Question

Using integration, find the area bounded by the curve \( y = x^2 \) and the line \( y = 2 \).

Hint:

Find the intersection points \( x = \pm\sqrt{2} \), then integrate \( 2 - x^2 \) between them.

Answer 1

Step 1: Find where the curves meet:
The curve is \( y = x^2 \) and the line is \( y = 2 \).
Put \( x^2 = 2 \). So \( x = \pm\sqrt{2} \).
The bounded region lies between \( x = -\sqrt{2} \) and \( x = \sqrt{2} \).
In this region the line is above the parabola.

Step 2: Write the area as an integral:
Area = (upper curve) - (lower curve), integrated over \( x \).
\[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (2 - x^2)\,dx \]
The integrand is even, so we can double the integral from 0 to \( \sqrt{2} \).
\[ A = 2\int_{0}^{\sqrt{2}} (2 - x^2)\,dx \]

Step 3: Integrate:
\[ A = 2\left[ 2x - \frac{x^3}{3} \right]_{0}^{\sqrt{2}} \]
Put the limits: \( 2x = 2\sqrt{2} \) and \( \frac{x^3}{3} = \frac{2\sqrt{2}}{3} \).
\[ A = 2\left( 2\sqrt{2} - \frac{2\sqrt{2}}{3} \right) = 2 \cdot \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \]

Final Answer:
The required area is \( \dfrac{8\sqrt{2}}{3} \) square units. \[ \boxed{\dfrac{8\sqrt{2}}{3}\ \text{sq. units}} \]