Expand the integrand:
[tex](2x - 3) (4x^2 + 1) = 8x^3 - 12x^2 + 2x - 3[/tex]
Then integrate using the power rule:
[tex]\displaystyle \int (8x^3-12x^2+2x-3) \, dx = 2x^4 - 4x^3 + x^2 - 3x + C[/tex]
By the fundamental theorem of calculus, the definite integral has a value of
[tex]\displaystyle \int_0^2 (2x-3)(4x^2+1) \, dx = \left(2x^4 - 4x^3 + x^2 - 3x\right)\bigg|_{x=2} - \left(2x^4 - 4x^3 + x^2 - 3x\right)\bigg|_{x=0}[/tex]
[tex]\displaystyle \int_0^2 (2x-3)(4x^2+1) \, dx = 32 - 32 + 4 - 6 = \boxed{-2}[/tex]
