Question

The area bounded by the parabola $y = x^2$ and the line $y = x$ is

• A. $\frac{1}{2}$ sq. units
• B. $\frac{1}{3}$ sq. units
• C. $\frac{2}{3}$ sq. units
• D. $\frac{1}{6}$ sq. units

Hint:

There is a standard shortcut formula for the area bounded between a parabola $x^2 = 4ay$ and a line $y = mx$: $\text{Area} = \frac{8a^2}{3m^3}$. Here, $4a = 1 \implies a = \frac{1}{4}$ and $m = 1$. Area = $\frac{8(1/16)}{3(1)} = \frac{1/2}{3} = \frac{1}{6}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the area of the region enclosed between the upward-facing parabola $y = x^2$ and the straight line $y = x$.

Step 2: Key Formula or Approach:
First, find the points of intersection by setting the two equations equal to each other. Then, integrate the difference between the upper curve and the lower curve between those intersection limits: $\text{Area} = \int_{a}^{b} (y_{\text{upper}} - y_{\text{lower}}) dx$.

Step 3: Detailed Explanation:
Equate the two functions to find the points of intersection:
$$x^2 = x$$ $$x^2 - x = 0$$ $$x(x - 1) = 0$$ The curves intersect at $x = 0$ and $x = 1$.
Between $x = 0$ and $x = 1$, the line $y = x$ lies above the parabola $y = x^2$ (for example, at $x = 0.5$, $0.5 > 0.25$).
Set up the definite integral for the area:
$$\text{Area} = \int_{0}^{1} (x - x^2) dx$$ Integrate term by term:
$$\text{Area} = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}$$ Substitute the upper and lower limits:
$$\text{Area} = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - (0 - 0)$$ $$\text{Area} = \frac{1}{2} - \frac{1}{3}$$ Find a common denominator (6) to subtract:
$$\text{Area} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$

Step 4: Final Answer:
The bounded area is $\frac{1}{6}$ sq. units, matching option (D).