Given fuction,`{(x^(2), "for" 0 le x lt1),(sqrtx, "for" 1le x le 2):}`
Evaluate `int_(0)^(2) f(x) d x `.

Answer

To evaluate the integral \( \int_{0}^{2} f(x) \, dx \) for the given piecewise function:

\[ f(x) = \begin{cases} x^2 & \text{for } 0 \leq x < 1 \\ \sqrt{x} & \text{for } 1 \leq x \leq 2 \end{cases} \]

we will break the integral into two parts according to the definition of the function.

### Step 1: Break the integral into two parts
We can express the integral from 0 to 2 as the sum of two integrals:

\[ \int_{0}^{2} f(x) \, dx = \int_{0}^{1} f(x) \, dx + \int_{1}^{2} f(x) \, dx \]

### Step 2: Substitute the function in the integrals
Now, we substitute the function \( f(x) \) in each integral:

\[ \int_{0}^{1} f(x) \, dx = \int_{0}^{1} x^2 \, dx \]
\[ \int_{1}^{2} f(x) \, dx = \int_{1}^{2} \sqrt{x} \, dx \]

### Step 3: Evaluate the first integral
Now, we evaluate the first integral:

\[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \]

### Step 4: Evaluate the second integral
Next, we evaluate the second integral:

\[ \int_{1}^{2} \sqrt{x} \, dx = \int_{1}^{2} x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{2} = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{2} \]

Calculating the limits:

\[ = \frac{2}{3} \left( 2^{3/2} - 1^{3/2} \right) = \frac{2}{3} \left( 2\sqrt{2} - 1 \right) \]

### Step 5: Combine the results
Now we combine the results from both integrals:

\[ \int_{0}^{2} f(x) \, dx = \frac{1}{3} + \frac{2}{3} \left( 2\sqrt{2} - 1 \right) \]

### Step 6: Simplify the expression
Now, we simplify the expression:

\[ = \frac{1}{3} + \frac{2}{3} \cdot 2\sqrt{2} - \frac{2}{3} \]
\[ = \frac{1 - 2}{3} + \frac{4\sqrt{2}}{3} = \frac{-1 + 4\sqrt{2}}{3} \]

### Final Answer
Thus, the final result is:

\[ \int_{0}^{2} f(x) \, dx = \frac{4\sqrt{2} - 1}{3} \]