`int_0^3 |x^(3)+x^(2)+3x|dx` is equal to

Options

A. `(171)/(2)`
B. `(171)/(4)`
C. `(170)/(4)`
D. `(170)/(3)`

Answer

To solve the integral \( \int_0^3 |x^3 + x^2 + 3x| \, dx \), we will follow these steps:

### Step 1: Define the function
Let \( f(x) = x^3 + x^2 + 3x \).

### Step 2: Determine the sign of the function in the interval [0, 3]
We need to check if \( f(x) \) is positive or negative in the interval from 0 to 3.

- Calculate \( f(0) \):
\[ f(0) = 0^3 + 0^2 + 3 \cdot 0 = 0 \]

- Calculate \( f(1) \):
\[ f(1) = 1^3 + 1^2 + 3 \cdot 1 = 1 + 1 + 3 = 5 > 0 \]

- Calculate \( f(2) \):
\[ f(2) = 2^3 + 2^2 + 3 \cdot 2 = 8 + 4 + 6 = 18 > 0 \]

- Calculate \( f(3) \):
\[ f(3) = 3^3 + 3^2 + 3 \cdot 3 = 27 + 9 + 9 = 45 > 0 \]

Since \( f(x) \) is positive for \( x \) in the interval [0, 3], we can remove the absolute value.

### Step 3: Rewrite the integral without the absolute value
Thus, we can rewrite the integral as:
\[ \int_0^3 |f(x)| \, dx = \int_0^3 (x^3 + x^2 + 3x) \, dx \]

### Step 4: Integrate the function
Now we will integrate the function:
\[ \int (x^3 + x^2 + 3x) \, dx = \frac{x^4}{4} + \frac{x^3}{3} + \frac{3x^2}{2} \]

### Step 5: Evaluate the integral from 0 to 3
Now we need to evaluate this from 0 to 3:
\[ \left[ \frac{x^4}{4} + \frac{x^3}{3} + \frac{3x^2}{2} \right]_0^3 \]

Calculating at \( x = 3 \):
\[ = \frac{3^4}{4} + \frac{3^3}{3} + \frac{3 \cdot 3^2}{2} = \frac{81}{4} + 9 + \frac{27}{2} \]

Converting \( 9 \) and \( \frac{27}{2} \) to a common denominator of 4:
\[ 9 = \frac{36}{4}, \quad \frac{27}{2} = \frac{54}{4} \]

So,
\[ \frac{81}{4} + \frac{36}{4} + \frac{54}{4} = \frac{171}{4} \]

Calculating at \( x = 0 \):
\[ = 0 \]

Thus, the value of the integral is:
\[ \frac{171}{4} - 0 = \frac{171}{4} \]

### Final Answer
The integral \( \int_0^3 |x^3 + x^2 + 3x| \, dx \) is equal to \( \frac{171}{4} \).

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