Proof with Darboux integral - question.

  • Thread starter Thread starter peripatein
  • Start date Start date
  • Tags Tags
    Integral Proof
peripatein
Messages
868
Reaction score
0
Hello,

Homework Statement


I was asked to prove that for f(x), Riemann integrable and bounded from above by real number M:
∫[a,b] f(x) < M(b-a)


Homework Equations





The Attempt at a Solution


Since f(x) is Riemann integrable, ∫[a,b] f(x) must be equal to both the lower and upper Darboux sums. Therefore: ∫[a,b] f(x) = supf(x) [a,b]Ʃ[i=1,n]Δxi = supf(x) [a,b](b-a) <= M(b-a)
I am really not sure this is rigorous enough, or even how it ought to be approached. I'd appreciate some guidance please.
 
Physics news on Phys.org
I think this can be seen as a consequence of the integral mean value theorem.
 
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