Question

\[\int_{-1}^{1} x|x| \,dx\]

• A. 1
• B. \(\frac12\)
• C. 0
• D. \(\frac23\)

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the integral $\int_{-1}^1 x |x| dx$.

1. Split the Integral:
Since $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$, we have:
$x |x| = x^2$ for $x \ge 0$ and $x |x| = -x^2$ for $x < 0$.
Thus:
$\int_{-1}^1 x |x| dx = \int_{-1}^0 -x^2 dx + \int_0^1 x^2 dx$.

2. Evaluate the Integrals:
Compute each part:
$\int_{-1}^0 -x^2 dx = -\int_{-1}^0 x^2 dx = -\left[ \frac{x^3}{3} \right]_{-1}^0 = -\left( \frac{0^3}{3} - \frac{(-1)^3}{3} \right) = -\left( 0 + \frac{1}{3} \right) = -\frac{1}{3}$.
$\int_0^1 x^2 dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$.
Sum:
$-\frac{1}{3} + \frac{1}{3} = 0$.

3. Alternative Approach (Symmetry):
Define $f(x) = x |x|$. Check if it’s odd:
$f(-x) = (-x) |-x| = (-x) |x| = -x |x| = -f(x)$, so $f(x)$ is odd.
For an odd function, $\int_{-a}^a f(x) dx = 0$.
Thus, $\int_{-1}^1 x |x| dx = 0$.

Final Answer:
The value of the integral is $0$.