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

Question

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

mertsj · Answer

$$g(x)=\frac{1}{2}\sin 2x-\frac{1}{3}\cos 3x+C$$

mertsj · Answer

$$\frac{1}{2}\sin 2\pi-\frac{1}{3}\cos 3\pi+C=2$$

mertsj · Answer

$$0-\frac{1}{3}(-1)+C=2$$

mertsj · Answer

$$C=2-\frac{1}{3}=\frac{5}{3}$$

anonymous · Answer

Thank you!

mertsj · Answer

yw