- #1
njama
- 216
- 1
The 2nd part of fundamental theorem of calculus says:
Over what open interval does the formula
[tex]F(x)=\int_{1}^{x}\frac{dt}{t}[/tex]
represent antiderivative of f(x)=1/x ?
By looking at the theorem I would say that f(x) is continuous only for [tex]x \neq 0[/tex]
So I would say that F(x) is defined on [tex](-\infty, 0) U (0, +\infty)[/tex] that is F'(x)=f(x) for all numbers [tex]x \neq 0[/tex].
But there is problem. F(x) actually equals Log(x) , because (Log(x))' = 1/x.
Why this theorem does not work?
Now there is problem with the fundamental theorem of calculus part 1:
[tex]\int_{0}^{2}|2x-3|dx=\frac{|2x-3|x(x-3)}{2x-3}|_{0}^{2}[/tex]
But again the area is not -2 it is actually 5/2.
One more thing:
[tex](\frac{|2x-3|x(x-3)}{2x-3})' = |2x-3|[/tex] where [itex]x \neq 3/2[/itex] because F(x) is not defined at 3/2
Which is not same as |2x-3| which is defined everywhere.
Thanks in advance.
If f is continuous on an interval I, then f has an anti derivative on I. In particular, if a is any number in I, then the function F defined by
[tex]F(x)=\int_{a}^{x}f(t)dt[/tex]
is an antiderivative of f on I; that is F'(x) = f(x) for each x in I.
Over what open interval does the formula
[tex]F(x)=\int_{1}^{x}\frac{dt}{t}[/tex]
represent antiderivative of f(x)=1/x ?
By looking at the theorem I would say that f(x) is continuous only for [tex]x \neq 0[/tex]
So I would say that F(x) is defined on [tex](-\infty, 0) U (0, +\infty)[/tex] that is F'(x)=f(x) for all numbers [tex]x \neq 0[/tex].
But there is problem. F(x) actually equals Log(x) , because (Log(x))' = 1/x.
Why this theorem does not work?
Now there is problem with the fundamental theorem of calculus part 1:
Here is the counterexample:If f is continuous on [a,b] and F is any anti derivative of f on [a,b], then
[tex]\int_{a}^{b}f(x)dx=F(b)-F(a)[/tex]
[tex]\int_{0}^{2}|2x-3|dx=\frac{|2x-3|x(x-3)}{2x-3}|_{0}^{2}[/tex]
But again the area is not -2 it is actually 5/2.
One more thing:
[tex](\frac{|2x-3|x(x-3)}{2x-3})' = |2x-3|[/tex] where [itex]x \neq 3/2[/itex] because F(x) is not defined at 3/2
Which is not same as |2x-3| which is defined everywhere.
Thanks in advance.