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

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

First we need the points of intersection of the graphs.

x^3 = x^2

=> x^3 - x^2 = 0

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

=> x = 0 and x = 1

The integral of the function x^2 between x = 0 and x = 1 is

1^3/3 - 0 = 1/3

The integral of the function x^3 between x = 0 and x = 1 is

1^4/4 - 0 = 1/4

The enclosed area is (1/3 - 1/4) = 1/12

The area bounded by x^3 and x^2 is 1/12

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

We’ve answered 318,996 questions. We can answer yours, too.

Ask a question