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**Unformatted text preview: **(u − 0)2 |≤ .
2!
2
328 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series Observe that the integral is over the interval [0, 0.1]. If 0 ≤ x ≤ 0.1, then squaring gives
us that 02 ≤ x2 ≤ (0.1)2 , or equivalently, that
0 ≤ x2 ≤ 0.01
Multiplying each term in the above inequality by −1 reverses the inequality to show that
if 0 ≤ x ≤ 0.1, then
−0.01 ≤ −x2 ≤ 0
This is helpful because if we choose any x ∈ [0, .1] and then let u = −x2 , the above
inequality shows that
−0.01 ≤ u = −x2 ≤ 0.
However, for such a u, we know that
| eu − (1 + u) |≤ u2
.
2 Substituting −x2 for u in the above inequality, we see that if x ∈ [0, 0.1], then
2 | e−x − (1 + (−x2 )) |≤
or
2 | e−x − (1 − x2 ) |≤
Recall that b | (−x2 )2
2
x4
.
2 b g (x) dx |≤
a | g (x) | dx
a for any function g (x) and that if f (x) ≤ g (x), then for a ≤ b
b b f (x) dx ≤
a g (x) dx.
a Using these rules of integration and the above inequality, we get 0 .1 0.1 2 e−x dx − |
0 0 .1 2 e−x − (1 − x2 ) dx | (1 − x2 ) dx | = |
0 0
0.1 2 | e−x − (1 −...

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