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Solve g(x) if g'(x) = sin 3x + cos 2x and g(pi) = 2 I get the following: G(pi) = (cos 3pi / 3) - (sin 2pi / 2) + C = 2 G(pi) = (cos pi) - (sin pi) + C = 2 G(pi) = -1 - 0 + C = 2 C = 3 In the correct answer, C = 5/3
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\[g(x)=\frac{1}{2}\sin 2x-\frac{1}{3}\cos 3x+C\]
\[\frac{1}{2}\sin 2\pi-\frac{1}{3}\cos 3\pi+C=2\]
\[0-\frac{1}{3}(-1)+C=2\]
\[C=2-\frac{1}{3}=\frac{5}{3}\]
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