# What is the area between the line #y = x# and the curve #y = x^3#?

##### 2 Answers

Area

#### Explanation:

Graphs of the functions

graph{(y-x^3)(y-x)=0 [-3.465, 3.464, -1.733, 1.73]}

Both

#x^3=x => x^3-x=0 => x(x^2-1)=0 => x-0,+-1#

The area bounded by the

#A_1 = 1/2(1)(1) = 1/2#

The area bounded by the

#A_2 = int_0^1 \ x^3 \ dx #

# \ \ \ \ =[1/4x^4]_0^1 #

# \ \ \ \ =1/4 #

Then the total bounded area that we seek (LHS and RHS) is given by:

#A=2(A_1-A_2)#

# \ \ \ =2(1/2-1/4)#

# \ \ \ =1/2#

#### Explanation:

Start by finding the points of intersection by solving the system

So, we have to find the area in the interval

Here is the graph:

As you can see, in the interval

**Evaluating #int_-1^0 (x^3 - x) dx#**

Integrate using the rule

Evaluate using

**Evaluating #int_0^1x - x^3dx#**

Use the same process as above:

Add the two areas to find the total area.

Hence, the area between

Hopefully this helps!