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

The key idea is that area is the integral of the absolute value of the function, because the curve goes below the x‑axis.

Step 1: Identify where $\sin x$ changes sign.

$\sin x \ge 0$ on $[0, \pi]$ and $\sin x \le 0$ on $[\pi, 2\pi]$.

Step 2: Write area as sum of two integrals with correct sign:

Step 3: Evaluate each. …

The key insight here is that "area bounded by the curve" means geometric area — the physical region enclosed between the curve and the x-axis. When a curve dips below the x-axis, the definite integral gives a negative value, but area is always positive. So we cannot simply integrate $\sin x$ from $0$ to $2\pi$ and call it done.

Let’s look at the graph of $y = \sin x$ between $0$ and $2\pi$. It starts at $0$, rises to $1$ at $x = \pi/2$, comes back to $0$ at $x = \pi$, then goes negative to $-1$ at $x = 3\pi/2$, and returns to $0$ at $x = 2\pi$. The curve crosses the x-axis at $x = 0, \pi, 2\pi$. The region above the axis (from $0$ to $\pi$) and the region below the axis (from $\pi$ to $2\pi$) are mirror images in shape. So the total geometric area is twice the area of one lobe.

1. Set up the integrals with absolute value. The area $A$ is given by

Since $\sin x \ge 0$ on $[0, \pi]$ and $\sin x \le 0$ on $[\pi, 2\pi]$, we split:

2. Evaluate the first integral.

3. Evaluate the second integral.

Let’s do it carefully: $-\cos x$ evaluated from $\pi$ to $2\pi$ gives $(-\cos 2\pi) - (-\cos \pi) = (-1) - (1) = -2$. Then the negative sign outside makes it $-(-2) = 2$. …

Method: Area of a curve that crosses the x-axis (take the modulus)

Use this whenever "area bounded by the curve" is asked over an interval where the curve goes both above and below the $x$-axis (e.g. a full period of $\sin x$). Geometric area is always positive.

Steps

Step 1: Find where the curve crosses the axis.

Solve $f(x) = 0$ inside $[a,b]$. These roots divide the interval into pieces that are entirely above or entirely below the axis.

Step 2: Determine the sign on each piece.

On a piece where $f \ge 0$ integrate $f$; where $f \le 0$ integrate $-f$ (or take the magnitude of the result).

Step 3: Sum the magnitudes. …

Common Mistakes

Mistake 1: Reporting the plain signed integral.

Why it's wrong: $\int_0^{2\pi}\sin x\,dx = 0$ because the positive lobe $[0,\pi]$ and the negative lobe $[\pi,2\pi]$ cancel — but the region genuinely encloses area. Correct approach: integrate $|\sin x|$, i.e. split at $x=\pi$ and add magnitudes, giving $2+2=4$.

Mistake 2: Not splitting where the curve crosses the axis. …