Q.The area of the region bounded by the curve $y = \sin x$ between the ordinates $x = 0$, $x = \frac{\pi}{2}$ and the x-axis is
(A) $2$ sq units
(B) $4$ sq units
(C) $3$ sq units
(D) $1$ sq units

The key idea is that the area under a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the definite integral $\int_a^b |f(x)| \, dx$. Since $\sin x \ge 0$ on $[0, \frac{\pi}{2}]$, the absolute value is unnecessary.

Step 1: Set up the integral for the area:

Step 2: Integrate:

Step 3: Evaluate the definite integral: …

The problem asks for the area bounded by the curve $y = \sin x$, the x-axis, and the vertical lines $x = 0$ and $x = \pi/2$. This is a classic "area under a curve" problem — but the key is to remember that area is always a positive quantity. When the curve lies above the x-axis (which $\sin x$ does on $[0, \pi/2]$), the area is simply the definite integral of the function.

Why does the integral give area? Because the definite integral $\int_a^b f(x)\,dx$ sums up infinitely many infinitesimally thin rectangles of height $f(x)$ and width $dx$. When $f(x) \ge 0$, each rectangle's height is positive, so the sum is exactly the geometric area.

Here, $\sin x$ is non-negative on $[0, \pi/2]$ — it starts at 0, rises to 1 at $x = \pi/2$, and never dips below zero. So no absolute value or splitting is needed.

1. Set up the integral The area $A$ is given by

2. Evaluate the antiderivative The antiderivative of $\sin x$ is $-\cos x$. So

3. Apply the limits At the upper limit $x = \pi/2$: $\cos(\pi/2) = 0$, so $-\cos(\pi/2) = 0$. At the lower limit $x = 0$: $\cos 0 = 1$, so $-\cos 0 = -1$. Therefore …

Method: Area under a curve that stays above the x-axis

Use this for the area bounded by $y = f(x)$, the $x$-axis, and two vertical ordinates, when $f(x) \ge 0$ across the whole interval (as $\sin x$ is on $[0,\tfrac{\pi}{2}]$).

Steps

Step 1: Confirm the curve does not dip below the axis on $[a,b]$.

If $f(x) \ge 0$ throughout, the signed integral already equals the geometric area — no absolute value or splitting is needed. Check the sign at a sample point (or from the known shape of the graph).

Step 2: Write the definite integral.

$\text{Area} = \int_a^b f(x)\, dx$ …

Common Mistakes

Mistake 1: Using $+\cos x$ as the antiderivative of $\sin x$.

Why it's wrong: $\int \sin x\,dx = -\cos x$, not $\cos x$. Dropping the minus sign gives $\cos(\tfrac{\pi}{2}) - \cos 0 = -1$, a negative "area." Correct approach: use $[-\cos x]_0^{\pi/2} = 0-(-1) = 1$.

Mistake 2: Mishandling the lower-limit subtraction. …