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

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(C) 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

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

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(C) 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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