Luas daerah yang dibatasi oleh kurva y = x^2 - 8x + 12, sumbu X, x = 3 dan x = 6 adalah....satuan luas
a. 2
b. 5
c. 7
d. 8
e. 9
a. 2
b. 5
c. 7
d. 8
e. 9
Takamori37
Karena kurva berada di bawah sumbu x, maka batasa atas dan batas bawah dibalik
![\displaystyle \int_6^3 x^2-8x+12$ $dx=\frac{x^3}{3}-4x^2+12x]_6^3 \\ \\ (\frac{3^3}{3}-4.3^2+12.3)-(\frac{6^3}{3}-4.6^2+12.6) \\ (9-36+36)-(72-144+72) \\ 9](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cint_6%5E3+x%5E2-8x%2B12%24+%24dx%3D%5Cfrac%7Bx%5E3%7D%7B3%7D-4x%5E2%2B12x%5D_6%5E3+%5C%5C++%5C%5C+%28%5Cfrac%7B3%5E3%7D%7B3%7D-4.3%5E2%2B12.3%29-%28%5Cfrac%7B6%5E3%7D%7B3%7D-4.6%5E2%2B12.6%29+%5C%5C+%289-36%2B36%29-%2872-144%2B72%29+%5C%5C+9 "\displaystyle \int_6^3 x^2-8x+12$ $dx=\frac{x^3}{3}-4x^2+12x]_6^3 \\ \\ (\frac{3^3}{3}-4.3^2+12.3)-(\frac{6^3}{3}-4.6^2+12.6) \\ (9-36+36)-(72-144+72) \\ 9")
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