Question

Using integration, find the area of the region
{(x,y) : 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x≤3}

Answer 1

Area Bounded by the Curves

Given:

Find the area of the region bounded by the curves: \[ y = x^2,\quad y = x,\quad \text{from } x = 0 \text{ to } x = 3 \]

Step 1: Find points of intersection

Equate \( y = x \) and \( y = x^2 \): \[ x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x - 1) = 0 \Rightarrow x = 0,\ x = 1 \] So the region of intersection lies between \( x = 0 \) and \( x = 1 \), where: \[ y = x \text{ is above } y = x^2 \]

Step 2: Set up the integral

Area between curves from \( x = a \) to \( x = b \) is: \[ \int_a^b \left[\text{Upper} - \text{Lower}\right] dx \] Here, from \( x = 0 \) to \( x = 1 \): \[ \text{Upper} = x,\quad \text{Lower} = x^2 \Rightarrow \text{Area} = \int_0^1 (x - x^2)\,dx \]

Step 3: Evaluate the integral

\[ \int_0^1 (x - x^2)\,dx = \int_0^1 x\,dx - \int_0^1 x^2\,dx \] \[ = \left[\frac{x^2}{2}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1 = \left(\frac{1}{2} - 0\right) - \left(\frac{1}{3} - 0\right) = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \]

✅ Final Answer:

\[ \boxed{\text{Area} = \frac{1}{6}} \text{ square units} \]