Final answer:
To find the area below the curve y=8x−x2 and above the x-axis, we need to integrate the equation with respect to x and evaluate the definite integral using the limits of integration. The area is found to be 256/3 square units.
Step-by-step explanation:
To find the area below the curve y=8x−x2 and above the x-axis, we need to integrate the equation with respect to x. First, let's set up the integral:
A = ∫(8x - x^2) dx. To find the limits of integration, we need to determine the x-values where the curve intersects the x-axis. Solving 8x - x^2 = 0, we find x=0 and x=8. Therefore, our limits of integration are from x=0 to x=8:
A = ∫(8x - x^2) dx from x=0 to x=8.
Now, we can integrate the equation: A = [4x^2 - (1/3)x^3] from x=0 to x=8.
Plugging in the limits of integration, we have:
A = [4(8)^2 - (1/3)(8)^3] - [4(0)^2 - (1/3)(0)^3],
A = [256 - (512/3)] - [0 - 0],
A = [256 - (512/3)],
A = (768 - 512)/3,
A = 256/3.
Therefore, the area below the curve y=8x−x2 and above the x-axis is 256/3 square units.