What is the total area bounded by the curves y = x^3 and y = x^7 .?

Question

I'm not getting any of the numerical answers from the text book.

a. 1/40
b. 1/24
c. 1/20
d. 1/12
e. 1/8
f. 1/6
g. 1/4
h. 1/2
i. none of these

Answer 1

1st find the point of intersection

x^3 = x^7

x^7-x^3 = 0
x^3(x^4-1) =0

so x= 0 x = -1 or x = 1

so there are 2 regions -1 to 0 and 0 to 1

the total area is sum of both areas and both areas are same

x^3 > x^7 in 0 to 1

x^3-x^7
integrate and get x^4/4-x^8/8

at 0 value = 0
at 1 value = 1/4-1/8 = 1/8
area in the region 0 1o 1 = 1/8
area in region -1 to 0 = 1/8

total area = 2* 1/8 or 1/4

Answer 2

Step 1: Determine the bounds of integration. To do this, you must equate the functions to each other, and solve for x.

x^7 = x^3

Solve for x by factoring.

x^7 - x^3 = 0
x^3 (x^4 - 1) = 0
x^3 (x^2 - 1) (x^2 + 1) = 0
x^3 (x - 1) (x + 1) (x^2 + 1) = 0

We equate each of those factors to 0.
x^3 = 0
x - 1 = 0
x + 1 = 0
x^2 + 1 = 0

And then solve for each.
x^3 = 0 implies x = 0
x = 1
x = -1
x^2 = -1 [this cannot happen, so we reject this factor]

So we have three points of intersection: -1, 0, and 1. Now, we have to determine which is the higher function from -1 to 0, and then determine the higher function from 0 to 1.

Test -1/2 in both functions: (-1/2)^3 = (-1/8) and (-1/2)^7 = (-1/64). x^7 is higher because -1/128 is greater than -1/8.

So the first part of our area calculation goes as follows

A1 = Integral (-1 to 0, [x^7 - x^3]dx)

Now, let's test a value between 0 and 1 to determine the higher function. Let's test x=1/2. then (1/2)^3 = 1/8 and (1/2)^7 = 1/128. In this case, x^3 is the higher function, so the second part of our area calculation goes as follows:

A2 = Integral (0 to 1, [x^3 - x^7]dx)

Our total area is equal to A1 + A2, but let's solve for the areas individually first.

A1 = Integral (-1 to 0, [x^7 - x^3]dx)
A1 = [(x^8)/8 - (x^4)/4] {evaluated from -1 to 0)
A1 = [0 - 0] - { [(-1)^8 / 8] - [(-1)^4 / 4] }
A1 = 0 - { 1/8 - 1/4 }
A1 = 1/8

A2 = Integral (0 to 1, [x^3 - x^7]dx)
A2 = [(x^4)/4 - (x^8)/8] evaluated from 0 to 1
A2 = [1/4 - 1/8] - [0 - 0]
A2 = 1/8

Therefore

A = A1 + A2
A = 1/8 + 1/8 = 2/8 = 1/4