I have an integral equation containing dot product
$$\int_{0}^{L} \left(\frac{a}{L}.b(s)\right)\mathrm ds\tag 1$$
Data Given
Question
Can we write $$\int_{0}^{L} \left(\frac{a}{L}.b(s)\right)\mathrm ds = a . \int_{0}^{L} b(s)\mathrm ds\tag 2$$ Is it right to write like this? I am not familiar with calculus on dot product
On
Let me expand it from where helpful answer of @shawn stopped
$ \int_0^L (1/L)\mathbf a \cdot \mathbf b(s) \ ds = (1/L) \int_0^L \sum_{i=1}^3 a_i f_i (s) \ ds \tag1$
$(1/L) \sum_{i=1}^3 \int_0^L a_i f_i(s) \ ds.= \frac{a_1}{L} \int_0^Lf_1(s) \ ds+\frac{a_2}{L} \int_0^Lf_2(s) \ ds+\frac{a_3}{L} \int_0^Lf_3(s) \ ds\tag 2$
By using definition 126 in this(link) article. We can rewrite the equation as
$ =(\frac{a_1}{L}i+\frac{a_2}{L}j+\frac{a_3}{L}k).(\int_0^Lf_1(s)\ ds\hspace{.2cm}i+\int_0^Lf_2(s)\ ds\hspace{.2cm}j+\int_0^Lf_3(s)\ ds\hspace{.2cm}k) \tag3$
which is equal to
$\frac{a}{L} . \int_{0}^{L} b(s)\mathrm ds\tag 4$
I would say the right-hand side has some type issues.
For simplicity let's assume all vectors live in a vector space of at least dimension two.
The left-hand side makes sense since $f(s):= \mathbf a \cdot \mathbf b(s)$ is a real-valued function.
What is $\int_0^L \mathbf b(s) \ ds$? For instance, what does it mean to integrate the constant vector $\mathbf b = (1, 1)$ over the interval $[0, 1]$? If this is a number, then you are taking the dot product of a vector with a number, which is not defined in this case. If we are to interpret this integral as another vector, what should it be?
If you have access to the component functions of $\mathbf b(s)$ then can simplify things a bit. Suppose that $\mathbf a = (a_1, a_2, a_3)$ (where $a_i \in \mathbb R$) and $\mathbf b(s)=(f_1(s), f_2(s), f_3(s))$ (where $f_i \colon \mathbb R \to \mathbb R$ is a function). Then of course $$ \int_0^L (1/L)\mathbf a \cdot \mathbf b(s) \ ds = (1/L) \int_0^L \sum_{i=1}^3 a_i f_i (s) \ ds = (1/L) \sum_{i=1}^3 \int_0^L a_i f_i(s) \ ds. $$