Example 4: Area bounded by $y = \cos x$ from $x = 0$ to $x = 2\pi$

Basics you need

Solution

1. Find where the curve crosses the axis, and sketch

   In $[0, 2\pi]$⁠, $\cos x = 0$ exactly at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$⁠. The sign of $\cos x$⁠:

   $\cos x \ge 0$ on $\left[0, \frac{\pi}{2}\right]$ (first quadrant angles), $\cos x \le 0$ on $\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ (second and third quadrants), and $\cos x \ge 0$ on $\left[\frac{3\pi}{2}, 2\pi\right]$ (fourth quadrant).

   Key points: $A(0, 1)$⁠, $B\left(\frac{\pi}{2}, 0\right)$⁠, $C(\pi, -1)$⁠, $D\left(\frac{3\pi}{2}, 0\right)$⁠, $E(2\pi, 1)$⁠, $F(2\pi, 0)$⁠.

   The region consists of three pieces, $OABO$⁠, $BCDB$ and $DEFD$⁠, and the middle one lies below the $x$-axis. The crossings $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ are where we must split.
   On the unit circle, $\cos x$ is the $x$-coordinate: positive in the first and fourth quadrants, negative in the second and third.

2. Split the area into three parts

   $$\begin{aligned} \text{Area} &= \text{area } OABO + \text{area } BCDB \\ &\quad + \text{area } DEFD \\ &= \int_0^{\pi/2}\cos x\,dx \\ &\quad + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| \\ &\quad + \int_{3\pi/2}^{2\pi}\cos x\,dx. \end{aligned}$$
   On the first and last pieces $\cos x \ge 0$⁠, so the integrals are the areas. On the middle piece $\cos x \le 0$⁠, so the integral is negative and its absolute value is the area.
   Remark, Section 8.2: below the axis, take the absolute value. Lesson: below the axis, take the absolute value

3. Integrate $\cos x$

   An antiderivative of $\cos x$ is $F(x) = \sin x$⁠, since $\frac{d}{dx}\sin x = \cos x$⁠.

   $\cos x$ is continuous on all of $[0, 2\pi]$⁠, so Theorem 2 applies on each piece.
   Formula (ii), Section 7.2: $\int\cos x\,dx = \sin x + C$⁠. Basics: a definite integral is a number

4. Evaluate each part

   $$\begin{aligned} \int_0^{\pi/2}\cos x\,dx &= \sin\frac{\pi}{2} - \sin 0 \\ &= 1 - 0 = 1, \\ \int_{\pi/2}^{3\pi/2}\cos x\,dx &= \sin\frac{3\pi}{2} - \sin\frac{\pi}{2} \\ &= -1 - 1 = -2, \\ \int_{3\pi/2}^{2\pi}\cos x\,dx &= \sin 2\pi - \sin\frac{3\pi}{2} \\ &= 0 - (-1) = 1. \end{aligned}$$
   The middle value is negative, as it must be for a piece below the axis; its area is $|-2| = 2$⁠. The other two are positive and are areas as they stand.
   $\sin\frac{3\pi}{2} = \sin\left(\pi + \frac{\pi}{2}\right) = -\sin\frac{\pi}{2} = -1$⁠.

5. Add the areas

   $$\begin{aligned} \text{Area} &= 1 + |-2| + 1 \\ &= 1 + 2 + 1 = 4 \text{ sq. units}. \end{aligned}$$
   Every piece of an area counts as positive. The plain integral $\int_0^{2\pi}\cos x\,dx = \sin 2\pi - \sin 0 = 0$ adds $1 - 2 + 1$ and is not the area.
   Split at every crossing, then add the sizes. Lesson: when the curve dips below the axis, Examples 3–4

6. Check with the shape of the graph

   The graph from 0 to $2\pi$ is made of four congruent quarter-waves: from 0 to $\frac{\pi}{2}$⁠, $\frac{\pi}{2}$ to $\pi$⁠, $\pi$ to $\frac{3\pi}{2}$⁠, and $\frac{3\pi}{2}$ to $2\pi$⁠. Each has area $\int_0^{\pi/2}\cos x\,dx = 1$⁠, so the total is $4 \times 1 = 4$⁠.

   For $0 \le u \le \frac{\pi}{2}$⁠: $|\cos(\pi - u)| = |\cos(\pi + u)| = |\cos(2\pi - u)| = \cos u$⁠, because $\cos(\pi \mp u) = -\cos u$ and $\cos(2\pi - u) = \cos u$⁠. So the strips of each later quarter-wave have the same lengths as those of the first one (in reverse order for $x = \pi - u$ and $x = 2\pi - u$⁠), and the areas are equal. This matches $BCDB$ having area 2: two quarter-waves.
   Strips of equal lengths over intervals of equal width add up to equal areas.

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