5.2.11-eg-1 - ) x, i.e., a simpler expression which does...

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Example Find the area under the curve y = f ( x ) = 100 - 3 x 2 , from x = 1 to x = 5, using n equal-width rectangles and right endpoints, and taking the limit as n → ∞ . Solution: The region is graphed below: The steps for computing the shaded area are as follows: (a) The interval [1 , 5] is divided into n equal subintervals of length Δ x , where Δ x = 5 - 1 n = 4 n . (b) We will use the notation c k to represent the right-hand endpoint of the k th subinterval. Since c 1 = 1 + Δ x = 1 + 4 n , c 2 = 1 + 2Δ x = 1 + 2 4 n , c 3 = 1 + 3Δ x = 1 + 3 4 n , . . . . . . c k = 1 + k Δ x = 1 + k 4 n , so it follows that c k = 1 + k 4 n . (c) The area is then given by Area = lim n →∞ n X k =1 f ( c k x, where, using the fact that f ( x ) = 100 - 3 x 2 and the formula for c k given above, f ( c k ) = 100 - 3 c 2 k = 100 - 3 ± 1 + k 4 n ² 2 = 100 - 3 ± 1 + 2 k 4 n + k 2 16 n 2 ² = 97 - 24 n k - 48 n 2 k 2 ,
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so it follows that f ( c k x = ± 97 - 24 n k - 48 n 2 k 2 ²± 4 n ² . (d) We wish to obtain a closed-form expression for n X k =1 f ( c k
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Unformatted text preview: ) x, i.e., a simpler expression which does not involve the summation symbol. Since n X k =1 f ( c k ) x = n X k =1 97-24 n k-48 n 2 k 2 4 n = 4 n n X k =1 97-24 n k-48 n 2 k 2 = 388 n n X k =1 1-96 n 2 n X k =1 k-192 n 3 n X k =1 k 2 , we may use standard summation formulas to simplify, leading to n X k =1 f ( c k ) x = 388 n n-96 n 2 1 2 n 2 + 1 2 n -192 n 3 1 3 n 3 + 1 2 n 2 + 1 6 n = 388-48-48 n-64-96 n-32 n 2 = 276-144 n-32 n 2 , a closed-form representation of the summation. (e) Finally, the area is given by Area = lim n n X k =1 f ( c k ) x = lim n n X k =1 276-144 n-32 n 2 = 276 . 2...
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5.2.11-eg-1 - ) x, i.e., a simpler expression which does...

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