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

2 Answers
Jan 4, 2017

Answer:

Area #= 1/2 \ "unit"^2 #

Explanation:

Graphs of the functions #y=x^3# and #y=x# are:
graph{(y-x^3)(y-x)=0 [-3.465, 3.464, -1.733, 1.73]}

Both #y=x^3# and #y=x# are odd functions so by symmetry we only need to consider the RHS and double it. We can see from the graph that the (positive) intersection points are #(0,0)# and #(1,1)# which we can verify algebraical:

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

The area bounded by the #x#-axis, #y#-axis and #y=x# is that of a triangle with area:

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

The area bounded by the #x#-axis, #y#-axis and #y=x^3# is given by:

#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#

Jan 4, 2017

Answer:

#1/2# unit#"s"^2#

Explanation:

Start by finding the points of intersection by solving the system #{(y = x^3), (y = x):}#

#x^3 = x#

#x^3 - x = 0#

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

#x(x + 1)(x - 1) = 0#

#x = 0, 1 and -1#

So, we have to find the area in the interval #0 ≤ x ≤ 1# and in the interval #-1 ≤ x ≤ 0#. If you trace the graphs of the two functions, you will find that the area in #-1 ≤ x ≤ 0# lies beneath the y-axis while the area in #0 ≤ x ≤1# will lie above the y-axis. We simply add the two areas up in this case.

Here is the graph:

enter image source here

As you can see, in the interval #0 ≤ x ≤ 1#, the line #y = x# lies above #y = x^3# and in the interval #-1 ≤ x ≤ 0#, #y = x^3# lies above #y = x#. Our integrals will therefore be #int_-1^0 (x^3 - x)# and #int_0^1 (x - x^3)#

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

Integrate using the rule #int(x^n)dx = x^(n + 1)/(n + 1)#:

#= [1/4x^4 - 1/2x^2]_-1^0#

Evaluate using #int_a^bF(x)dx = f(b) - f(a)#, where #f'(x) = F(x)#.

#= 1/4(0)^4 - 1/2(0)^2 - (1/4(-1)^4 - 1/2(-1)^2)#

#=-1/4 + 1/2#

#= 1/4#

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

Use the same process as above:

#=[1/2x^2 - 1/4x^4]_0^1#

#=1/2(1)^2 - 1/4(1)^2 - (1/2(0)^2 - 1/4(0)^4)#

#= 1/4#

Add the two areas to find the total area.

#A_"total" = 1/4 + 1/4 = 1/2#

Hence, the area between #y= x^3# and #y = x# is #1/2# unit#"s"^2#.

Hopefully this helps!