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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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49./- 1 - dx = -ln{4- x) + C and<br />

4-x<br />

I I I 1 1 1 1 00<br />

(X)" 1/ oo x" 1 oo xn+1<br />

4 - X . 4 1 - x/4 4 n=O 4 4 n=O 4" . 4 n=O 4"(n + 1} ·<br />

--dx = - - - dx = - L: - dx = - E - dx = - E + C So<br />

CHAPTER 11 REVIEW D 103<br />

1 00 xn+1 00 xn+1 00 x" .<br />

ln{4 - x)=-- 4<br />

E ,.( 4 1 ) +C= - E 4 ,.+ 1( + ) +C = - 1<br />

E n +C.Puttmg x=O,wegetC = ln4.<br />

n=O n+ n=O n n=1 n 4<br />

oo x"<br />

. Thus, f(x) = ln{4- x) = ln4- E<br />

rt=1 n 4<br />

Another solution:<br />

" . The series converges for Jx/ 41 < 1

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