Constant when integrating

Question

I was integrating $(t+2)^2$ by using the substitution method and by expanding the function. By expanding it then integrating I got the same answer as the book ($\frac{1}{3}(t^3+6t^2+12t)+C$), but when I substituted I got $\frac{1}{3}\cdot(t+2)^3 + C = \frac{1}{3} \cdot (t^3+6t^2+12t+8) + C$. Am I missing something here? Got an 8 at the end when substituting, but no constant when expanding.

Answer 1

Both of your answers are correct. Recall that the antiderivative of a function $f$

$$\int f(x) dx$$

is not a function. It's the collection of all functions $g$ so that $g' = f$. So whenever you write

$$ \int f(x) dx = (\text{some function}) + C$$

it is understood that $C$ is any constant. So, going back to your case,

$$ \frac 13 (t+2)^3 +C \ \ \text{and} \ \ \frac 13 (t^3 + 6t^2 + 12t + 8) +C$$

represents the same collections of functions. So both of them are the same.

Answer 2

Your answers differ by a constant factor, so the values of $C$ will be different in these 2 methods.

In other words, $$ \frac{t^3+6t^2+12t}{3} + C= \frac{t^3 + 6t^2 + 12t +8}{3} + K $$ just $C = K + 8/3$...

Answer 3

No worries,

$$\left(\frac{1}{3}(t^3+6t^2+12t)+C\right)'=t^2+4t+4$$ and $$\left(\frac{1}{3}(t+2)^3 + C \right)'=(t+2)^2.$$