∫₀π sin x dx = [-cos x]₀π = 2 square units.

Area Under y = sin x from 0 to π

Question

Find the area under the curve $y = \sin x$ from $x = 0$ to $x = \pi$ .

This is a direct NCERT exercise and appears regularly in CBSE Class 12 board exams. The setup is simple, but students trip up on the evaluation more than you’d expect.

Solution — Step by Step

The area under a curve $y = f(x)$ from $x = a$ to $x = b$ is:

A = \int_a^b f(x)\, dx

Since $\sin x \geq 0$ on $[0, \pi]$ , the curve lies entirely above the x-axis. No sign correction needed.

A = \int_0^{\pi} \sin x\, dx

The antiderivative of $\sin x$ is $-\cos x$ . Keep the negative sign — this is where most errors happen.

A = \left[ -\cos x \right]_0^{\pi}

Substitute the upper limit first, then subtract the lower limit:

A = (-\cos \pi) - (-\cos 0)

A = (-(-1)) - (-(1))

A = 1 + 1 = 2

\boxed{A = 2 \text{ square units}}

The area enclosed between $y = \sin x$ and the x-axis from $0$ to $\pi$ is 2 square units.

Why This Works

The definite integral computes the net signed area between the curve and the x-axis. When the function is non-negative throughout the interval, the integral directly gives us the geometric area — no complications.

On $[0, \pi]$ , $\sin x$ stays non-negative (it’s zero at the endpoints, positive in between). The curve forms a single arch above the x-axis, and we’re computing exactly the area of that arch.

This result is elegant: the arch of $\sin x$ over one half-period has area exactly 2. It’s worth memorising because it’s a benchmark — the arch over $[\pi, 2\pi]$ gives $-2$ when integrated directly, meaning you need $|\int_{\pi}^{2\pi} \sin x\, dx| = 2$ for area.

Alternative Method — Symmetry Shortcut

$\sin x$ is symmetric about $x = \pi/2$ on $[0, \pi]$ . So we can compute the area from $0$ to $\pi/2$ and double it.

A = 2 \int_0^{\pi/2} \sin x\, dx = 2 \left[-\cos x\right]_0^{\pi/2} = 2\left[(-\cos \frac{\pi}{2}) - (-\cos 0)\right] = 2[0 + 1] = 2

Same answer, and useful when the full integral feels intimidating. In competitive exams where time matters, spotting symmetry saves 30 seconds.

Common Mistake

Writing $\int \sin x\, dx = \cos x$ (dropping the negative sign).

The correct antiderivative is $-\cos x$ , not $\cos x$ . Students who differentiate $\cos x$ and get $-\sin x$ know this — but under exam pressure, the sign vanishes.

If you write $[\cos x]_0^{\pi}$ , you get $\cos\pi - \cos 0 = -1 - 1 = -2$ . An area of $-2$ is physically impossible, which should immediately signal an error. Always verify: area cannot be negative.

A quick self-check: sketch $\sin x$ on $[0, \pi]$ and confirm the curve is above the x-axis. If your answer is negative, recheck the antiderivative sign.

Question

Solution — Step by Step

Why This Works

Alternative Method — Symmetry Shortcut

Common Mistake

Try These Next