Calculate the area bounded by the curve y = cos² x

Calculate the area bounded by the curve $y = \cos^2 x$, the lines $x = 0$, $x = \pi$, and the segment $[0, \pi]$ on the $x$-axis.

We will evaluate the integral using the power-reduction identity for cosine.

Step 1: Understand the region

The region is bounded by:

• The curve: $y = \cos^2 x$
• Left boundary: $x = 0$
• Right boundary: $x = \pi$
• Bottom boundary: $y = 0$ (the $x$-axis)

Step 2: Verify that the curve is above the $x$-axis

For all $x$:

\[\cos^2 x \geq 0\]

So the curve is always above or on the $x$-axis. ✓

Step 3: Set up the integral

\[A = \int_0^\pi \cos^2 x \, dx\]

Step 4: Use the power-reduction formula

Recall the identity:

\[\cos^2 x = \frac{1 + \cos 2x}{2}\]

Therefore:

\[A = \int_0^\pi \frac{1 + \cos 2x}{2} \, dx = \frac{1}{2}\int_0^\pi (1 + \cos 2x) \, dx\]

\[= \frac{1}{2}\int_0^\pi 1 \, dx + \frac{1}{2}\int_0^\pi \cos 2x \, dx\]

Step 5: Evaluate each integral

First integral:

\[\frac{1}{2}\int_0^\pi 1 \, dx = \frac{1}{2}[x]_0^\pi = \frac{1}{2}(\pi - 0) = \frac{\pi}{2}\]

Second integral:

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

Step 6: Combine the results

\[A = \frac{\pi}{2} + \frac{1}{2} \cdot 0 = \frac{\pi}{2}\]

\[\boxed{A = \frac{\pi}{2} \text{ square units}}\]

Step 7: Numerical value

\[A = \frac{\pi}{2} \approx 1.571 \text{ square units}\]

Alternative method: Direct integration

We can also integrate directly using:

\[\int \cos^2 x \, dx = \frac{x}{2} + \frac{\sin 2x}{4} + C\]

Then:

\[A = \left[\frac{x}{2} + \frac{\sin 2x}{4}\right]_0^\pi\]

At $x = \pi$:

\[\frac{\pi}{2} + \frac{\sin 2\pi}{4} = \frac{\pi}{2} + 0 = \frac{\pi}{2}\]

At $x = 0$:

\[\frac{0}{2} + \frac{\sin 0}{4} = 0 + 0 = 0\]

Therefore:

\[A = \frac{\pi}{2} - 0 = \frac{\pi}{2} \quad \checkmark\]

Interpretation

The function $\cos^2 x$ oscillates between 0 and 1 on the interval $[0, \pi]$:

• At $x = 0$: $\cos^2 0 = 1$
• At $x = \frac{\pi}{2}$: $\cos^2 \frac{\pi}{2} = 0$
• At $x = \pi$: $\cos^2 \pi = 1$

The average value of $\cos^2 x$ over $[0, \pi]$ is:

\[\bar{y} = \frac{1}{\pi}\int_0^\pi \cos^2 x \, dx = \frac{1}{\pi} \cdot \frac{\pi}{2} = \frac{1}{2}\]

This makes sense because $\cos^2 x$ oscillates symmetrically around the value $\frac{1}{2}$.

Summary

The area bounded by the curve $y = \cos^2 x$, the lines $x = 0$, $x = \pi$, and the $x$-axis is:

\[\boxed{A = \frac{\pi}{2} \approx 1.571 \text{ square units}}\]

Key steps:

1. Set up integral: $A = \int_0^\pi \cos^2 x \, dx$
2. Use power-reduction: $\cos^2 x = \frac{1 + \cos 2x}{2}$
3. Split: $\frac{1}{2}\int_0^\pi 1 \, dx + \frac{1}{2}\int_0^\pi \cos 2x \, dx$
4. Evaluate: $\frac{\pi}{2} + 0 = \frac{\pi}{2}$

Key insight:

The integral of $\cos 2x$ over a complete period (or multiple periods) equals zero, leaving only the constant term $\frac{1}{2}$ to contribute to the integral.