`int|x|^(3) dx ` is equal to

Options

A. `(1)/(4) |x|^(4) +C`
B. `(1)/(4) x^(4) + C`
C. `- ( 1)/( 4) x^(4) + C`
D. none of these

Answer

To solve the integral \(\int |x|^3 \, dx\), we need to consider the definition of the absolute value function, which behaves differently depending on whether \(x\) is positive or negative.

### Step-by-Step Solution:

1. **Split the Integral Based on the Absolute Value**:
The absolute value function \(|x|^3\) can be expressed as:
- \(|x|^3 = x^3\) when \(x \geq 0\)
- \(|x|^3 = (-x)^3 = -x^3\) when \(x < 0\)

Therefore, we can split the integral into two parts:
\[ \int |x|^3 \, dx = \int_{0}^{a} x^3 \, dx + \int_{-a}^{0} (-x^3) \, dx \]
where \(a\) is a positive constant.

2. **Integrate for \(x \geq 0\)**:
For the first part, where \(x \geq 0\):
\[ \int x^3 \, dx = \frac{x^4}{4} + C_1 \]

3. **Integrate for \(x < 0\)**:
For the second part, where \(x < 0\):
\[ \int -x^3 \, dx = -\left(\frac{x^4}{4}\right) + C_2 = -\frac{x^4}{4} + C_2 \]

4. **Combine the Results**:
We can express the results from both intervals as:
- For \(x \geq 0\): \(\frac{x^4}{4} + C_1\)
- For \(x < 0\): \(-\frac{x^4}{4} + C_2\)

To combine these, we can write:
\[ \int |x|^3 \, dx = \begin{cases} \frac{x^4}{4} + C & \text{if } x \geq 0 \\ -\frac{x^4}{4} + C & \text{if } x < 0 \end{cases} \]

5. **Final Expression**:
We can express the final result in a compact form:
\[ \int |x|^3 \, dx = \frac{|x|^4}{4} + C \]

### Final Answer:
\[ \int |x|^3 \, dx = \frac{|x|^4}{4} + C \]