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Unformatted text preview: x2 )  dx ≤
0
0.1 ≤
0 x4
dx
2 x 5 0.1
=

5(2) 0
(0.1)5
=
10
= 10−6 DE Math 128 329 (B. Forrest)2 4.5. Introduction to Taylor Series CHAPTER 4. Sequences and Series
0 .1 2 This tells us that if we approximate the integral 0 e−x dx by the much simpler integral
0.1
(1 − x2 ) dx, the estimate will have an error of no more than 10−6 which is exactly what
0
we wanted.
Therefore, we get that
0.1 0 .1 2
e−x dx ∼
= (1 − x2 ) dx 0 0 x 3 0 .1
)
30
0.001
= 0.1 −
3 = (x − with an error of no more than 10−6 .
There are two more important observations we can make with respect to this example.
The ﬁrst observation is that for every u, 1 + u ≤ eu . This is clear from the diagram below,
but it also follows from the fact that y = 1 + u is the tangent line to the graph of y = eu at
u = 0 and the graph of eu is always concave up. However, if a function has a graph that is
always concave up, then every tangent line sits below the function.
Since 1 + u ≤ eu for every u, given x ∈ [0.0.1], when we let u = −x2 we get
2 1 + (−...
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