F'(x)=sin((pi (e^x)) /2) and f(0)=1

  • Thread starter Punkyc7
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  • #1
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Main Question or Discussion Point

f'(x)=sin((pi (e^x)) /2) and f(0)=1 then f(2)= ?

so i integrate and get -(1/((pi/2)*e^x)) cos((pi (e^x)) /2)+ 1=1

when i plug two into that i dont get any of the answers listed

a)-1.819
b) -.843
c.) -.819
d) .157
e) 1.157
 

Answers and Replies

  • #2
CompuChip
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Are you supposed to do this analytically? Because I can't really find a primitive function for sin(pi e^x / 2). I can do the integration numerically though, and get one of the listed answers.
 
  • #3
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What he said. Your "integrate" is incorrect. Instead of finding an exact formula, can you tell by looking at the sign and approximate size of the RHS that all but one of the answers is ruled out?
 
  • #4
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If you have a calculator (which I can assume by the answer choices), try Euler's method.
 
  • #5
dextercioby
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Use the Taylor series around 0 (aka MacLaurin), mate.

http://mathworld.wolfram.com/TaylorSeries.html

Up to third derivative I get

1+2-(pi^2 / 3) approx = -0.290.

Then u compute up to 5th derivative and see what you get.
 

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