Unformatted text preview: ≤ x ≤ 2. Answer: If we split up the interval 0 ≤ x ≤ 2 into n pieces, each piece will be 2n = 2 n long. The endpoints of the ﬁrst subinterval will be 0 and 0 + 2 n = 2 n . The endpoints of the second subinterval will be 2 n and (0+ 2 n )+ 2 n = 2 2 n .The endpoints of the third subinterval will be 2 2 n and (2 2 n ) + 2 n = 3 2 n . If we label the endpoints x ,x 1 ,...,x n , then we have x i = i 2 n . The point on the curve y = x 2 corresponding to x = 0 is 0. The point corresponding to x 1 = 2 n is ‡ 2 n · 2 . And the point corresponding to x 2 = 2 2 n is ‡ 2 2 n · 2 . Thus R n = n X i =1 ± i 2 n ¶ 2 2 n = ± 2 n ¶ 3 n X i =1 i 3 . Thus an expression for the area under the curve is Area = lim n →∞ ± 2 n ¶ 3 n X i =1 i 3 . 1...
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 Spring '08
 Justin
 Calculus, Derivative, Continuous function, H° akan Nordgren

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