Prove that ∫aaf(x)dx=0.

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Lesson Parent: 
Integration III: Definite Integrals As Upper and Lower Sums
Problem: 

Prove that
∫aaf(x)dx=0.

Answer: 

It is true that ∫aaf(x)dx=0.

SAMPLE SOLUTION

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Solution: 

From our definition of a definite integral we know that

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where m is the minimum of the function on [a,b] and M is the maximum value of the function on [a,b]. Furthermore, we know that all partitions P will satisfy

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If we set b=a then we know that <Sign in to see all the formulas> and <Sign in to see all the formulas> and we have

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Therefore,

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Before we end the proof we want to make one final comment. Our definition requires us to find a number for <Sign in to see all the formulas> such that

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for all possible partitions. However, in this case there are no partitions because there are no points between a andb (because a=b).