integral Dari integral Batas bawah 2 Batas atas 4(-x²+6x-8)dx

[tex] - \frac{1}{3} {x}^{3} + \frac{6}{2} {x}^{2} - 8x \\ = - \frac{1}{3} {x}^{3} + 3 {x}^{2} - 8x[/tex]
masukkan batasnya 4 dan 2
[tex]( - \frac{1}{3} . {4}^{3} + 3. {4}^{2} - 8.4) - ( - \frac{1}{3} {2}^{3} + 3. {2}^{2} - 8.2) \\ = ( - \frac{1}{3} .64 + 3.16 - 24) - ( - \frac{1}{3} .8 + 12 - 16) \\ = ( - \frac{64}{3} + 48 - 24) - ( - \frac{8}{3} + 12 - 16) \\ = ( - \frac{64}{3} + 24) - ( - \frac{8 }{3} - 4) \\ = ( \frac{ - 64 + 72}{3} ) - ( \frac{ - 8 - 12}{3} ) \\ = ( \frac{8}{3} ) - ( \frac{ - 20}{3} ) \\ = \frac{8}{3} + \frac{20}{3} \\ = \frac{28}{3} \\ = 9 \frac{1}{3} [/tex]