Final answer:
The solution to the initial value problem f'(x) = x^3 - 8x^2 + 16x + 1, with f(0) = 0, is found by integrating f'(x) term by term and applying the initial condition to solve for the constant of integration, which yields the function f(x) = 1/4x^4 - 2/3x^3 + 8x^2 + x.
Step-by-step explanation:
To solve the initial value problem f'(x) = x³ - 8x² + 16x + 1, with f(0) = 0, we integrate f'(x) to find f(x). The integral of f'(x) is the antiderivative, which involves finding a function whose derivative is the given function.
Integrating term by term, we get:
- Integral of x³ is ⅔x⁴
- Integral of -8x² is -8/3x³
- Integral of 16x is 8x²
- Integral of 1 is x
Combining these, we have f(x) = ⅔x⁴ - ¾ x³ + 8x² + x + C, where C is the constant of integration. To find C, we use the initial condition f(0) = 0. Substituting 0 into f(x), we get C = 0.
Therefore, the solution to the initial value problem is f(x) = ⅔x⁴ - ¾ x³ + 8x² + x.