# Area; functionsFind the area bounded by the graphs of the function f(x)=x^3 and g(x)=x^2, if x belongs to interval[0,1].

Asked on by jolenta

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The area between the curves f and g is the result of difference between the areas of the functions.

Since the difference must be positive and both areas are positive, you need to check what function is the minuend and what is subtrahend.

This is an easy task. You need to substitute x, in each function, by values from [0,1] and then compare the results.

x = 0 => f(x) = 0^3 = 0

x = 0 => g(x) = 0^2 = 0

x = 1/2 => f(x) = 1/8

x = 1/2 => g(x) = 1/4

Notice that for x = 1/2, ﻿g(x)>f(x)

That means that the curve representing f is under the curve representing g, over [0,1].

The minuend is g and the subtrahend is f. The region enclosed by f and g is g-f.

The area may be determined using Riemann's sum or definite integral.

int (x^2-x^3) = x^3/3-x^4/4 = (4x^3-3x^4)/12

Area =  ﻿(4-3)/12 - 0/12 = 1/12

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To calculate the area between the functions we have to calculate the definite integral from the difference between the functions. For this reason, we have to verify which function is bigger.

In our case, because x belongs to the interval [0,1], x^2>x^3, so the integral will be:

Integral (g(x)-f(x))dx = Integral(x^2-x^3)dx

Integral(x^2-x^3)dx = Integral x^2 - Integral x^3

Integral x^2 - Integral x^3 = x^3/3 - x^4/4

x^3/3 - x^4/4 = (1^3/3 - 0^3/3) - (1^4/4 - 0^4/4) = (1/3)-(1/4)

Integral(x^2-x^3)dx = (1/3)-(1/4)=1/12

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