Question

Find the area of the region that lies under the curve y = x^2 and above the x-axis for x between 0 and 1.

Answer 1

Answer:

Given y=2x−x 2 , x−axis

the points x-axis cuts the curves are 0=2−x2 =x(2−x)

x=0 or x=2

so the points are

A=(0,0) , B=(2,0)

Required area=∫ 02 2x−x 2 d=[x 2 − 3x 3 ]02 =2 2 − 32 3 −0=4− 38= 34

∴ Area= 34 sq units

Answer 2

The area under a curve between two points can be found by doing a definite integral between the two points. To find the area under the curve y = f(x) between x = 0 and x = 1, integrate y = x^2 between the limits of 0 and 1.

so after integrating y = x^2 you get

(x^3)/3 and then you substitute 1 in place of x - substitution of 0 in place of x .

(1^3)/3 - (0^3)/3

=1/3

there the area under the graph of y = x^2 with the x axis provided that the limits are 0 and 1 is

=1/3 units squared.