Q.Area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is
(A) $2$ sq units
(B) $4$ sq units
(C) $3$ sq units
(D) $1$ sq units

Concept: Area Under Curve — area is the integral of $|y|$ with respect to $x$ over the given interval, because area is always positive.

Steps:

1. The curve $y = \cos x$ is positive on $[0, \pi/2)$ and negative on $(\pi/2, \pi]$.
2. Area = $\int_0^{\pi} |\cos x| \, dx = \int_0^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$.
3. Compute: $\int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = 1$, …

The key idea here is that "area bounded by a curve" always means geometric area — the actual physical region, not the signed area. When a curve dips below the x-axis, the definite integral gives a negative value, but area is always positive. So we must split the interval wherever the curve changes sign.

For $y = \cos x$ between $0$ and $\pi$, the curve is positive from $0$ to $\pi/2$ and negative from $\pi/2$ to $\pi$. The total area is the sum of the absolute areas of these two parts.

Let’s work through it step by step.

1. Identify where the curve crosses the x-axis.

   Solve $\cos x = 0$ in $[0, \pi]$.

   $\cos x = 0$ at $x = \pi/2$. So the sign changes at this point.

2. Set up the area as a sum of absolute integrals.

   Area $= \int_{0}^{\pi/2} \cos x \, dx + \int_{\pi/2}^{\pi} (-\cos x) \, dx$

   The second integral uses $-\cos x$ because $\cos x$ is negative there, and we want the positive magnitude.

3. Evaluate the first integral.

   $\int_{0}^{\pi/2} \cos x \, dx = [\sin x]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.

4. Evaluate the second integral.

   $\int_{\pi/2}^{\pi} (-\cos x) \, dx = -[\sin x]_{\pi/2}^{\pi} = -(\sin \pi - \sin(\pi/2)) = -(0 - 1) = 1$.

5. Add the two parts.

   Total area $= 1 + 1 = 2$ square units. …

Method: Geometric area under a sign-changing curve

"Area bounded by the curve and the $x$-axis" always means positive, physical area. When the curve crosses the axis inside the interval, split at the crossing and take magnitudes.

Steps

Step 1: Solve $f(x) = 0$ on the interval to find the crossing points.

For $y = \cos x$ this happens at odd multiples of $\frac{\pi}{2}$; keep the ones lying inside your range.

Step 2: Split the interval at each crossing and fix the sign.

On stretches where the curve is above the axis keep $f(x)$; where it is below, integrate $-f(x)$ so the contribution stays positive. …

Common Mistakes

Mistake 1: Evaluating $\int_0^{\pi}\cos x\,dx$ directly.

Why it's wrong: this gives $[\sin x]_0^{\pi}=0$, the net signed area, because the curve is above the axis on $[0,\tfrac{\pi}{2}]$ and below on $[\tfrac{\pi}{2},\pi]$ and the two cancel. Correct approach: split at the zero $x=\tfrac{\pi}{2}$ and add the magnitudes, $1+1=2$.

Mistake 2: Forgetting the curve goes negative after $\tfrac{\pi}{2}$. …