Consider the function $$f(x) = \sin x + \frac{\sin 2x}{2} + \frac{\sin 3x}{3}$$ on the interval $0 \leqq x \leqq \pi$. We are to show that $f(x) > 0$ on $0 < x < \pi$, and to find the area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis.
(1) For $\mathbf{K}$, $\mathbf{N}$, $\mathbf{Q}$, $\mathbf{R}$ in the following sentences, choose the correct answer from the following two choices: (0) increasing, (1) decreasing, and for the other blanks, enter the correct number.
When we differentiate $f(x)$, we have $$f'(x) = (\mathbf{A}\cos^2 x - \mathbf{B})(\mathbf{C}\cos x + \mathbf{D}).$$ Hence, over the range $0 \leqq x \leqq \pi$, there are three $x$'s at which $f'(x) = 0$, and when they are arranged in ascending order, they are given accordingly.
Next, looking at whether $f(x)$ is increasing or decreasing, the behaviour is described accordingly.
Also, we have $$f(0) = 0, \quad f(\pi) = 0, \quad f\left(\frac{\mathbf{L}}{\mathbf{M}}\pi\right) = \frac{\sqrt{\mathbf{S}}}{\mathbf{T}} > 0.$$ Hence we see that $f(x) > 0$ on $0 < x < \pi$.
(2) The area $S$ of the region bounded by the graph of $y = f(x)$ and the $x$-axis is $$S = \frac{\mathbf{UV}}{\mathbf{W}}.$$