batas atas=2 batas bawah 1, (3-2x)^2 dx

[tex]\int_1^2(3-2x)^2$ dx$ \\ \int_1^2 (9-12x+4x^2)$ dx$ \\ 9x-6x^2+\frac{4}{3}x^3]_1^2 \\ (9.2-6.2^2+\frac{4}{3}2^3)-(9.1-6.1^2+\frac{4}{3}1^3) \\ (18-24+\frac{32}{3})-(9-6+\frac{4}{3}) \\ \frac{20}{3}-\frac{13}{3}=\frac{7}{3}[/tex]