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 Question:
We need to find the geometric area enclosed between the downward-opening parabola $y = 4x - x^2$ and the horizontal x-axis ($y = 0$).

Step 2: Key Formula or Approach:
1. Find the roots (x-intercepts) of the parabola to determine the limits of integration ($a$ and $b$).
2. Integrate the function $f(x)$ with respect to $x$ over the interval $[a, b]$.
$$\text{Area} = \int_{a}^{b} (4x - x^2) \, dx$$

Step 3: Detailed Explanation:
Find the x-intercepts by setting $y = 0$:
$$4x - x^2 = 0 \implies x(4 - x) = 0$$
The roots are $x = 0$ and $x = 4$.
Since the parabola opens downwards (the $x^2$ coefficient is negative), the curve lies entirely above the x-axis between $x=0$ and $x=4$.
Set up the definite integral:
$$\text{Area} = \int_{0}^{4} (4x - x^2) \, dx$$
Integrate term by term:
$$= \left[ \frac{4x^2}{2} - \frac{x^3}{3} \right]_0^4 = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4$$
Evaluate at the upper limit (the lower limit evaluates to 0):
$$= \left( 2(4)^2 - \frac{(4)^3}{3} \right) - 0$$
$$= 2(16) - \frac{64}{3} = 32 - \frac{64}{3}$$
Find a common denominator:
$$= \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$$

Step 4: Final Answer:
The area is 32/3 square units, matching option (a).