The area (in sq. units) bounded by the curve `y=max. (x^(3), x^(4))` and the x - axis from `x=0" to "x=1` is

Answer

To find the area bounded by the curve \( y = \max(x^3, x^4) \) and the x-axis from \( x = 0 \) to \( x = 1 \), we can follow these steps:

### Step 1: Identify the Functions
We have two functions:
- \( y_1 = x^3 \)
- \( y_2 = x^4 \)

### Step 2: Find the Intersection Points
To determine which function is greater in the interval from \( x = 0 \) to \( x = 1 \), we need to find the points where \( x^3 = x^4 \).

Set the equations equal to each other:
\[ x^3 = x^4 \]

Rearranging gives:
\[ x^4 - x^3 = 0 \]
Factoring out \( x^3 \):
\[ x^3(x - 1) = 0 \]

This gives us the solutions:
\[ x = 0 \quad \text{and} \quad x = 1 \]

### Step 3: Determine Which Function is Maximum
Now we need to check which function is greater in the interval \( (0, 1) \):
- For \( x = 0.5 \):
\[ y_1 = (0.5)^3 = 0.125 \quad \text{and} \quad y_2 = (0.5)^4 = 0.0625 \]
Since \( y_1 > y_2 \) for \( x = 0.5 \), we conclude that \( y = x^3 \) is the maximum function in the interval \( [0, 1] \).

### Step 4: Set Up the Integral for Area
The area \( A \) under the curve from \( x = 0 \) to \( x = 1 \) is given by:
\[ A = \int_0^1 (y_1 - y_2) \, dx \]
Since \( y_2 = 0 \) (the x-axis), we have:
\[ A = \int_0^1 x^3 \, dx \]

### Step 5: Calculate the Integral
Now we compute the integral:
\[ A = \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 \]
Evaluating the limits:
\[ A = \left( \frac{1^4}{4} - \frac{0^4}{4} \right) = \frac{1}{4} - 0 = \frac{1}{4} \]

### Final Answer
Thus, the area bounded by the curve and the x-axis from \( x = 0 \) to \( x = 1 \) is:
\[ \boxed{0.25} \text{ square units} \]