area question 2

Question

Find the area of the region between two curves below. (Hint: the answer is not 0)

 

y=x^3

y=x

Answer 1

The following gives an indication as to why there was a hint:

 

https://www.desmos.com/calculator/dtamgjakyg

 

We have two definite integrals to evaluate (or we can evaluate one and double it if we are convinced the 2 areas between the curves are equal.

 

∫ from (-1, 0) of (x-x³)dx + ∫ from (0 to 1) of (x³-x)dx

 

First integral:

x²/2 - x⁴/4 (evaluate from -1 to 0)

 

1/2 - 1/4 = 1/4

 

Second integral:

x⁴/4 - x²/2 evaluated from (0 to 1)

 

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

 

So the areas are equal, and 1/4 + 1/4 = 1/2.

 

By the way to find the points of intersection of the two functions substitute x for y in the first function giving:

 

x = x³

 

x³ - x = 0

 

x(x² - 1) = 0

 

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

 

So, here are three real roots (-1,0, 1).

Answer 2

On this you will do the following

 

set x³=x to find points of intersection. 

 

You will then get x³-x=0  Factor it and get x(x²-1)=0.  The points of intersection are -1, 0, 1.  You are correct in saying the answer is not 0, however you will get 0 if you perform an integration such as the following

 

∫^-1

 1     x³-x.  You will instead have to break up the integral by graphing.  

 

There is a portion where the graph of x³ is on the top and where x³ is on the bottom of the graph.  

 

It is on top from -1 to 0 and on bottom from 0 to 1. 

 

 

So set up your integral like the following: 

 

∫⁰   x³-x   The integral of this is 1/4 x⁴ - 1/2 x²   Plug in 0 and -1

 

You will get 0 -(1/4-1/2). 

 

This gives you an area of 1/. 

 

You will do the same with the other value from 0 to 1 but x will be on top so the integral will be set up like the following

 

∫¹    x-x³

 ⁰ 

 

 

You will in turn get 1/2 x² - 1/4 x⁴  Plug the values of 1 and 0 in and you will get 1/4.  

 

Add the 2 areas and your final answer is 1/2

Answer 3

First we need to know where there two functions intersect to find points of integration. Setting them equal to each other and solving for x we see they are equal at x=-1, 0, 1. Next we need to find the upper function and lower function. We can do this by picking a point in our interval (-1, 0) and (0, 1) and plugging in that number for x for x=x^3. Lets take -1/2 and 1/2, for -1/2 we see that -1/8 is larger than -1/2 so x^3 is our upper function for the interval from -1 to 0. Then for 1/2 we see that 1/2 is greater than 1/8 so y=x is our upper function for 0 to 1. Now we want to take the integral of the upper function subrtracted by the lower function for both intervals and add them together for the total area. 

integral of -x+x^3dx - integral of x-x^3dx

 

[((-x^2)/2)+((x^4)/4)] - [((x^2)/2)-((x^4)/4)]

[(1/2)+(1/4)]-[(1/2)-(1/4)]

[(3/4) - (1/4)]

 

=(1/2) 

 

Hope this helps