Find the area of the region bounded by the following curves
5. and the -axis on

Ask by Cervantes Horton. in the United States

Jan 21,2025

Question

Find the area of the region bounded by the following curves 5. $$ y=\cos x \quad y=\sin x $$ and the $$ x $$-axis on $$ [0, \pi] $$

UpStudy AI · Accepted Answer

To find the area of the region bounded by the curves $$ y = \cos x $$, $$ y = \sin x $$, and the $$ x $$-axis on the interval $$ [0, \pi] $$, we'll break the problem into manageable parts.

### Step 1: Identify Intersection Points
First, determine where $$ y = \cos x $$ and $$ y = \sin x $$ intersect:
$$
\cos x = \sin x \implies \tan x = 1 \implies x = \frac{\pi}{4} \quad (\text{within } [0, \pi])
$$

### Step 2: Analyze the Regions
The interval $$ [0, \pi] $$ can be divided into three regions based on the intersection points and where each function is above the $$ x $$-axis:
1. **From $$ 0 $$ to $$ \frac{\pi}{4} $$:**
   - $$ y = \cos x $$ is above $$ y = \sin x $$ and the $$ x $$-axis.
2. **From $$ \frac{\pi}{4} $$ to $$ \frac{\pi}{2} $$:**
   - $$ y = \sin x $$ is above $$ y = \cos x $$ and the $$ x $$-axis.
3. **From $$ \frac{\pi}{2} $$ to $$ \pi $$:**
   - $$ y = \sin x $$ remains above the $$ x $$-axis, while $$ y = \cos x $$ is below the $$ x $$-axis.

### Step 3: Compute the Area in Each Region

1. **Region 1 ($$ 0 $$ to $$ \frac{\pi}{4} $$):**
   $$
   A_1 = \int_{0}^{\frac{\pi}{4}} \cos x \, dx = \left[ \sin x \right]_0^{\frac{\pi}{4}} = \sin\left(\frac{\pi}{4}\right) - 0 = \frac{\sqrt{2}}{2}
   $$

2. **Region 2 ($$ \frac{\pi}{4} $$ to $$ \frac{\pi}{2} $$):**
   $$
   A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx = \left[ -\cos x - \sin x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (-0 - 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = \sqrt{2} - 1
   $$

3. **Region 3 ($$ \frac{\pi}{2} $$ to $$ \pi $$):**
   $$
   A_3 = \int_{\frac{\pi}{2}}^{\pi} \sin x \, dx = \left[ -\cos x \right]_{\frac{\pi}{2}}^{\pi} = (-(-1)) - (-0) = 1
   $$

### Step 4: Sum the Areas
$$
\text{Total Area} = A_1 + A_2 + A_3 = \frac{\sqrt{2}}{2} + (\sqrt{2} - 1) + 1 = \frac{3\sqrt{2}}{2}
$$

### Final Answer
$$
\frac{3\,\sqrt{2}}{2}
$$
The area of the region bounded by $$ y = \cos x $$, $$ y = \sin x $$, and the $$ x $$-axis on $$ [0, \pi] $$ is $$ \frac{3\sqrt{2}}{2} $$.

Find the area of the region bounded by the following curves
5. and the -axis on

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Find the area of the region bounded by the following curves 5. and the -axis on