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Explain how we get the inequality

fmhhxx+hf(t)dtfMhh

in the proof of the Second Fundamental Theorem of Calculus. Make sure you define mhand Mhclearly.

Short Answer

Expert verified

Since mh<f(x)<Mh, fmhhxx+hf(t)dtfMhh. This is how we get the required inequality.

Step by step solution

01

Step 1. Given information

fmhhxx+hf(t)dtfMhh.

02

Step 2. The objective is to explain how the given inequality is achieved.

The Extreme value Theorem states that if a real-valued function fis continuous in the closed and bounded interval a,b, then fmust attain a maximum and a minimum, each at least once. That is, there exist numbers cand din a,bsuch that:

f(c)f(x)f(d).

So,

Mh>f(x)>mhmk<f(x)<Mh

where Mhis the maximum value on a,band mhis the minimum value on a,b.

Hence, on x,x+hthe area of the rectangles using left and right sum can be written as, fmhhand fMhhrespectively.

Since,mk<f(x)<Mhso,fmhhxx+hf(t)dtfMhh.

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