I'm trying to do this question below, which is from an A Level textbook book.
I've said that the area enclosed by the curve, the $y-$axis, $y=2$ and $y=3$ is given by
\begin{equation} \int_{2}^{3} x \, dy = \int_{2}^{3} (y^2 - 2) \, dy \end{equation}
and so to get the area $CDE$ I just need to take off the area of the rectangle enclosed by the lines $DE$, the $y-$axis, $y=2$, and $y=3$. The area of the rectangle is 2, and so my final answer for the area of $CDE$ is
\begin{equation} \int_{2}^{3} (y^2 - 2) \, dy - 2. \end{equation}
Comparing my answer to their answer, it looks as if they have taken the $-2$ inside the integral, which I thought was odd. Sanity check: you can't just take additive constants in/ out an integral can you?
So, curiosity then lead me to investigate when you can take an additive constant in or out of an integral, and I found the following:
It looks like that when the limits of integration differ by 1, you can take in/out an additive constant.
This result does neither seem obvious or intuitive. If some disagrees, please could they enlighten me?
Do people therefore think 6(iii) is a reasonable question? Or, do you think it is possibly an overlooked typo?
@Flame Trap
Ah, so the area is actually
\begin{equation} \int_{2}^{3} (y^2 - 2) \, dy - \int_{2}^{3} 2 \, dy \end{equation}
Got it!