Find the area of the region between y=x1/2 and y=x1/4 for 0≤x≤1.

Answer 1

To find the area of the region between the two curves, we need to find the points of intersection first.

x^(1/2) = x^(1/4)

Squaring both sides, we get:

x = x^(1/2)^2 = x^(1/4)^4 = x^(1/2)

Subtracting x^(1/2) from both sides, we get:

0 = x^(1/2) - x

Squaring again, we get:

x = x - 2x^(3/2) + x^2

Simplifying, we get:

x^(3/2) - x^2 = 0

Factoring out x^(3/2), we get:

x^(3/2)(1 - x^(1/2)) = 0

Therefore, x = 0 or x = 1.

Now we can find the area of the region using integration:

∫(x^(1/2) - x^(1/4)) dx from x=0 to x=1

= [2/3x^(3/2) - 4/5x^(5/4)] from x=0 to x=1

= 2/3 - 4/5

= 2/15

Therefore, the area of the region between y=x^(1/2) and y=x^(1/4) for 0≤x≤1 is 2/15 square units.