计算∫√（x²-4x+4）dx在【0,3】的定积分

计算∫√（x²-4x+4）dx在【0,3】的定积分
计算∫√（x²-4x+4）dx在【0,3】的定积分

计算∫√（x²-4x+4）dx在【0,3】的定积分
∫√（x²-4x+4）dx在【0,3】的定积分
=∫√（x²-4x+4）dx在【0,2】的定积分+∫√（x²-4x+4）dx在【2,3】的定积分
=∫（2-x）dx在【0,2】的定积分+∫（x-2）dx在【0,2】的定积分
=(2x-0.5x^2)|【0,2】+(0.5x^2-2x)|【2,3】
=[(4-2)-0]+[(4.5-6)-(-2)]
=2.5

∫√（x²-4x+4）dx
=∫lx-2ldx
=∫[0,2)[2-x]dx+∫[2,3][x-2]dx=(2x-x²/2)l[0,2)+(x²/2+2x)l[2,3]=4-2+(9/2+6-4/2-4)=13/2