I have $N$ different linear functions defined by,
$$y_i = a_i - b_i Q_i, \forall i \in [1, 2, \cdots N]$$
Which I want to aggregate. Here, $a_i$, $b_i$ are constants, and are $\in \mathbb{R}^+$. I can represent the variable $Q_i$ as $\alpha \beta_i$, where, $\beta_i$ is also a constant.
Essentially, my aggregated function $Y(Q)$ and is defined as:
$$Y(Q) = \sum_{i=1}^{N} y_i = \sum_{i=1}^{N} a_i - \alpha \sum_{i=1}^{N} b_i \beta_i$$
And, I have defined, $Q$ to be $= \alpha\sum_{i=1}^{N}\beta_i$
Then, my question is can I write as follows?
$$\frac{d}{d\alpha} \int_0^{\alpha \sum_{i=1}^{N} \beta_i} Y(Q) dQ = \left( \sum_{i=1}^{N} \beta_i \right) Y( \alpha \sum_{i=1}^{N} \beta_i ) $$
$$ \alpha = \dfrac{Q}{\sum_{i=1}^N \beta_i} $$.
So, $$\frac{d}{d\alpha} \int_0^{\alpha \sum_{i=1}^{N} \beta_i} Y(Q) dQ = \Bigg(\sum_{i=1}^{N} \beta_i\Bigg)\frac{d}{d Q} \int_0^{Q} Y(Q) dQ = \left( \sum_{i=1}^{N} \beta_i \right) Y( \alpha \sum_{i=1}^{N} \beta_i )$$
The second equality follows because of the second fundamental theorem of calculus. Unless I am missing something, you are right.