`int_(0)^(3) [x^(3)-3x^(2) + 2x] dx`=

Options

A. `(9)/(4)`
B. `(7)/(4)`
C. `(3)/(4)`
D. none

Answer

To solve the integral \( \int_{0}^{3} (x^{3} - 3x^{2} + 2x) \, dx \), we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{3} (x^{3} - 3x^{2} + 2x) \, dx \]

### Step 2: Integrate the function
We will integrate each term separately:
1. The integral of \( x^{3} \) is \( \frac{x^{4}}{4} \).
2. The integral of \( -3x^{2} \) is \( -3 \cdot \frac{x^{3}}{3} = -x^{3} \).
3. The integral of \( 2x \) is \( 2 \cdot \frac{x^{2}}{2} = x^{2} \).

Putting it all together, we have:
\[ I = \left[ \frac{x^{4}}{4} - x^{3} + x^{2} \right]_{0}^{3} \]

### Step 3: Evaluate the integral at the limits
Now we will evaluate this expression at the upper limit \( x = 3 \) and the lower limit \( x = 0 \):
\[ I = \left( \frac{3^{4}}{4} - 3^{3} + 3^{2} \right) - \left( \frac{0^{4}}{4} - 0^{3} + 0^{2} \right) \]
Since the lower limit evaluates to 0, we only need to compute the upper limit:
\[ I = \left( \frac{81}{4} - 27 + 9 \right) \]

### Step 4: Simplify the expression
Now, we simplify the expression:
\[ I = \frac{81}{4} - 27 + 9 \]
Convert \( 27 \) and \( 9 \) to fractions with a common denominator of 4:
\[ 27 = \frac{108}{4}, \quad 9 = \frac{36}{4} \]
So,
\[ I = \frac{81}{4} - \frac{108}{4} + \frac{36}{4} = \frac{81 - 108 + 36}{4} = \frac{9}{4} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{9}{4} \]