Find the area bounded by the graphs of the following functions: f(x)=x^2, g(x)=x^3.
Given the functions:
f(x) = x^2
g(x) = x^3
We need to find the area bounded by both curves.
First we need to determine the intersection points of the functions.
==> f(x) = g(x).
==> x^3 = x^2
==> x^3 - x^2 = 0
==> x^2 ( x-1) = 0
==> x1= 0 and x2= 1
==> y1= 0 and y2 = 1
Then the intersection points are ( 0,0) and (1,1).
Now we need to find the area of f(x) between x=1 and x=0.
==> A1 = F(1) - F(0) such that F(x) = intg f(x).
==> F(x) = intg (x^2) dx
= x^3/3 + C
==> A = F(1) - F(0) = 1/3 - 0 = 1/3.
==> A1 = 1/3.
Now we will find the area bounded by g(x) and x= 1 and x= 0.
==> G(x) = intg g(x)
= intg x^3 dx
= x^4/4 + C
==>A2 = G(1) - G(0) = 1/4 - 0 = 1/4.
Then the area between f(x) and g(x) is:
A = A1 - A2 = 1/3 - 1/4 = (4-3)/12 = 1/12
Then, the area is (1/12) square units.
First, we'll find the intersection of the given graphs. We'll put:
y =x^2 and y = x^3
x^2 = x^3
We'll subtract x^2 both sides:
0 = x^3 - x^2
We'll use the symmetric property and we'll factorize by x^2:
x^2(x - 1 ) = 0
We'll put x^2 = 0
x1 = x2 = 0
x - 1 = 0
x3 = 1
So, the intercepting points of the 2 graphs are x = 0 and x = 1.
We'll check what function is larger over the range [0,1].
Now, we can calculate the area bounded by the 2 graphs using Leibniz Newton formula:
Int (x^3 - x^2)dx = F(1) - F(0)
We'll use the additivity of integrals:
Int (x^3 - x^2)dx = Int x^3dx - Intx^2dx
Int (x^2 - x^3)dx = x^3/3 - x^4/4
F(1) = 1/3 - 1/4
F(0) = 0
The area bounded by f(x) and g(x) is:
A = 1/3 - 1/4
A = 1/12 square units
To find the area bounded by the graphs of the following functions : f(x)=x^2, g(x)=x^3.
Let us first find the points where f(x) = y = x^2 and g(x)= y = x^3 intersect.
At tthe inersection y coordinates are equal. So x^2=x^3.
x^2-x^3 = 0. So x^2(1-x).
Therefore x = 0 and 1-x = 0. Or x= 1.
So the point of intersections are at x= 0 and x= 1.
Thefore the area under f(x) from x= 0 to x =1 is calculated now:
The area A of the function f(x) between x = a and x = b is given by:
A = F(b)-F(a), where F(x) = Int(x)dx.
Int f(x) dx = Int x^2 dx.
F(x) = (1/3)x^3.
Therefore Area = F(b) -F(a) = (1/3) (1^3-0^3) = 1/3.
Similarly area under g(x) is given by:
B = G(b)-G(a) , where G(x) = Int g(x) dx.
G(x) = Int x^3 dx = (1/4)x^4.
B = G(1)- G(0) = (1/4) (1^4-0^4) = 1/4.
Therefore the area enclosed betweeen f(x) and g(x) = |A-B| = (1/3-1/4) = 1/12 sq units.