$$C(x) = \int C'(x) dx = \int 4x\sqrt{x+3} dx$$
To solve this integral, we can use substitution. Let $u = x + 3$, then $du = dx$. When $x = 0$, $u = 3$. We can rewrite the integral in terms of $u$:
$$C(x) = \int_{3}^{u} 4(u-3)\sqrt{u} du$$
Now, we can
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