`f(x)={:{(1-x","" "0 le c le 1),(0","" "1 le x le 2" and "phi (x)=int_(0)^(x)f(t)" dt"."Then"),((2-x)^(2)","" "2 le x le 3):}`for any `x in [2,3],phi(x)` equals

Options

A. `((x-2)^(3))/(3)`
B. `(1)/(2)-((x-2)^(3))/(3)`
C. `(1)/(2)+((x-2)^(3))/(3)`
D. none of these

Answer

To solve the problem, we need to evaluate the function \( \phi(x) = \int_0^x f(t) \, dt \) for \( x \) in the interval \([2, 3]\) using the piecewise function \( f(x) \).

### Step 1: Define the function \( f(x) \)
The function \( f(x) \) is defined as:
\[ f(x) = \begin{cases} 1 - x & \text{for } 0 \leq x < 1 \\ 0 & \text{for } 1 \leq x < 2 \\ (2 - x)^2 & \text{for } 2 \leq x \leq 3 \end{cases} \]

### Step 2: Set up the integral for \( \phi(x) \)
Since we are interested in \( \phi(x) \) for \( x \) in \([2, 3]\), we can express \( \phi(x) \) as:
\[ \phi(x) = \int_0^2 f(t) \, dt + \int_2^x f(t) \, dt \]
Given that \( f(t) = 0 \) for \( 1 \leq t < 2 \), we can split the integral:
\[ \phi(x) = \int_0^1 f(t) \, dt + \int_2^x f(t) \, dt \]

### Step 3: Calculate \( \int_0^1 f(t) \, dt \)
For \( 0 \leq t < 1 \), \( f(t) = 1 - t \):
\[ \int_0^1 (1 - t) \, dt = \left[ t - \frac{t^2}{2} \right]_0^1 = \left(1 - \frac{1}{2}\right) - (0 - 0) = \frac{1}{2} \]

### Step 4: Calculate \( \int_2^x f(t) \, dt \) for \( 2 \leq x \leq 3 \)
For \( 2 \leq t \leq 3 \), \( f(t) = (2 - t)^2 \):
\[ \int_2^x (2 - t)^2 \, dt \]
We can compute this integral:
\[ \int (2 - t)^2 \, dt = \int (4 - 4t + t^2) \, dt = 4t - 2t^2 + \frac{t^3}{3} + C \]
Now, evaluate from \( 2 \) to \( x \):
\[ \left[ 4t - 2t^2 + \frac{t^3}{3} \right]_2^x \]
Calculating at \( t = x \):
\[ 4x - 2x^2 + \frac{x^3}{3} \]
Calculating at \( t = 2 \):
\[ 4(2) - 2(2^2) + \frac{2^3}{3} = 8 - 8 + \frac{8}{3} = \frac{8}{3} \]
Thus,
\[ \int_2^x (2 - t)^2 \, dt = \left( 4x - 2x^2 + \frac{x^3}{3} \right) - \frac{8}{3} \]

### Step 5: Combine results to find \( \phi(x) \)
Now we combine the results:
\[ \phi(x) = \frac{1}{2} + \left( 4x - 2x^2 + \frac{x^3}{3} - \frac{8}{3} \right) \]
Simplifying:
\[ \phi(x) = 4x - 2x^2 + \frac{x^3}{3} - \frac{8}{3} + \frac{1}{2} \]
To combine the constants:
\[ \frac{1}{2} - \frac{8}{3} = \frac{3}{6} - \frac{16}{6} = -\frac{13}{6} \]
Thus,
\[ \phi(x) = 4x - 2x^2 + \frac{x^3}{3} - \frac{13}{6} \]

### Step 6: Evaluate \( \phi(3) \)
Now we can evaluate \( \phi(3) \):
\[ \phi(3) = 4(3) - 2(3^2) + \frac{3^3}{3} - \frac{13}{6} \]
Calculating:
\[ = 12 - 18 + 9 - \frac{13}{6} = 3 - \frac{13}{6} = \frac{18}{6} - \frac{13}{6} = \frac{5}{6} \]

### Final Answer
Thus, for \( x \in [2, 3] \), \( \phi(x) = \frac{5}{6} \).