Proving the Limit of Integrable Functions on a Closed Interval

  • Thread starter Thread starter tomboi03
  • Start date Start date
  • Tags Tags
    Functions Limit
tomboi03
Messages
74
Reaction score
0
Prove: If f is integrable on [a , b] then
lim f =0
x\rightarrowa+

the integral goes from a to x.

How do i go about and prove this? I'm confused.
Please help me out!
Thank You
 
Last edited:
Physics news on Phys.org


Do you mean
\lim_{x\rightarrow a^+} \int_a^x f(t)dt= 0

The way you have written it, that the limit of f is 0, makes no sense- that certainly is not necessarily true.

My suggestion here is the same as to your other question: use the definition of integral in terms of Riemann sums.
 


I've never learn Riemann sum definition.
What is that?
 


First, is what I wrote what you mean. And if you have never learned Riemann sums, what definition of \int_a^b f(x)dx are you using?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top