Calculate the area bounded by the curves y = sin x, y =

Calculate the area bounded by the curves $y = \sin x$, $y = \cos x$, and the segment $\left[0, \frac{\pi}{2}\right]$ of the $x$-axis.

We will split the interval at the crossing point of the two curves.

Step 1: Find intersection points

The curves $y = \sin x$ and $y = \cos x$ intersect where:

\[\sin x = \cos x\]

Dividing by $\cos x$ (assuming $\cos x \neq 0$):

\[\tan x = 1\]

On the interval $\left[0, \frac{\pi}{2}\right]$:

\[x = \frac{\pi}{4}\]

Step 2: Determine which curve is above

For $x \in \left[0, \frac{\pi}{4}\right]$:

• At $x = 0$: $\cos 0 = 1 > \sin 0 = 0$
• So $\cos x > \sin x$ on $\left[0, \frac{\pi}{4}\right]$

For $x \in \left[\frac{\pi}{4}, \frac{\pi}{2}\right]$:

• At $x = \frac{\pi}{2}$: $\sin \frac{\pi}{2} = 1 > \cos \frac{\pi}{2} = 0$
• So $\sin x > \cos x$ on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$

Step 3: Set up the integral

The total area consists of two parts:

Part 1: Area between $y = \cos x$ and $y = \sin x$ on $\left[0, \frac{\pi}{4}\right]$:

\[A_1 = \int_0^{\pi/4} (\cos x - \sin x) \, dx\]

Part 2: Area between $y = \sin x$ and $y = \cos x$ on $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$:

\[A_2 = \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx\]

Total area:

\[A = A_1 + A_2\]

Step 4: Evaluate $A_1$

\[A_1 = \int_0^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_0^{\pi/4}\]

At $x = \frac{\pi}{4}$:

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

At $x = 0$:

\[\sin 0 + \cos 0 = 0 + 1 = 1\]

Therefore:

\[A_1 = \sqrt{2} - 1\]

Step 5: Evaluate $A_2$

\[A_2 = \int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}\]

At $x = \frac{\pi}{2}$:

\[-\cos \frac{\pi}{2} - \sin \frac{\pi}{2} = 0 - 1 = -1\]

At $x = \frac{\pi}{4}$:

\[-\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}\]

Therefore:

\[A_2 = -1 - (-\sqrt{2}) = -1 + \sqrt{2} = \sqrt{2} - 1\]

Step 6: Calculate total area

\[A = A_1 + A_2 = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)\]

\[\boxed{A = 2(\sqrt{2} - 1) = 2\sqrt{2} - 2 \text{ square units}}\]

Step 7: Numerical value

\[A = 2(\sqrt{2} - 1) \approx 2(1.414 - 1) = 2(0.414) \approx 0.828 \text{ square units}\]

Alternative approach: Single integral

We can also write:

\[A = \int_0^{\pi/2} |\cos x - \sin x| \, dx\]

Since $\cos x - \sin x$ changes sign at $x = \frac{\pi}{4}$:

\[A = \int_0^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/2} -(\cos x - \sin x) \, dx\]

which gives the same result.

Verification

Notice that $A_1 = A_2 = \sqrt{2} - 1$ by symmetry. The curves $y = \sin x$ and $y = \cos x$ are related by:

\[\cos x = \sin\left(\frac{\pi}{2} - x\right)\]

So there's a symmetry about $x = \frac{\pi}{4}$ that makes the two areas equal.

Summary

The area bounded by $y = \sin x$, $y = \cos x$, and the segment $\left[0, \frac{\pi}{2}\right]$ is:

\[\boxed{A = 2\sqrt{2} - 2 \approx 0.828 \text{ square units}}\]

Key steps:

1. Find intersection: $\sin x = \cos x \implies x = \frac{\pi}{4}$
2. Determine which curve is above in each region
3. Split at $x = \frac{\pi}{4}$: $A = A_1 + A_2$
4. Calculate: $A_1 = \sqrt{2} - 1$ and $A_2 = \sqrt{2} - 1$
5. Total: $A = 2(\sqrt{2} - 1)$

Key insight:

The curves intersect at $x = \frac{\pi}{4}$ where both equal $\frac{\sqrt{2}}{2}$, and by symmetry, the areas on either side of this point are equal.