an interesting integral

[@cosmiccake791]

If an apple falling on someone's head caused this insanity imagine a coconut falling😂

[@ZazaNugget_]

I think I just fell out of a coconut tree
Also I’m currently being raised in a middle class family

[@MASHabibi-d2d]

Thanks for an other video.

[@ahmedsabri-ng2xx]

an interesting integral

[@holyshit922]

I would use by parts twice

[@physics_rev]

the limits should have changed man, when you converted the
(1/x^2) to - d(1/x)

[@glady.]

how does that work? im a highschooler and im curious cuz we never learnt this

[@physics_rev]

​@@glady. whenever you want to make an integral easy, you use a substitution like
x = f(t) some other function and dx = f'(t) and since the integral's previous limits where based on x, and you have a integral in t, you've gotta change the limits for ex: if the limits where 0 to 1 on x , and hade made a substitution: x = sin(t) then i would get the new limits of the integral to be, i) x=0; sint=0; " t=0 " ii) x=1; sint=1; " t=pi/2 "
hence new limits 0 to pi/2
hope you understood :)

[@physics_rev]

​@@glady. also, mate it's a point to be noted that the limits aregiven for an integral based on what is after the letter "d" like, if its dx then the limits are for x, and lets say if its d(x^-2) then the limits are for (x^-2)

[@glady.]

@@physics_rev got itt thanks dude!

[@UD__07]

Nowadays people are playing with integrals like they know everything limits should've been changed your answer is approximated to 0.066626312481 while the actual answer to this is 0.0125665018627

[@seegeeaye]

Unless there’s an error, my answer should be right

[@maxvangulik1988]

sign error on the first term

[@maxvangulik1988]

I(a)=int[1,2](x^a)dx
I'(a)=int[1,2](x^a•ln(x))dx
I''(a)=int[1,2](x^a•ln^2(x))dx
I''(-2)=I
I(a)=(2^(a+1)-1)/(a+1)
I'(a)=2^(a+1)•ln(2)/(a+1)-(2^(a+1)-1)/(a+1)^2
I''(a)=2^(a+1)•ln^2(2)/(a+1)-2^(a+2)•ln(2)/(a+1)^2+(2^(a+2)-2)/(a+1)^3
I''(-2)=-ln^2(2)/2-ln(2)+1
I=1-ln(2)-ln^2(2)/2