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Calculus and Beyond Homework Help
Using Upper, Lower Sum, Prove the Following
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[QUOTE="HallsofIvy, post: 4306844, member: 637751"] As long as a and b are positive, [itex]x^2[/itex] is an increasing function so the maximum value on each interval, [itex]M_i[/itex], is at the right end and the minimum value on each interval, [itex]m_i[/itex], is at the left end. Further, if you divide [a, b] into n intervals, each interval has length [itex]t_i- t_{i-1}= (b- a)/n[/itex], the left endpoint is a+ (b- a)i/n for i= 0 to n-1, and the right endpoint is a+ (b- a)i/n for i= 1 to n. That is, [itex]M_i= (a+ (b-a)i/n)^2[/itex], for i= 1 to n and [itex]m_i= (a+ (b-a)i/n)^2[/itex] for i= 0 to n-1. [itex]M_i(t_i- t_{i-1})= ((b-a)/n)(a+ (b-a)i/n)^2[/itex] for i= 1 to n and [itex]m_i(t_i- t_{i-1})= ((b- a)/n)(a+ (b- a)i/n)^2[/itex] Multiply those out and sum. It will help if you know that [itex]\sum_{i=1}^n c= nc[/itex], [itex]\sum_{i=1}^n i= n(n+ 1)/2[/itex] and [itex]\sum_{i=1}^n i^2= [n(n+1)(2n+1)]/6[/itex]. Those should be in your text. [/QUOTE]
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Using Upper, Lower Sum, Prove the Following
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