MCQ
$\int|\text{x}|^3\text{ dx}$ is equal to:
- A$\frac{-\text{x}^4}{4}+\text{C}$
- B$\frac{|\text{x}|^4}{4}+\text{C}$
- C$\frac{\text{x}^4}{4}+\text{C}$
- ✓none of these.
✓
Answer
Correct option: D.
none of these.
$\int|\text{x}|^3\text{ dx}$
$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$
Case I:
When $\text{x}\geq0$
$\therefore\ \int|\text{x}|^3\text{ dx}$
$=\int\text{x}^3\text{ dx}$
$=\frac{\text{x}^4}{4}+\text{C}$
Case II:
$\text{x}<0$
$\int|\text{x}|^3\text{ dx}$
$=-\int\text{x}^3\text{ dx}$
$=\frac{-\text{x}^4}{4}+\text{C}$
$|\text{x}|=\begin{cases}\text{x},\text{ x}\geq0\\-\text{x},\text{ x}<0\end{cases}$
Case I:
When $\text{x}\geq0$
$\therefore\ \int|\text{x}|^3\text{ dx}$
$=\int\text{x}^3\text{ dx}$
$=\frac{\text{x}^4}{4}+\text{C}$
Case II:
$\text{x}<0$
$\int|\text{x}|^3\text{ dx}$
$=-\int\text{x}^3\text{ dx}$
$=\frac{-\text{x}^4}{4}+\text{C}$
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