求由抛物线y=(1/4)x^2与直线3x-2y=4所围成的图形的面积
问题描述：
求由抛物线y=(1/4)x^2与直线3x-2y=4所围成的图形的面积
答
直线为y=(3/2)x-2与抛物线交天点(2,1)、(4,4).
所求面积=积分[2,4][(3/2)x-2-(1/4)x^2]dx
=[2,4][(3/4)x^2-2x-(1/12)x^3]
=[(3/4)*16-2*4-(1/12)*64]-[(3/4)*4-2*2-(1/12)*8]
=1/3