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OpenStudy (anonymous):

If f'(x) = sin [(π×e^x) / 2] and f(0) = 1, then f(2) = (A) -1.819 (B) -0.843 (C) -0.819 (D) 0.157 (E) 1.157

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OpenStudy (anonymous):

integrate f'(x) and substitute 0 in it

OpenStudy (anonymous):

How do I integrate it though?

OpenStudy (anonymous):

w8 wirting the solution

OpenStudy (anonymous):

u know the substitution method?

OpenStudy (anonymous):

Yes i do. i think i figured it out. Thanks!

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OpenStudy (anonymous):

substitute u=\[\frac{ \pi*e^x }{ 2 }\]

OpenStudy (anonymous):

\[\frac{ du }{ dx }=\frac{ x }{ 2 \pi*e^x }\]

OpenStudy (anonymous):

\[dx=\frac{ 2u }{ xpi*e^x }\]

OpenStudy (anonymous):

(-2 cos((1/2 pi e^x)/(x pi e^x))) / (x pi e^x)

OpenStudy (anonymous):

That's great. I got it. Thank you!!

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