Find the area of y=cute root of (2x), y=(1/8)x^2, 0 less than or equal to x less than or equal to 6

Answer 1

I assume you mean the area between the curves. In that case, there are two possibilities which you need to consider. If you want the algebraic area, it is just
∫[0,6] (2x)^(1/3) - 1/8 x^2 dx = 9 ((3/2)^(1/3) - 1) ≈ 1.3024

However, since the curves cross at x=4, if you want the geometric area, that would be

∫[0,4] (2x)^(1/3) - 1/8 x^2 dx + ∫[4,6] 1/8 x^2 - (2x)^(1/3) dx

when you figure out what you want, crank it out.

Answer 2

how did you you get 9 ((3/2)^(1/3) - 1) steve?

Answer 3

well, I evaluated the integral. It's the simple power rule, right?
∫[0,6] (2x)^(1/3) - 1/8 x^2 dx
Let u = 2x so du = 2 dx. Now you get
(1/2)(3/4)(2x)^(4/3) - (1/8)(1/3 x^3) = 3/8 (2x)^(4/3) - 1/24 x^3
evaluate at 6 and 0, to get
(3/8 (12)^(4/3) - 1/24 6^3)-(3/8 (0)^(4/3) - 1/24 0^3)
= 3/8 * 12 * ∛12 - 9
= 9/2 ∛12 - 9
= 9(1/2 ∛12 - 1)
= 9(∛(12/8) - 1)
= 9(∛(3/2) - 1)