\left(\frac{x}{\sqrt{2}}\right)^2+1\right)\)
(Factoring the 2 out in a clever way).

So this particular integral would probably give us something like,
\(\large\rm \sqrt{2}~arctan\left(\frac{x}{\sqrt{2}}\right)+C\)

Answer 9

ohhh okay now it makes sense

Answer 10

so then the answer would be tan^-1

Answer 11

If you decided to take the `normal approach`,
you would end up with some arctangent,
but that's BEFORE taking your derivative.

Derivative would bring you back to the 1/stuff squared or whatever.
So no, the answer contains no arctangent.
FTC1 did everything we needed. All in one step.

Answer 12

It basically means, whenever there is some random variable at the upper limit 'b' and a number in the bottom limit 'a' -- if the function has a variable that is different than the upper limit -- you just switch variables.

Answer 13

oh so you mean itll look like this ?