Question

k

Answer 1

The reason the 1 is put in as the value of a is because they want to evaluate the function at 1. When you evaluate the function, you're plugging in a number for x, so if you had this integral:

\[\int\limits_{1}^{1}x^2dx\] Then obviously when you evaluate from 1 to 1, the integral will always be 0.

That's perfect, now the arbitrary constant of integration is just 5, since the rest of it is 0. How do we know? Well, look at it like this: \[f(1)=5\] was given, and we know: \[f(x)=C+\int\limits f(x)dx\] so now we can see that when we plug in 1, the integral becomes zero:\[f(1)=C+0\] So now you can easily solve for C by plugging in f(1)=5,\[5=C\]