Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the curve $f(x)=\max \{\sin x,\cos x\},-\pi \le x\le \pi$ and the $x$-axis is

We are analyzing the interval $[-\pi ,\pi ]$. If you sketch $\sin x$ and $\cos x$ on the same coordinate plane, you will see them crossing each other.

The function $f(x)=\max \{\sin x,\cos x\}$ acts as a path that always chooses the higher ground. Whenever $\sin x>\cos x$, we follow the sine curve; when $\cos x>\sin x$, we follow the cosine curve. This creates a composite, jagged boundary.

To determine where to switch our path, we solve $\sin x=\cos x$, which simplifies to $\tan x=1$.

Within the interval $[-\pi ,\pi ]$, this equality occurs at:

These are the critical junctions where our function changes its identity. Mark these points clearly; they serve as the signposts for our integration.

We are calculating the area enclosed by the curve and the $x$-axis. Because the area must be positive, we must be vigilant about regions where the function dips below the $x$-axis.

The function $\cos x$ crosses the $x$-axis at $x=-\frac{\pi }{2}$. To ensure the area is positive, we must take the absolute value of the function, splitting our journey into four distinct intervals:

For the first two intervals, the function lies below the $x$-axis, so we integrate the negative of the function. For the last two, the function is above the axis, so we integrate normally.

The total area $A$ is the sum of four integrals:

Evaluating these yields: 1. ${\int }_{-\pi }^{-\frac{3\pi }{4}}(-\sin x)dx=1-\frac{1}{\sqrt{2}}$ 2. ${\int }_{-\frac{3\pi }{4}}^{-\frac{\pi }{2}}(-\cos x)dx=1-\frac{1}{\sqrt{2}}$ 3. ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{4}}\cos xdx=1+\frac{1}{\sqrt{2}}$ 4. ${\int }_{\frac{\pi }{4}}^{\pi }\sin xdx=1+\frac{1}{\sqrt{2}}$

When we sum these areas, the terms involving $\frac{1}{\sqrt{2}}$ cancel each other out perfectly:

The complexity of the trigonometric functions dissolves into a simple, clean integer. The final area is: $A=4$