- $\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
Solution:
$\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
According to the additivity property of integrals,
$\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{c}_\text{a}\text{f}(\text{x})+\int\limits^\text{b}_\text{c}\text{f}(\text{x})\text{dx},$ where a < c < b
Using this property,
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}\ ....(\text{i})$
Now, consider the integral, $\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}$
Let x = 2a - t Then dx = d (2a - t), dx = - dt
Also, x = a, t = a and x = 2a, t = 0
Therefore, $\int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=-\int\limits^0_\text{a}\text{f}(2\text{a}-\text{t})\text{dt}$
$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{t})\text{dt}$
$\Rightarrow \int\limits^{2\text{a}}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$
Substituting this in equation (i) we get,
$\int\limits^{2\text{a}}_0\text{f}(\text{x})\text{dx}=\int\limits^\text{a}_0\text{f}(\text{x})\text{dx}+\int\limits^\text{a}_0\text{f}(2\text{a}-\text{x})\text{dx}$