∫√（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