You need to evaluate the definite integral, hence, you need to use the property of linearity of integral, such that:

`int_1^a (a - 4x)dx = int_1^a a dx - int_1^a 4xdx`

Since a is a constant, like 4 is, you can take it out, such that:

`int_1^a (a - 4x)dx = a*int_1^a dx - 4*int_1^a x dx`

`int_1^a (a - 4x)dx = a*x|_1^a - 4x^2/2|_1^a`

`int_1^a (a - 4x)dx = a*x|_1^a - 2x^2|_1^a`

Using the fundamental theorem of calculus, yields:

`int_1^a (a - 4x)dx = a*(a - 1) - 2(a^2 - 1)`

The problem provides the information that `int_1^a (a - 4x)dx = 6 - 5a` , hence, you need to replace `6 - 5a` for `int_1^a (a - 4x)dx` , such that:

`6 - 5a = a^2 - a - 2a^2 + 2`

Hence, you need to move all the terms to one side and you need to solve for a the quadratic equation, such that:

`a^2 - 4a + 4 = 0 => (a - 2)^2 = 0 => a - 2 = 0 => a = 2`

**Hence, evaluating a, under the given conditions, yields `a = 2` .**