Question

The area bounded by the curve \( y = x|x| \), x-axis and the ordinates \( x = -1 \) and \( x = 1 \) is given by

• A. \( 0 \)
• B. \( \frac{1}{3} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{3} \)

Hint:

Area is a physical quantity and can never be zero or negative.
In area problems, if you get zero, you likely forgot to take the absolute value of the function before integrating.

Answer 1

The Correct Option is C

Concept:
• The area bounded by a curve \( y = f(x) \), the x-axis, and lines \( x=a, x=b \) is given by \( \int_{a}^{b} |y| dx \).
• Absolute value function definition: \( |x| = x \) for \( x \geq 0 \) and \( |x| = -x \) for \( x < 0 \).

Step 1: Redefine the function to handle the absolute value
The given curve is \( y = x|x| \).
For \( x \geq 0 \), \( y = x(x) = x^2 \).
For \( x < 0 \), \( y = x(-x) = -x^2 \). Thus: \[ y = \begin{cases} -x^2 & \text{if } x \\ x^2 & \text{if } x \geq 0 \end{cases} \]

Step 2: Set up the integral for area
The area required is the integral of the absolute value of \( y \) from \( -1 \) to \( 1 \): \[ \text{Area} = \int_{-1}^{1} |x|x|| dx \] Since \( |x|x|| = x^2 \) for all real \( x \) (because \( x^2 \) is always non-negative): \[ \text{Area} = \int_{-1}^{1} x^2 dx \]

Step 3: Evaluate the definite integral
Since \( x^2 \) is an even function, we can simplify the integral: \[ \text{Area} = 2 \int_{0}^{1} x^2 dx \] \[ \text{Area} = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ \text{Area} = 2 \left( \frac{1^3}{3} - \frac{0^3}{3} \right) \] \[ \text{Area} = 2 \left( \frac{1}{3} \right) = \frac{2}{3} \]