To find the area of the smaller region bounded by the curves \( y = \cos x \), \( x + y = 1 \), and \( y = 0 \), we will follow these steps:
### Step 1: Identify the curves and their intersections
We have three curves:
1. \( y = \cos x \)
2. \( x + y = 1 \) (which can be rewritten as \( y = 1 - x \))
3. \( y = 0 \) (the x-axis)
Next, we need to find the points of intersection of these curves.
### Step 2: Find the intersection points
To find the intersection of \( y = \cos x \) and \( y = 1 - x \):
Set \( \cos x = 1 - x \).
This equation does not have a straightforward algebraic solution, so we can find the intersection points graphically or numerically. However, we can identify that:
- At \( x = 0 \), \( \cos(0) = 1 \) and \( 1 - 0 = 1 \) (they intersect at \( (0, 1) \)).
- At \( x = \frac{\pi}{2} \), \( \cos(\frac{\pi}{2}) = 0 \) and \( 1 - \frac{\pi}{2} \) is negative (no intersection here).
Thus, the relevant intersection point is \( (0, 1) \).
### Step 3: Determine the area of the region
The area we want to calculate is bounded by:
- The curve \( y = \cos x \)
- The line \( y = 1 - x \)
- The x-axis from \( x = 0 \) to the point where \( y = \cos x \) intersects \( y = 1 - x \).
### Step 4: Set up the integral
The area \( A \) of the region can be calculated using the integral:
\[
A = \int_{0}^{\frac{\pi}{2}} (\cos x - (1 - x)) \, dx
\]
This integral represents the area between the curve \( y = \cos x \) and the line \( y = 1 - x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \).
### Step 5: Calculate the integral
First, we simplify the integrand:
\[
\cos x - (1 - x) = \cos x + x - 1
\]
Now, we compute the integral:
\[
A = \int_{0}^{\frac{\pi}{2}} (\cos x + x - 1) \, dx
\]
This can be split into three separate integrals:
\[
A = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \int_{0}^{\frac{\pi}{2}} x \, dx - \int_{0}^{\frac{\pi}{2}} 1 \, dx
\]
Calculating each integral:
1. \( \int \cos x \, dx = \sin x \) evaluated from \( 0 \) to \( \frac{\pi}{2} \):
\[
\sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1
\]
2. \( \int x \, dx = \frac{x^2}{2} \) evaluated from \( 0 \) to \( \frac{\pi}{2} \):
\[
\frac{\left(\frac{\pi}{2}\right)^2}{2} - 0 = \frac{\pi^2}{8}
\]
3. \( \int 1 \, dx = x \) evaluated from \( 0 \) to \( \frac{\pi}{2} \):
\[
\frac{\pi}{2} - 0 = \frac{\pi}{2}
\]
Putting it all together:
\[
A = 1 + \frac{\pi^2}{8} - \frac{\pi}{2}
\]
### Step 6: Final area calculation
Now, we need to simplify:
\[
A = 1 + \frac{\pi^2}{8} - \frac{4\pi}{8} = 1 + \frac{\pi^2 - 4\pi}{8}
\]
This gives us the area of the smaller region bounded by the curves.
### Conclusion
The area of the smaller region bounded by \( y = \cos x \), \( x + y = 1 \), and \( y = 0 \) is:
\[
\text{Area} = 1 - \frac{1}{2} = \frac{1}{2} \text{ square units.}
\]
