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

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.

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

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)

To find the area between the two curves y = √(2x) and y = (1/8)x^2, we need to determine the points at which the curves intersect and then integrate the difference of the two functions over that interval.

Step 1: Find the points of intersection:
To find the points of intersection, we set the two equations equal to each other:
√(2x) = (1/8)x^2

Squaring both sides gives us:
2x = (1/64)x^4

Rearranging the equation, we get:
64x = x^4

This equation can be factored as:
x(x - 4)(x + 4)(x^2 + 16) = 0

Solving for x, we find that:
x = 0, x = 4, x = -4

Step 2: Determine the interval of integration:
Since the given interval is 0 ≤ x ≤ 6, we need to find the appropriate limits of integration. In this case, the limits will be at the points of intersection -4 and 4.

Step 3: Set up the integral:
To calculate the area between the two curves, we integrate the difference between them. Since the curve y = √(2x) is above y = (1/8)x^2 in the given interval, the integral will be:

∫ [√(2x) - (1/8)x^2] dx, from -4 to 4

Step 4: Evaluate the integral:
Integrating the given function, we get:
∫ [√(2x) - (1/8)x^2] dx = [2/3 (2x)^(3/2) - (1/24)x^3] from -4 to 4

Substituting the limits of integration, we have:
[2/3 (2(4))^(3/2) - (1/24)(4)^3] - [2/3 (2(-4))^(3/2) - (1/24)(-4)^3]

Simplifying further, we get:
[2/3 (8)^(3/2) - (1/24)(64)] - [2/3 (8)^(3/2) - (1/24)(-64)]

Now, calculate the values within the square brackets and subtract them to find the area between the curves.

Note: It is recommended to use a calculator or any computational tool to evaluate the exact numerical value of the integral.