Q.Find the area bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$

The key idea is that area is always positive, so we must split the interval wherever the curve crosses the x‑axis.

Step 1: Find where $\cos x = 0$ in $[0, 2\pi]$.

$\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

Step 2: On $[0, \frac{\pi}{2}]$, $\cos x \ge 0$; on $[\frac{\pi}{2}, \frac{3\pi}{2}]$, $\cos x \le 0$; on $[\frac{3\pi}{2}, 2\pi]$, $\cos x \ge 0$.

Step 3: Total area = …

Why area under a curve isn’t always “just integrate”

When a student first sees “area bounded by the curve”, the reflex is to compute $\int_0^{2\pi} \cos x \, dx$. That integral evaluates to $0$, which is true for the signed area — but the question asks for the geometric area, the actual region enclosed between the curve and the x‑axis. The curve $y = \cos x$ crosses the x‑axis at $x = \pi/2$ and $x = 3\pi/2$, so parts of it lie below the axis. Area is always positive, so we must take the absolute value of each piece.

Step‑by‑step solution

1. Identify the zeroes of $\cos x$ in $[0, 2\pi]$

$\cos x = 0$ when $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$. These split the interval into three parts:

$[0, \pi/2]$, $[\pi/2, 3\pi/2]$, $[3\pi/2, 2\pi]$.

2. Determine the sign of $\cos x$ on each subinterval

• On $[0, \pi/2]$: $\cos x \ge 0$ (starts at $1$, falls to $0$).
• On $[\pi/2, 3\pi/2]$: $\cos x \le 0$ (goes from $0$ to $-1$ and back to $0$).
• On $[3\pi/2, 2\pi]$: $\cos x \ge 0$ (rises from $0$ to $1$).

3. Write the area as a sum of absolute integrals

The total geometric area $A$ is:

4. Evaluate each integral

• First piece: $\int_0^{\pi/2} \cos x \, dx = \left[ \sin x \right]_0^{\pi/2} = \sin(\pi/2) - \sin 0 = 1 - 0 = 1$. …

Method: Area of a periodic curve over a full period (modulus)

Use this for "area bounded by the curve" over an interval where a periodic function (like $\cos x$ over $[0,2\pi]$) is above the axis on some sub-intervals and below on others.

Steps

Step 1: Find all axis crossings in the interval.

Solve $f(x) = 0$; for $\cos x$ on $[0,2\pi]$ the crossings are $x = \tfrac{\pi}{2}$ and $x = \tfrac{3\pi}{2}$, splitting the range into three pieces.

Step 2: Fix the sign on each sub-interval.

Integrate $f$ where it is positive and $-f$ where it is negative, so every contribution is positive.

Step 3: Sum the pieces. …

Common Mistakes

Mistake 1: Concluding the area is zero from $\int_0^{2\pi}\cos x\,dx = 0$.

Why it's wrong: that integral is the net signed value; the positive lobes and the negative lobe cancel, but real area is enclosed. Correct approach: integrate $|\cos x|$ by splitting at the zeros $x=\tfrac{\pi}{2},\tfrac{3\pi}{2}$, giving $1+2+1=4$.

Mistake 2: Splitting at only one crossing. …