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 theseries for e* given in Table 1 in the book. Thus for all values of x,ex² = Σ= 1∞o (-x²)"n=0 n!1!4Sex²dx = (1-= C + X-=Now we integrate term by term.= Σ (-1)^n=0x²1!2x³2(1)2!e-² dx = [x -x2nn!1/24 +1/24 +X2!++63!+5x²5 2 7 3!+x7(-1)^2n+1(-1) x²n!2n+1This series converges for all x because the original series for e*² converges for all x.(b) The fundamental Theorem of Calculus gives6.³2n + 1+ ... 10.5X).n!5.2! 7.3!1/5376 + 1/110592- 1/5376 + 1/110592 0.4278 xThe Alternating Series Estimation Theorem shows that the error involved in this approximation is less than 0.0001.Activate WindowsGo to Settings to activate Windows.

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.

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.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

Problem 67E

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