A problem with evaluating an integral

1. Oct 7, 2011

mahmoud2011

$\int^{\pi}_{0} f(x) dx$ where ,

f(x) = sin x if $0 \leq x < \frac{\pi}{2}$
and f(x) = cos(x) if $\frac{\pi}{2} \leq x \leq \pi$

The problem is that the version I am using of Fundamental theorem is if f is continuous on some closed interval , I wrote the integral as

$\int^{\pi / 2}_{0} f(x) dx + \int^{\pi}_{\pi /2} f(x) dx$

but I have in the first integral f still is not continuous on $[0,\pi/2]$

I tried to open some references and reached another version for the theorem there f is integrable on f , and g' =f , but I couldn't do any thing

Thanks

2. Oct 7, 2011

Erland

That's no problem. This integral is the same as the integral of sin x over all of [0,pi/2], since the two functions are equal except in a single point. And if you change an integrable function in a single point (or finitely many points) then the function is still integrable with the same integral. This can be seen by using upper and lower sums.

3. Oct 8, 2011

mahmoud2011

The problem is that the text book I am using doesn't introduce the integrals by first introducing the upper Darboux sums and the lower Darboux sum , it uses another approach which introduce the area as the limit of sums of approximating rectangles.

4. Oct 8, 2011

mahmoud2011

Thanks , Now I am reading in another textbook Which deals with this point precisely .

5. Oct 11, 2011

mahmoud2011

now I justified my solution as following , the first definite integral exist because f is integrable on [0,pi] (since it is piece-wise continuous ) , now since f is a function on [0,pi\2] where f= sin (x) ,0 ≤x≤ pi/2 , except at pi/2 , their definite integral is the same and then we can use the FTC , or I can use the version of the theorem which apply to any integrable function which is not necessary continuous we know that f is integrable on [0,pi/2] so we want to find a continuous function g such that g' = f on (0,pi/2) and g is continuous on [0,pi/2] so we see that this is g is sin (x) , ( I think all of this will be easy in practice ) , but what I am wandering of is why this is an exercise in my text-book , although it doesn't concentrate at this point yet .

Thanks

Last edited: Oct 11, 2011
6. Oct 11, 2011

Erland

Yes, use FTC to sin x over [0,pi/2] and observe that the given function has the same integral over this interval since it differs only at one point.

I think your textbook should have mentioned thia problem,

7. Oct 11, 2011

mahmoud2011

no , it hasn't mentioned it yet , it uses a definition of integral which depends on dividing the interval into sub intervals equal in length , so I began to read a bout the theory of integral in Advanced Calculus by agnus Taylor is this good for me .

Last edited: Oct 11, 2011
8. Oct 21, 2011

mahmoud2011

So after I have read Integration it some Advanced Calculus books , I had only one question to ask , this only to confidence myself with only a part of of a proof ( you Know that the proof in Advanced books depend leaving details for you ) , it wrote that if f is integrable , then there exist partitions P' and P'' , such that given ε >0
U(f,P') < U(f) + ε/2 and L(f,P'') > L(f) -ε/2

Here I began to justify this to my self as following , I would only now justify the first the second one is similar , we will assume it is not true, then we have

U(f,P') ≥ U(f) + ε/2 > U(f)

and this contradicts that U(f) is the grates lower bound of of all Darboux sums
But I see that this justification is not good at all because of there is P' such that U(f,P') = U(f) + ε/2 , so I began to make another one which I see is a better one assume that
U(f,P') > U(f) + ε/4 , then this gives a contradiction , so We have

U(f,P') ≤ U(f) + ε/4
and since ε/4 < ε/2
so we have U(f,P') ≤ U(f) + ε/4 < U(f) + ε/2

Is all of What I have said is true or not

Thanks