How to perform integration for questions involving cumulative distribution functions?

Question

From a textbook, the following example is shown. I do not understand why $2(x-1)$ integrates to become $(x-1)^2$. How come it isn't $x^2-2x$? Is it something to do with the constant since $x^2-2x (+1)$ would then equal $(x-1)^2$. Also, isn't this a definite integral with limits $k$ and $1$ so isn't the constant not included? Any clarification would be appreciated.

Answer 1

The antiderivatives $(x-1)^2 +C$ and $x^2-2x+C$ are both correct, they're equivalent.

When you expand the square, you get $x^2 - 2x + 1 + C$, which is the same as $x^2 + 2x + C_2$ where $C_2 = C + 1$. The constants are arbitrary so these are two ways of addressing the same set of functions.

The fact that this is a definite integral isn't a concern because both functions satisfy $F'(x) = 2(x-1)$, so the fundamental theorem of calculus applies equally to both. Equivalently you can see this as the fact that any constant will cancel out when you do $F(k) - F(1)$.

Answer 2

You can use $x^2-2x$ if you wish; you'll get the same answer. Since it is a definite integral, the constant will cancel.

$$[x^2-2x]_{x=1}^k = k^2 - 2k - (1^2 - 2) = k^2 - 2k + 1 = (k-1)^2.$$