Could someone please help me with this question, I would really appreciate it. Thank you!
You can do this by reasoning with transformations and how you expect this to affect the area - personally, I find that annoying, and instead I use a substitution. Note that I'm about to do is an integration technique taught in specialist - it's not hard to do, you won't be frowned upon for using it (particularly in multiple choice), but if you haven't seen it before, that's why.
So, let's start with breaking up the integral they want you to do:
+3\:dx\right)=2\int_{x=0}^{x=5a}f\left(\frac{x}{5}\right)\:dx+2\int_{x=0}^{x=5a}3\:dx)
The second integral is pretty trivial, and comes to 30a. The first one is our issue - so, we make a substitution.
If we let

, then

So,

. Note that when we put this into the new integral, we also need to change the bounds (as indicated about, the bounds are in terms of x - we need them in terms of u). Putting this into the original function, we get that when x=0, u=0 and when x=5a, u=a. So, we now have:
\:dx+2\int_{x=0}^{x=5a}3\:dx=2\int_{u=0}^{u=a}f\left(u\right)\:5\:du+30a=10\int_{0}^{a}f\left(u\right)\:du+30a)
Now, the final integral is just like the one they gave you, which is equal to a, so our final answer becomes 10a+30a=40a, so the answer is D.