Find the area of the region bounded by y=x and y=x2 in the first quadrant.?

Answer 1

Intersect when x = x²
x.(x - 1) = 0
x = 0, x = 1
Area = ∫ x - x² dx-----lims 0 to 1
Area = x² / 2 - x³ / 3-----lims 0 to 1
Area = 1 / 2 - 1 / 3
Area = 1 / 6 units²

Answer 2

Let f(x) = x and g(x) = x^2.

To determine the area of the region bounded by x, x^2, and the first quadrant, we want to find the area to the right of the vertical line x = 0 (since the first quadrant begins at 0).

Our first step is to find our bounds of integration. This is done by equating f(x) and g(x).

x = x^2

Solve for x.

0 = x^2 - x
0 = x(x - 1), which implies x = {0, 1}.
Our bounds for integration are 0 and 1.

Our next step is to determine which of f(x) or g(x) is greater, on the interval 0 to 1. After all, the area will be calculated by

A = Integral (0 to 1, {higher curve} - {lower curve} dx )

Test x = 1/2.
f(1/2) = 1/2
g(1/2) = (1/2)^2 = 1/4

1/2 is greater than 1/4, so f(x) is greater on that interval. Therefore,

A = Integral (0 to 1, [ f(x) - g(x) ]dx )

A = Integral (0 to 1, [ x - x^2 ] dx )

And solve as normal.

A = [ (1/2)x^2 - (1/3)x^3 ] {evaluated from 0 to 1}
A = [ (1/2)1^2 - (1/3)1^3] - [ (1/2)0^2 - (1/3)0^3 ]
A = [ (1/2) - (1/3) ] - [ 0 - 0 ]
A = [ (3/6) - (2/6) ] - 0
A = 1/6

Answer 3

If x^2 - x = 0, then:
x(x - 1) = 0
x = 0 or 1.

The graphs meet where x = 0 and x = 1.

The area between them is:
Int (0 to 1)( x - x^2 )dx
= [ x^2 / 2 - x^3 / 3 ] (0 to 1)
= 1 / 2 - 1 / 3
= ( 3 - 2 ) / 6
= 1 / 6.

Answer 4

It's the integral from 0 to 1 of (x - x^2)dx

Answer 5

integral (0 to 1) [ x-x^2 ] dx = 1/6

Answer 6

300.78