**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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