Evaluate the indefinite integral: 8x-x^2.

I got this but I the homework system says its wrong:sqrt((-x-8)x)/(2*sqrt(x-8)*sqrt(x))*(((sqrt(x-8)*(x-4)*sqrt(x))-32*log(sqrt(x-8)+sqrt(x))

How on earth did you get that mess? Just add the separate integrals of 8x and -x^2.


The answer is
4x^2 - (x^3)/3

To verify, differentiate it and see if you get 8x-x^2 back again.

Sorry, it is evaluate the indefinite integral: sqrt(8x-x^2)

2y*y'=x/sqrt(x^2-16)

To evaluate the indefinite integral of 8x - x^2, we can follow these steps:

1. Separate the terms: The integral of 8x - x^2 can be written as the sum of two integrals: ∫(8x) dx - ∫(x^2) dx.

2. Evaluate the integrals separately:

- For the first integral, we use the power rule of integration. The power rule states that ∫(x^n) dx = (x^(n+1))/(n+1) + C, where C is the constant of integration. Applying this to the first integral, we get:
∫(8x) dx = (8/2)x^2 + C1 = 4x^2 + C1.

- For the second integral, we again use the power rule, but with n = 2:
∫(x^2) dx = (1/3)x^3 + C2.

3. Combine the results: Since we are looking for an indefinite integral, we do not need to worry about specific limits of integration. We can simply add the constants of integration to get the final answer:
∫(8x - x^2) dx = 4x^2 + (1/3)x^3 + C.

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