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Riemann integral

  • Thread starter dan
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dan

Hi there, I have a problem and I was wondering if anyone can help me this one.

Q)Suppose f:[a,b]->R is (Riemann) integrable and satisfies m<=f(x)<=M for all element x is a member of set [a,b]. Prove from the defintion of the Riemann integral that

m(b-a)<=int[f(x).dx]<=M(b-a).

where the upper limit = b, the lower limit = a.

I can tell you know that your help is much appreciated.
 

HallsofIvy

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To define the Riemann sums, divide the interval into n pieces of length deltax_i (may be different for i= 1 to n). Define x_i to be an x value in the ith interval. The rectangle with base the ith interval and height f(x_i) has area f(x_i)deltax_i. The sum of all those areas is an approximation to the area and it can be shown that IF f(x) is Riemann integrable over the interval [a,b] then the limit as n goes to infinity (no matter how you choose the intevals or x_i in each interval) exists and is the integral.

In this problem you know that m<= f(x)<= M for all x so, in particular, m<= f(x_i)<= M and so
m(deltax_i)<= f(x_i)(deltax_i)<= M(deltax_i)

Adding all of these for all sub intervals of [a,b] gives
sum (m deltax_i)<= sum(f(x_i)(deltax_i))<= sum M(deltax_i)

which, factoring m and M out, since they are constant) gives
m * sum(deltax_i)<= sum(f(x,i)(deltax_i))<= M*sum(deltax_i)

Of course, sum(deltax_i)= b-a, the length of the interval [a,b]
so we have

m*(b-a)<= sum(f(x,i)(deltax_i)<= M*(b-a)

Taking the limit as n-> infinity, we have

m*(b-a)<= int(x=a to b) f(x)dx)<= M*(b-a)
 

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