Find the average value `mu` of the function `f(x) = root(3)(x)` over the interval [0, 1]

Answer

To find the average value \( \mu \) of the function \( f(x) = \sqrt[3]{x} \) over the interval \([0, 1]\), we will use the formula for the average value of a function over an interval \([a, b]\):

\[ \mu = \frac{1}{b-a} \int_a^b f(x) \, dx \]

### Step 1: Identify the interval and function
Here, the function is \( f(x) = \sqrt[3]{x} = x^{1/3} \) and the interval is \([0, 1]\). Thus, \( a = 0 \) and \( b = 1 \).

### Step 2: Set up the integral
Using the average value formula, we have:

\[ \mu = \frac{1}{1 - 0} \int_0^1 x^{1/3} \, dx \]

This simplifies to:

\[ \mu = \int_0^1 x^{1/3} \, dx \]

### Step 3: Calculate the integral
To compute the integral \( \int_0^1 x^{1/3} \, dx \), we use the power rule for integration:

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]

For our case, \( n = \frac{1}{3} \):

\[ \int x^{1/3} \, dx = \frac{x^{1/3 + 1}}{1/3 + 1} = \frac{x^{4/3}}{4/3} = \frac{3}{4} x^{4/3} \]

### Step 4: Evaluate the definite integral
Now we evaluate the definite integral from 0 to 1:

\[ \int_0^1 x^{1/3} \, dx = \left[ \frac{3}{4} x^{4/3} \right]_0^1 \]

Calculating this gives:

\[ = \frac{3}{4} (1^{4/3}) - \frac{3}{4} (0^{4/3}) = \frac{3}{4} - 0 = \frac{3}{4} \]

### Step 5: Conclusion
Thus, the average value \( \mu \) of the function \( f(x) = \sqrt[3]{x} \) over the interval \([0, 1]\) is:

\[ \mu = \frac{3}{4} \]