# 1-7 - January 7 2005 Today 5.2 The Definite Integral...

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January 7, 2005 Today § 5.2 The Definite Integral: definition and properties. 1

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When solving the area problem we en- countered, Riemann sums , which are expressions of the form n i =1 f ( x * i x = f ( x * 1 x + f ( x * 2 x + · · · + f ( x * n x. We also studied their limits, i.e. lim n →∞ n i =1 f ( x * i x = lim n →∞ [ f ( x * 1 x + f ( x * 2 x · · · + f ( x * n x ] .
Definition: Definite Integral Let f be a continuous function defined on the interval [ a, b ]. Divide the interval [ a, b ] into n -subintervals of equal width Δ x = b - a n . Let x 0 = a, x n = b, x i +1 = x i + Δ x Let x * i [ x i - 1 , x i ] be sample points. The definite integral of f from a to b is b a f ( x ) dx = lim n →∞ n i =1 f ( x * i x.

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Proposition: n i =1 i = n ( n + 1) 2 n i =1 i 2 = n ( n + 1)(2 n + 1) 6 n i =1 i 3 = n ( n + 1) 2 2 Proof (of the first): Note that n i =1 i = 1 + 2 + · · · + n n i =1 i = n + ( n - 1) + · · · + 1 Adding we have 2 n i =1 i = ( n + 1) + ( n + 1) + · · · + ( n + 1) = n ( n + 1) .
To compute n i =1 i 2 we use a telescoping sum . Note that ( i + 1) 3 - i 3 = 3 i 2 + 3 i + 1 and n i =1 [( i + 1) 3 - i 3 ] = [2 3 - 1 3 ] + [3 3 - 2 3 ] + · · · + [( n + 1) 3 - n 3 ] = ( n + 1) 3 - 1 = n 3 + 3 n 2 + 3 n. Thus n i =1 [( i + 1) 3 - i 3 ] = n i =1 3 i 2 + 3 i + 1 = 3 n i =1 i 2 + 3 n i =1 i + n i =1 1 = 3 n i =1 i 2 + 3 n ( n + 1) 2 + n.

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Combining the 2 equalities above we have 3 n i =1 i 2 = n 3 + 3 n 2 - 3 n ( n + 1) 2 + 2 n = n 3 + 3 2 n 2 + 1 2 n = 2 n 3 + 3 n 2 + n 2 = n ( n + 1)(2 n + 1) 2
Properties of sums

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