FInd the area of the region between y=x^(1/2) and y=x^(1/5) for 0<x<1.
∫x^1/5 - x^1/2 dx = 5/6 x^6/5 - 2/3 x^3/2[0,1]
= (5/6 - 2/3) - (0-0) = 1/6
To find the area of the region between two curves, you need to integrate the difference between the upper and lower curves with respect to x within the given interval.
In this case, the upper curve is y = x^(1/2) and the lower curve is y = x^(1/5). We need to find the x-values where the two curves intersect to determine the interval of integration.
Setting the two equations equal to each other, we have:
x^(1/2) = x^(1/5)
To get rid of the exponents, we can raise both sides of the equation to the power of 10:
(x^(1/2))^10 = (x^(1/5))^10
x^5 = x^2
x^5 - x^2 = 0
Factoring out x^2, we get:
x^2(x^3 - 1) = 0
Solving for x, we have two possible solutions:
x^2 = 0 --> x = 0 (extraneous solution)
x^3 - 1 = 0 --> x = 1
So, the interval of integration is 0 < x < 1.
To find the area, we evaluate the integral of the difference between the upper and lower curves over the interval [0, 1]:
A = ∫[0,1] (x^(1/2) - x^(1/5)) dx
Integrating the difference of the curves, we get:
A = ∫[0,1] (x^(1/2) - x^(1/5)) dx
= [2/3 * x^(3/2) - 5/6 * x^(6/5)] evaluated from 0 to 1
= (2/3 - 5/6) - (0)
= 1/6
Therefore, the area of the region between y = x^(1/2) and y = x^(1/5) for 0 < x < 1 is 1/6 square units.