# AreaFind the area between y=x and f(x)=x^2.

### 3 Answers | Add Yours

You need to evaluate the limits of integration, hence, you need to solve the following equation, such that:

`x^2 = x => x^2 - x = 0 => x(x - 1) = 0 => {(x = 0),(x - 1 = 0):}`

`{(x = 0),(x = 1):}`

You need to evaluate the area of the region bounded by the given curves, hence, you need to evaluate the following definite integral, such that:

`A = int_0^1(x - x^2)dx` (` y = x > y = x^2` over `x in [0,1]` )

You need to use the property of linearity of integral, such that:

`A = int_0^1 x dx - int_0^1 x^2 dx`

`A = (x^2/2 - x^3/3)|_0^1`

`A = (1/2 - 0/2 - 1/3 + 0/3) => A = (1/2 - 1/3)`

`A = 1/6`

**Hence, evaluating the area of the region bounded by the given curves yields **`A = 1/6.`

The area enclosed between the curves y=x and f(x)=x^2 has to be determined.

y = x is the equation of a straight line and y = x^2 is the equation of a parabola. First, find the points of contact between the two. This is done by solving y = x and y = x^2

x = x^2

x^2 - x = 0

x(x -1) = 0

x = 0 and x = 1

The required area is the absolute value of the integral :

`int_0^1 x^2 - x dx`

= `int_0^1 x^2 dx - int_0^1 x dx`

= `(x^3/3)_0^1 - (x^2/2)_0^1`

= `1/3 - 1/2`

= `-1/6`

The area is the absolute value of this or 1/6 square units.

You've given only the limit curves but you did not gave the limit lines.

We'll calculate the area located between the 2 given curves, using Leibniz-Newton formula:

Int (x^2 - x)dx = F(b) - F(a), where x = a and x = b are the limit lines.

We'll determine the indefinite integral first:

Int (x^2 - x)dx = Int x^2 dx - Int x dx

Int (x^2 - x)dx = x^3/3 - x^2/2

F(b) = b^3/3 - b^2/2

F(a) = a^3/3 - a^2/2

Int (x^2 - x)dx = b^3/3 - b^2/2 - a^3/3 + a^2/2