Then since h u eu we get t 1 0u e0 e0 u 0

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Unformatted text preview: lus to evaluate integrals involving this function. This is somewhat problematic since integrating functions similar to f (x) is very often necessary in the statistical analysis of data, as well as many other applications. As such we are left to use numerical methods to approximate these integrals. In the next example, we will see how we can use Taylor polynomials and Taylor’s Theorem to aid us with this approximation process. Example. Estimate 0 .1 − x2 e 0 dx with an error of no more than 10−6 . We begin by using linear approximation to approximate the function h(u) = eu on the interval [−0.01, 0]. (The reason we chose this interval will be clear very soon). Then, since h (u) = eu , we get T 1, 0(u) = e0 + e0 (u − 0) = 1 + u and that h (c) (u − 0)2 | 2! where −.01 ≤ u < c < 0. But h (u) = eu and since eu is positive and increasing, | eu − (1 + u) |=| R1,0 (u) |=| 0 < e−0.01 < ec < e0 = 1. Therefore, for any u ∈ [−0.01, 0], Taylor’s Theorem shows that | eu − (1 + u) |=| DE Math 128 h (c) u2...
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This document was uploaded on 03/23/2013.

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