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

[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]

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)

 

[M98Ranger]

To prove this statement, we must first understand the definition of the Riemann integral. The Riemann integral is defined as the limit of a sum of rectangles that approximate the area under the curve of a function. In other words, we divide the interval [a,b] into smaller subintervals and approximate the area under the curve by using rectangles whose height is determined by the function at a particular point in the subinterval. As we make the subintervals smaller and smaller, the sum of these rectangles approaches the true area under the curve, which is the Riemann integral.

Now, let's consider the lower sum and upper sum of the function f on the interval [a,b]. The lower sum is the sum of the areas of rectangles whose heights are determined by the minimum value of f on each subinterval. Similarly, the upper sum is the sum of the areas of rectangles whose heights are determined by the maximum value of f on each subinterval. Since we know that m<=f(x)<=M for all x in [a,b], it follows that the minimum value of f on any subinterval is m and the maximum value is M.

Therefore, the lower sum is given by m(b-a) and the upper sum is given by M(b-a). Since the Riemann integral is the limit of the upper and lower sums, we can conclude that m(b-a)<=int[f(x).dx]<=M(b-a).

This can also be understood geometrically. Since f(x) is bounded between m and M, the area under the curve of f(x) must lie between the areas of rectangles with heights m and M and base (b-a). This is illustrated in the diagram below:

[Diagram showing a curve bounded by two horizontal lines with base (b-a)]

Therefore, the area under the curve, which is the Riemann integral, must also lie between m(b-a) and M(b-a).

In conclusion, from the definition of the Riemann integral, we can prove that m(b-a)<=int[f(x).dx]<=M(b-a) for any function f that is Riemann integrable and satisfies m<=f(x)<=M for all x in [a,b].