To find the area bounded by \( y = \max(\sin^2 x, \sin^4 x) \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) with the x-axis, we will follow these steps:
### Step 1: Determine the maximum function
We need to compare \( \sin^2 x \) and \( \sin^4 x \) over the interval \( [0, \frac{\pi}{2}] \).
- For \( x = 0 \):
\[
\sin^2(0) = 0, \quad \sin^4(0) = 0
\]
- For \( x = \frac{\pi}{2} \):
\[
\sin^2\left(\frac{\pi}{2}\right) = 1, \quad \sin^4\left(\frac{\pi}{2}\right) = 1
\]
- For \( x = \frac{\pi}{4} \):
\[
\sin^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}, \quad \sin^4\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^4 = \frac{1}{4}
\]
Since \( \sin^2 x \) is greater than \( \sin^4 x \) for all \( x \) in \( [0, \frac{\pi}{2}] \), we have:
\[
\max(\sin^2 x, \sin^4 x) = \sin^2 x
\]
### Step 2: Set up the integral for the area
The area \( A \) under the curve from \( x = 0 \) to \( x = \frac{\pi}{2} \) is given by:
\[
A = \int_0^{\frac{\pi}{2}} \sin^2 x \, dx
\]
### Step 3: Use the identity for \( \sin^2 x \)
We can use the identity:
\[
\sin^2 x = \frac{1 - \cos(2x)}{2}
\]
Thus, we rewrite the integral:
\[
A = \int_0^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx
\]
### Step 4: Simplify the integral
Now, we can separate the integral:
\[
A = \frac{1}{2} \int_0^{\frac{\pi}{2}} (1 - \cos(2x)) \, dx
\]
\[
= \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} 1 \, dx - \int_0^{\frac{\pi}{2}} \cos(2x) \, dx \right)
\]
### Step 5: Evaluate the integrals
1. The first integral:
\[
\int_0^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_0^{\frac{\pi}{2}} = \frac{\pi}{2}
\]
2. The second integral:
\[
\int_0^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_0^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0
\]
### Step 6: Combine the results
Putting it all together:
\[
A = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}
\]
Thus, the area bounded by \( y = \max(\sin^2 x, \sin^4 x) \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:
\[
\boxed{\frac{\pi}{4}}
\]
