integral batas 0 dan 2. (x + 1) kuadrat dx

∫ 0→2 (x+1)² dx
[∫(x+1)² dx] 2 - [(x+1)² dx]0
[ ∫(x² + 2x+1)dx ]2 - [ ∫(x² + 2x+1)dx]0
= (1/3 x^3 + x² + x)2 - (1/3 x^3 + x² + x)0
= 1/3 (8) + 4 +2 -0
= 8/3 +6
= 26/3
= 8 2/3