Question

The area bounded by the curve $y = 4x - x^2$ and X-axis in square units, is \dots

• A. 32/3
• B. 16
• C. 32
• D. 21 1/3

Hint:

For a parabolic segment bounded by the x-axis, there is a shortcut formula if the roots are $\alpha$ and $\beta$: $\text{Area} = \frac{|a|}{6}(\beta - \alpha)^3$, where $a$ is the leading coefficient of $x^2$. Here, $a=-1, \alpha=0, \beta=4$. Area = $\frac{1}{6}(4-0)^3 = \frac{64}{6} = \frac{32}{3}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The area is $\int_{a}^{b} y \, dx$. We find the limits $a$ and $b$ by setting $y=0$ (where the curve hits the X-axis).

Step 2: Formula Application:
$4x - x^2 = 0 \implies x(4-x) = 0$. Limits are $x=0$ and $x=4$.

Step 3: Explanation:
Area $= \int_{0}^{4} (4x - x^2) dx = [2x^2 - x^3/3]_0^4$ $= 2(16) - 64/3 = 32 - 64/3 = (96 - 64)/3 = 32/3$.

Step 4: Final Answer:
The area is 32/3 square units.