# Find the area bounded by the graphs of the following functions: f(x)=x^2, g(x)=x^3.

hala718 | Certified Educator

Given the functions:

f(x) = x^2

g(x) = x^3

We need to find the area bounded by both curves.

First we need to determine the intersection points of the functions.

==> f(x) = g(x).

==> x^3 = x^2

==> x^3 - x^2 = 0

==> x^2 ( x-1) = 0

==> x1= 0  and x2= 1

==> y1= 0 and y2 = 1

Then the intersection points are ( 0,0) and (1,1).

Now we need to find the area of f(x) between x=1 and x=0.

==> A1 = F(1) - F(0) such that F(x) = intg f(x).

==> F(x) = intg (x^2) dx

=  x^3/3 + C

==> A = F(1) - F(0) = 1/3 - 0 = 1/3.

==> A1 = 1/3.

Now we will find the area bounded by g(x) and x= 1 and x= 0.

==> G(x) = intg g(x)

= intg x^3 dx

= x^4/4 + C

==>A2 = G(1) - G(0) = 1/4 - 0 = 1/4.

Then the area between f(x) and g(x) is:

A = A1 - A2 = 1/3 - 1/4 = (4-3)/12 = 1/12

Then, the area is (1/12) square units.

giorgiana1976 | Student

First, we'll find the intersection of the given graphs. We'll put:

y  =x^2 and y = x^3

x^2 = x^3

We'll subtract x^2 both sides:

0 = x^3 - x^2

We'll use the symmetric property and we'll factorize by x^2:

x^2(x - 1 ) = 0

We'll put x^2 = 0

x1 = x2 = 0

x - 1 = 0

x3  = 1

So, the intercepting points of the 2 graphs are x = 0 and x = 1.

We'll check what function is larger over the range [0,1].

f(0.5)>g(0.5)

Now, we can calculate the area bounded by the 2 graphs using Leibniz Newton formula:

Int (x^3 - x^2)dx = F(1) - F(0)

We'll use the additivity of integrals:

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

Int (x^2 - x^3)dx = x^3/3 - x^4/4

F(1) = 1/3 - 1/4

F(0) = 0

The area bounded by f(x) and g(x) is:

A = 1/3 - 1/4

A = 1/12 square units

neela | Student

To find the area bounded by the graphs of the following functions : f(x)=x^2, g(x)=x^3.

Let  us first find the points where f(x) = y  = x^2  and g(x)= y = x^3 intersect.

At tthe inersection y coordinates are equal. So x^2=x^3.

x^2-x^3 = 0. So x^2(1-x).

Therefore x = 0 and 1-x = 0. Or x= 1.

So the point of intersections are at x= 0  and x= 1.

Thefore the area under f(x)  from x= 0 to x =1 is calculated now:

The area A of the function f(x) between x = a and x = b is given by:

A = F(b)-F(a), where F(x) = Int(x)dx.

Int f(x) dx = Int x^2 dx.

F(x)  =  (1/3)x^3.

Therefore  Area = F(b) -F(a)  =  (1/3) (1^3-0^3) = 1/3.

Similarly area under g(x)  is given by:

B = G(b)-G(a) , where G(x) = Int g(x) dx.

G(x) = Int x^3 dx = (1/4)x^4.

B = G(1)- G(0) = (1/4) (1^4-0^4) = 1/4.

Therefore the area enclosed betweeen f(x) and g(x) = |A-B| = (1/3-1/4) = 1/12 sq units.