Integration Question: Solutions & Tips

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The integral of 1/x with respect to x is ln|x| + C, so if you took the integral from a to b, you would get
ln|b|-ln|a|
and as a goes to 0, this would be infinity.

(Note that you shouldn't say you integrate 1/x from 0 to x. You can't have the limit be the variable you are integrating. If you think about what that would mean, you will see that the integral 0f f(x) from 0 to x is impossible to make sense of, since x would vary from 0 to x, whatever that means)
 
Yer I did abuse the notation a little, sorry for that.

And sorry for the question as soon as I posted it I realized how stupid it was, hence the delete.

I was just reading a paper on zero and infinity and it got me thinking...the paper didn't really like infinity.
 
What I was questioning is, because the curve never reaches the y-axis (x=0), except for at infinity, then the little area between the curve and the y axis, whilst getting smaller and smaller never quite ends.
 
KingBigness said:
What I was questioning is, because the curve never reaches the y-axis (x=0), except for at infinity, then the little area between the curve and the y axis, whilst getting smaller and smaller never quite ends.

The same could be said for 1/x2, but the definite integral does have a value (assuming the lower limit isn't at 0).
 
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