<s>[INST] area between two curves 10 Find the area of the following curves: 
$$y=7\cos(2x) ,\quad y=7-7\cos(2x) ,\quad [0,\pi/2].$$
I found the boundaries (calculations not shown but they are correct) 
$7\cos(2x) > 7-7\cos(2x), \quad[0,\frac{\pi}{6}]$
$7-7\cos(2x) > 7\cos(2x),\quad [\frac{\pi}{6},\frac{5\pi}{6}]$
\begin{align*}
&\int 7 \cos(2x) - 7-7\cos(2x)~dx\\
=& \frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x).
\end{align*}
\begin{align*}
&\left[{\frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x)}\right]_{0}^{\frac{\pi}{6}}\\
=& -\frac{7\pi}{2} - 0 =-\frac{7\pi}{2}.
\end{align*}
\begin{align*}
&\left[7x-\frac{7}{2} \sin(2x)- \frac{7}{2}\sin(2x)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\\
=& \frac{35\pi}{6}+\frac{7\sqrt{3}}{2}-\frac{7\pi}{6}-\frac{7\sqrt{3}}{2}.
\end{align*}
\begin{align*}
&\left[{\frac{7}{2}\sin(2x)-7x-\frac{7}{2} \sin(2x)}\right]_{0}^{\frac{\pi}{6}}+\left[7x-\frac{7}{2} \sin(2x)- \frac{7}{2}\sin(2x)\right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\\
=&-\frac{7\pi}{2} + \frac{35\pi}{6}+\frac{7\sqrt{3}}{2}-\frac{7\pi}{6}-\frac{7\sqrt{3}}{2}\\
=& \frac{7\pi}{2}.
\end{align*}
This answer was wrong. I double checked my work and have no clue why its not right. To me it seems correct. Can anyone verify? 
 [/INST] Check your signs:
$$\int 7 \cos(2x) - \Big(7 - 7\cos(2x) \Big)$$
$$= \int 7 \cos(2x) - 7 \color{red}{+} 7\cos(2x)$$
$$= \frac{7}{2} \cos(2x) - 7x \color{red}{+} \frac{7}{2} \sin(2x)$$
</s>