A limit of an integral question

[transgalactic]

http://img225.imageshack.us/my.php?image=29117598dh7.png

i know that the integral of 0 to 0 interval equals to 0.

i get 0/0 form

how do i make the derivative of this integral

i know the it should cancel out the integral sign
but there is another variable "t"

what should i do in this
case??

another way that i thought of is
to solve this integral and then solve the limit

 

[sutupidmath]

http://img225.imageshack.us/my.php?image=29117598dh7.png

i know that the integral of 0 to 0 interval equals to 0.

i get 0/0 form

how do i make the derivative of this integral

i know the it should cancel out the integral sign
but there is another variable "t"

what should i do in this
case??

another way that i thought of is
to solve this integral and then solve the limit

Yeah just take the derivative of the integral, and the result will be the exact function under the integral sign, but now instead of the variable t, the result should contain the variable x. this is a well known theorem, and it can be proven, but it requres a little work.

So when u take the derivative of the integral on the denominator the result will be
ln(1+x^{2})

 

[sutupidmath]

By the way, you cannot actually solve the integral, because it does not have an elementary function, you just can express in terms of some gamma function, or just find an approximation for it. But there is no close form of it.

 

[Rainbow Child]

By the way, you cannot actually solve the integral, because it does not have an elementary function, you just can express in terms of some gamma function, or just find an approximation for it. But there is no close form of it.

Actually this is not true. Integrating by parts you can calulate the integral

\int \ln(1+x^2)\,d\,x=x\,\ln(1+x^2)-2\int \frac{x^2}{1+x^2}\,d\,x=x\,\ln(1+x^2)-2\int \frac{1+x^2}{1+x^2}\,d\,x+2\,\int\frac{d\,x}{1+x^2}=x\,\ln(1+x^2)-2\,x+2\,\arctan x

But for the OP, you don't have to do this!

 

[sutupidmath]

Actually this is not true. Integrating by parts you can calulate the integral

\int \ln(1+x^2)\,d\,x=x\,\ln(1+x^2)-2\int \frac{x^2}{1+x^2}\,d\,x=x\,\ln(1+x^2)-2\int \frac{1+x^2}{1+x^2}\,d\,x+2\,\int\frac{d\,x}{1+x^2}=x\,\ln(1+x^2)-2\,x+2\,\arctan x

But for the OP, you don't have to do this!

Pardone me! I was thinking of completely something else! Yeah, you are so right, it actually is just an easy thing to do, but i had something else in my mind, i mean another function, when i said that. My bad lol, i appologize for not paying so much attention to things!
Thnx for pointing this out by the way.

 

[Rainbow Child]

... i appologize for not paying so much attention to things!

Do not! That's what makes us humans!

 

[Gib Z]

Yeah just take the derivative of the integral, and the result will be the exact function under the integral sign, but now instead of the variable t, the result should contain the variable x. this is a well known theorem, and it can be proven, but it requires a little work.

I believe that's a little something called the Fundamental Theorem of Calculus =]

 

[sutupidmath]

I believe that's a little something called the Fundamental Theorem of Calculus =]

Well maybe it is, but i am not used with English notation or naming of theorems. But i know how we call it in my native language, and we did not call it The Fundamental Theorem of Calculus, i mean even if we translated this into my language.

 

[Gib Z]

O sorry my bad! I didn't know english was not your first language. Your english is very good =]

 

[sutupidmath]

O sorry my bad! I didn't know english was not your first language. Your english is very good =]

Well no need to appologize, and thnx for commenting on my English!

 

[transgalactic]

i solved it like you told me

is this solution correct??

http://img153.imageshack.us/my.php?image=img8218hw4.jpg

 

[Gib Z]

That indeed is correct.

 

[transgalactic]

thanks

 

[coomast]

i solved it like you told me

is this solution correct??

http://img153.imageshack.us/my.php?image=img8218hw4.jpg

No, it isn't. You forgot a factor -1/2 with the derivative of the square root. Your solution is not 1. Can you check it again please?

 

[Gib Z]

Do you think they'll ever develop a drug that cures my stupidity? I was specifically checking that for errors and I didn't see that. Sigh

 

[transgalactic]

damn chain rule