Integration - Find the area under the curve. Im not sure how to do this at all. Need help walking through the steps on how to get the answer. So below is the question. I need any and all help.
Find the area under the curve:
$$ y=2\sin(3x-\pi/3)$$
between $x=0$ and $x=\pi/2$, and give your answer to $4$ decimal places.
 A: I assume by "area under the curve," you mean area between the graph of $y=f(x)$ and the $x$-axis. The area under the curve between $x=x_1$ and $x=x_2$ is the definite integral of the function $f(x)$ with bounds $x_1$ and $x_2$. 
Here is a picture of this area from Wolfram Alpha (note that when the function goes below the $x$-axis, that area counts as negative):

The reason that this is the area in question is that this integral is defined as the limit as $\Delta x \rightarrow 0$ of the area under the curve approximated by rectangles of width $\Delta x$ and height $f(x)$ (the smaller $\Delta x$, the better the approximation).
In this case, the requested area is: $$
\begin{align}
&\int_{0}^{\frac{\pi}{2}}{2 \sin (3x - \dfrac{\pi}{3}) \ dx} \\
&= -\dfrac{2}{3} \cos (3x - \dfrac{\pi}{3}) |_{0}^{\frac{\pi}{2}} \\
&= -\dfrac{2}{3} \cos (\dfrac{3\pi}{2} - \dfrac{\pi}{3}) + \dfrac{2}{3} \cos (0 - \dfrac{\pi}{3}) \\
&= -\dfrac{2}{3} \cos (\dfrac{7\pi}{6}) + \dfrac{2}{3} \cos ( - \dfrac{\pi}{3}) \\
&\approx -\dfrac{2}{3} \cdot (-0.8660) + \dfrac{2}{3} \cdot(\dfrac{1}{2}) \\
&\approx 0.9107
\end{align}
$$
A: $$\int_{0}^{\frac{\pi}{2}}2\sin\left(3x-\frac{\pi}{3}\right)\text{ }dx$$
$$2\int_{0}^{\frac{\pi}{2}}\sin\left(3x-\frac{\pi}{3}\right)\text{ }dx$$
$$2\cdot\left(-\frac{1}{3}\cos\left({3\cdot\frac{\pi}{2}}-\frac{\pi}{3}\right)\right)+C$$
$$\frac{\sqrt{3}}{3}+C$$
Note: this is the algebraic area under the curve. 
A: If you meant the geometric are then you have to do (with $\;f(x)=$ the integrand function):
$$-\int_0^{\pi/9} f(x)\,dx+\int_{\pi/9}^{4\pi/9} f(x)\,dx-\int_{4\pi/9}^{\pi/2}f(x)\,dx$$
For the antiderivative itself:
$$\int\sin\left(3x-\frac\pi3\right)dx=-\frac13\cos\left(3x-\frac\pi3\right)+C$$
