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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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integrate f'(x) and substitute 0 in it
How do I integrate it though?
w8 wirting the solution
u know the substitution method?
Yes i do. i think i figured it out. Thanks!
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substitute u=\[\frac{ \pi*e^x }{ 2 }\]
\[\frac{ du }{ dx }=\frac{ x }{ 2 \pi*e^x }\]
\[dx=\frac{ 2u }{ xpi*e^x }\]
(-2 cos((1/2 pi e^x)/(x pi e^x))) / (x pi e^x)
That's great. I got it. Thank you!!
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