If f'(x) = x^2 + 5 and f(0) = -1, then f(x) = | vedclass

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(A) Given that $f'(x) = x^2 + 5$.To find $f(x)$, we integrate $f'(x)$ with respect to $x$:$f(x) = \int (x^2 + 5) dx = \frac{x^3}{3} + 5x + C$.We are g

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$\frac{x^3}{3} + 5x - 1$

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(A) Given that $f'(x) = x^2 + 5$.To find $f(x)$, we integrate $f'(x)$ with respect to $x$:$f(x) = \int (x^2 + 5) dx = \frac{x^3}{3} + 5x + C$.We are given the condition $f(0) = -1$.Substituting $x = 0$ into the expression for $f(x)$:$f(0) = \frac{0^3}{3} + 5(0) + C = -1$.This gives $C = -1$.Therefore, the function is $f(x) = \frac{x^3}{3} + 5x - 1$.

If $f'(x) = x^2 + 5$ and $f(0) = -1$, then $f(x) = $

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