Que- find the area enclosed by y = x and y=x² dal find the intersections: x=x²⇒ x(x-1)=0 So x = 0,1. Betuwen these points, y=x is above y A= ∫₀¹ (x-x²)dx A= [x²/2 - x³/3]¹₀ = ½ - ⅓ A= 1/6 Square units

Step-by-step solution

1. 1

   Given Curves

   We are given two curves: $y_1 = x$ and $y_2 = x^2$.

   $$y_1 = x, y_2 = x^2$$
2. 2

   Area Formula

   Area between curves $f(x)$ and $g(x)$ from $a$ to $b$ is $\int_a^b |f(x) - g(x)| dx$.

   $$A = \int_a^b |f(x) - g(x)| dx$$
3. 3

   Find Intersection Points

   Set $y_1 = y_2$ to find the points of intersection.

   $$x = x^2$$
4. 4

   Solve for x

   Rearrange and factorize the equation to find $x$ values.

   $$x^2 - x = 0 \Rightarrow x(x-1) = 0$$
5. 5

   Intersection Points

   The intersection points are $x=0$ and $x=1$.

   $$x=0, x=1$$
6. 6

   Identify Upper Curve

   Between $x=0$ and $x=1$, $y=x$ is above $y=x^2$.

   $$x \ge x^2 \text{ for } x \in [0,1]$$
7. 7

   Set up Integral

   The area integral is $\int_0^1 (x - x^2) dx$.

   $$A = \int_0^1 (x - x^2) dx$$
8. 8

   Integrate and Evaluate

   Integrate term by term and apply the limits.

   $$A = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1$$
9. 9

   Final Area

   Substitute limits and simplify to get the area.

   $$A = \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) = \frac{1}{2} - \frac{1}{3} = \frac{3-2}{6} = \frac{1}{6}$$