Find the indefinite integral and check the result by differentiation. (Remember the constant of integration.)33 − 4x2 (−8x) dx

The integral of the function 33 - 4x^2 * (-8x) dx can be simplified first.

The function simplifies to 33 - (-32x^3) which is 33 + 32x^3.

Now, we can integrate term by term:

∫33 dx + ∫32x^3 dx

The integral of a constant (like 33) is just that constant times x. The integral of x^n (where n ≠ -1) is (1/(n+1))x^(n+1).

So, the integral of 33 is 33x and the integral of 32x^3 is (32/4)x^4 = 8x^4.

Therefore, the indefinite integral of the function is 33x + 8x^4 + C, where C is the constant of integration.

To check the result, we can differentiate this function:

d/dx (33x + 8x^4 + C) = 33 + 32x^3

which is the original function, confirming that our integral is correct.