求由y = x,y = x² 和 y = x²所围成的平面图形的面积?
交点：(0,0),(1,1),(2,4)
A = ∫(0→1) [(2x) - (x)] dx + ∫(1→2) [(2x) - (x²)] dx
= ∫(0→1) x dx + ∫(1→2) (2x - x²) dx
= [x²/2]：[0→1] + [x² - x³/3]：[1→2]
= 1/2 + [4/3 - 2/3]
= 7/6
2x - tan(x - y) = ∫(0→x) sec²t dt,两边求导,这个应该是下限为0,上限为x的定积分?
2 - sec²(x - y) · (1 - y') = sec²x
2 - sec²x = sec²(x - y) + y' · sec²(x - y)
2 - sec²x - sec²(x - y) = y' · sec²(x - y)
dy/dx = [2 - sec²x - sec²(x - y)]cos²(x - y)