EXAMPLE 10 (a) Evaluate the integral below as an infinite series. fe-z² (b) Evaluate the integral below correct to within an error of 0.0001. foot € SOLUTION (a) First we find the Maclaurin series for f(x) = ex. Although it's possible to use the direct method, let's find it simply by replacing x with -x² in the series for e given in Table 1 in the book. Thus for all values of x, ex² = 1 ∞o (-x²)^ n=0 n! x2 1! 0.5 1.0³ 4 fex dx = (1- = C + X- Now we integrate term by term. = Σ (-1)^ n=0 x² 1! 2x 2(1) 2! 2n e-² dx = x- n! X 2! x6 3! + ... x5 (b) The fundamental Theorem of Calculus gives 5.2 7.3! x7 This series converges for all x because the original series for e -2 + x5 (-1) 2n+1 x² (-1)^ 2n + 1 n! 10.5 x²n+1 + ... 18. X converges for all x. ).n! 5 2! 7.3! - 1/24 + - 1/5376 + 1/110592 - ... - 1/24 + 1/5376 + 1/110592 0.4278 X The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 0.0001. Activate Windows Go to Settings to activate Windows.

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Author:James Stewart
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EXAMPLE 10 (a) Evaluate the integral below as an infinite series.
Je-z²
(b) Evaluate the integral below correct to within an error of 0.0001.
6.0³.
SOLUTION (a) First we find the Maclaurin series for f(x) = ex. Although it's possible to use the direct method, let's find it simply by replacing x with -x² in the
series for e* given in Table 1 in the book. Thus for all values of x,
ex² = Σ
= 1
∞o (-x²)"
n=0 n!
1!
4
Sex²dx = (1-
= C + X-
=
Now we integrate term by term.
= Σ (-1)^
n=0
x²
1!
2x³
2(1)
2!
e-² dx = [x -
x2n
n!
1/24 +
1/24 +
X
2!
+
+6
3!
+5
x²
5 2 7 3!
+
x7
(-1)^2n+1
(-1) x²
n!
2n+1
This series converges for all x because the original series for e*² converges for all x.
(b) The fundamental Theorem of Calculus gives
6.³
2n + 1
+ ... 10.5
X
).n!
5.2! 7.3!
1/5376 + 1/110592
- 1/5376 + 1/110592 0.4278 x
The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 0.0001.
Activate Windows
Go to Settings to activate Windows.
Transcribed Image Text:EXAMPLE 10 (a) Evaluate the integral below as an infinite series. Je-z² (b) Evaluate the integral below correct to within an error of 0.0001. 6.0³. SOLUTION (a) First we find the Maclaurin series for f(x) = ex. Although it's possible to use the direct method, let's find it simply by replacing x with -x² in the series for e* given in Table 1 in the book. Thus for all values of x, ex² = Σ = 1 ∞o (-x²)" n=0 n! 1! 4 Sex²dx = (1- = C + X- = Now we integrate term by term. = Σ (-1)^ n=0 x² 1! 2x³ 2(1) 2! e-² dx = [x - x2n n! 1/24 + 1/24 + X 2! + +6 3! +5 x² 5 2 7 3! + x7 (-1)^2n+1 (-1) x² n! 2n+1 This series converges for all x because the original series for e*² converges for all x. (b) The fundamental Theorem of Calculus gives 6.³ 2n + 1 + ... 10.5 X ).n! 5.2! 7.3! 1/5376 + 1/110592 - 1/5376 + 1/110592 0.4278 x The Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 0.0001. Activate Windows Go to Settings to activate Windows.
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