Help with Proof on Integration

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    Integration Proof
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I was assigned this problem in class. My instructor said it was a very popular theorem, but I cannot find it in my book or online. I am clueless on what to do. I would appreciate the help.

Let f(x) be bounded and integrable on [a, b]. Assume that g(x) differs from f(x) on only finitely many points in the domain. Show that g(x) is integrable. Moreover, show that ∫f(x)dx = ∫g(x)dx (Both integrals are from b to a).
 
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welcome to pf!

hi tomhawk24! welcome to pf! :smile:

start with the definition

which definition of integral (or integrable) are you using?
 
Well we are working mainly on the Fundamental Theorem of Calculus right now.
 
ok, we'll start with that, then …

what does the fundamental theorem of calculus say? :smile:
 
Starting from the area interpretation of the integral, answer this question: If I take finitely many points out of the graph of a curve f(x) and place them at some other y-coordinate, would the function still be integrable? What would be its integral?

Tip: Does a point have dimensions, or does a line have width? What is the area of a rectangle?
 
First show
∫(f(x)-g(x))dx =0
then use linearity
 
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...
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