The Area of bounded functions

The Area and Volume of bounded functions: 


This is not the most informative blog, but for those who have a rough idea of whats going on should be able to grasp and understand the examples shown.


Find the area between
y = 2 - x^2 and y = (1/2)x.

from x = 0  to x = 1 .









 Now as seen above we use integration to determine this. Basically you find the integral of the equation shown and solve for the domain of it. Which in this case is  x = 0 to x = 1.

Now the example jumps right to the answer being 17/12. The actual steps behind the seen is

= [ 2x - (x^3 / 3) - (x^2 / 4)] from x = 0 to x = 1

=[ (2(1) - (1^3 / 3) - ( 1^2 / 4) - (2(0) - (0^3)/3 - 0^2 / 4) ]

= [ 2 - 1/3 - 1/4 - 0 - 0 - 0]

= [ 2 - 1/3 - 1/4]

=  [ 6/3 - 1/3 - 1/4]  = [ 5/3 - 1/4] = [20/12 - 3/12] = 17/12 units squared. YAH!

So after these steps we have found the area between these two functions from x = 0 to x  = 1.


Find the area:

y = x^(1/3) , and y = x^3 from x = 0 to x = 2.

First thing we should notice about this is that they intersect at  x= 1. Therefore we need to split this up into two different integrals. One from x = 0 to 1, and the other from x = 1 to 2.







             

= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 0 to x = 1

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

= 3/4 - 1/4 - 0 - 0 = 3/4 - 1/4 = 2/4 = 1/2 YAH!



= [ 3/4 x ^(4/3) - 1/4 x ^4] from x = 1 to x =2

= [ 3/4 (2) ^(4/3) - 1/4 (2) ^4 -  3/4 (2) ^(4/3) - 1/4 (2)^4]

= 4 - 3/2(2)^1/3 - 1/4 - 3/4

= 4.5 - 1.5(2)^1/3

So add up the values of both these integrals to get

. 5 + 4.5 - 1.5(2)^1/3 = 3.11 approximate units squared.

This is how you find the area between two functions bounded.