Q.Find the area bounded by the curve $y = 2\cos x$ and the x-axis from $x = 0$ to $x = 2\pi$.

Concept: Area under a curve — integrate $y$ with respect to $x$ over the given interval, taking absolute value where the curve crosses the axis.

The curve $y = 2\cos x$ crosses the x-axis when $\cos x = 0$, i.e. at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ within $[0, 2\pi]$. The function is positive on $(0, \frac{\pi}{2})$ and $(\frac{3\pi}{2}, 2\pi)$, and negative on $(\frac{\pi}{2}, \frac{3\pi}{2})$.

Area is the sum of absolute values of the integrals over these intervals: …

When you're asked for the area bounded by a curve and the x-axis, you must take the absolute value of the function. The curve $y = 2\cos x$ dips below the x-axis for part of $[0, 2\pi]$, so simply integrating $2\cos x$ from $0$ to $2\pi$ would give zero — the positive and negative parts cancel. That's not the area; that's the net signed area. The actual physical area is the sum of the magnitudes of each region.

The key insight: find where the curve crosses the x-axis, split the interval at those points, integrate $|2\cos x|$ over each subinterval, and add.

1. Find the x-intercepts in $[0, 2\pi]$.

   Set $2\cos x = 0 \implies \cos x = 0$.

   In $[0, 2\pi]$, $\cos x = 0$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$.

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

   • On $(0, \frac{\pi}{2})$: $\cos x > 0$, so $2\cos x > 0$.
   • On $(\frac{\pi}{2}, \frac{3\pi}{2})$: $\cos x < 0$, so $2\cos x < 0$.
   • On $(\frac{3\pi}{2}, 2\pi)$: $\cos x > 0$, so $2\cos x > 0$.

   So the curve is above the axis on $[0, \frac{\pi}{2}]$ and $[\frac{3\pi}{2}, 2\pi]$, and below on $[\frac{\pi}{2}, \frac{3\pi}{2}]$.

3. Set up the area as the sum of absolute integrals.

The middle integral uses $-2\cos x$ because $2\cos x$ is negative there — taking the negative makes it positive.

4. Evaluate each integral.

   Recall: $\int \cos x \, dx = \sin x + C$.

   • First integral:

• Second integral:

 $\sin\frac{3\pi}{2} = -1$, $\sin\frac{\pi}{2} = 1$, so …

Method: Area between a curve and the x-axis when the curve changes sign

Use this whenever a graph (often a sine or cosine) rises above and dips below the $x$-axis across the given interval — a plain integral would let the positive and negative parts cancel and understate the true area.

Steps

Step 1: Find where the curve meets the x-axis.

Solve $f(x) = 0$ inside the interval. These roots are where the region flips from above the axis to below it, and they become the break-points of the integral.

Step 2: Determine the sign of $f(x)$ on each sub-interval.

Test one convenient point in each piece. Where $f(x) \ge 0$ the area element is $f(x)$; where $f(x) < 0$ it is $-f(x)$.

Step 3: Add the absolute areas. …

Common Mistakes

Mistake 1: Integrating $2\cos x$ straight from $0$ to $2\pi$.

Why it's wrong: $\int_0^{2\pi}2\cos x\,dx=[2\sin x]_0^{2\pi}=0$ — the positive and negative lobes cancel, giving the net signed area, not the physical area. Correct approach: locate where $\cos x=0$ (at $x=\tfrac{\pi}{2}$ and $\tfrac{3\pi}{2}$), split there, and add the magnitudes.

Mistake 2: Forgetting to flip the sign on the middle piece. …