Calculus

∫(x - √(25 - x^2)) dx, where x = -5..5
= [∫x dx, where x = -5..5] - [∫√(25 - x^2) dx, where x = -5..5]
= [∫x dx, where x = 0..5] - [∫x dx, where x = -5..0] - [∫√(25 - x^2) dx, where x = -5..5]
= [(area of triangle with base 5 and height 5) - (area of triangle with base 5 and height 5)] - [area of semicircle with radius 5]
= [5*5/2 - 5*5/2] - [(1/2)TT(5)^2]
= -25TT/2

題目看不清，請在補充發問那裏張貼更新的連結。