Find the area of the region.
f(x) = x^3
g(x) = x^4
In the graph f is above g and they both intersect at (1,1)
just integrate x^3-x^4 from 0 to 1.
To find the area of the region between the two curves, f(x) = x^3 and g(x) = x^4, where f(x) is above g(x) and they both intersect at (1,1), we can use the definite integral.
First, let's find the bounds of integration, which are the x-values where the two curves intersect.
Setting f(x) equal to g(x), we have:
x^3 = x^4
Rearranging the equation, we get:
x^4 - x^3 = 0
Factoring out x^3, we have:
x^3(x - 1) = 0
This equation is satisfied when either x = 0 or x - 1 = 0. Thus, the two curves intersect at x = 0 and x = 1.
Next, we integrate the difference of the two curves with respect to x and evaluate it within the interval [0, 1] to find the area.
The area A can be calculated using the formula:
A = ∫[a, b] (f(x) - g(x)) dx
In this case, a = 0 and b = 1. Therefore, the area A is given by:
A = ∫[0, 1] (x^3 - x^4) dx
Integrating the expression (x^3 - x^4) with respect to x, we get:
A = [1/4 * x^4 - 1/5 * x^5] evaluated from 0 to 1
Evaluating this expression at x = 1 and subtracting the value at x = 0, we have:
A = [1/4 * (1)^4 - 1/5 * (1)^5 - (1/4 * (0)^4 - 1/5 * (0)^5)]
Simplifying further, we get:
A = [1/4 - 1/5]
Now, we can evaluate this expression:
A = [5/20 - 4/20] = 1/20
Therefore, the area of the region between the curves f(x) = x^3 and g(x) = x^4, where f(x) is above g(x) and they both intersect at (1,1), is 1/20 square units.