Therefore it would be reasonable to ask why we used

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Unformatted text preview: x2 ) = 1 − x2 ≤ e−x . It follows immediately that 0 .1 0 .1 (1 − x ) dx ≤ 0 2 e−x dx. 2 0 As such, the estimate we obtained above is actually less than the true value of the integral 0.1 2 e−x dx. 0 The second observation is a little more subtle. You will notice that we began with the 2 linear approximation for eu rather than e−x , the function we were asked to integrate. We then used this to build the approximation 2 e−x ∼ 1 − x2 = by simply substituting −x2 for u. At the same time we used Taylor’s Theorem applied to 2 eu rather than e−x to bound the error in this approximation. It turns out that 1 − x2 is 2 not the linear approximation to e−x . This is easy to see since 1 − x2 is a second degree polynomial and linear approximations are at most first degree. However, 1 − x2 is actually 2 both the second and the third degree Taylor polynomial of e−x at x = 0. Therefore, it would be reasonable to ask why we used the error in linear approximation for eu rather 2 than applying Taylor’s The...
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