Question

求解一道高等数学的积分问题

求解一道高等数学的积分问题如图，请写出详细步骤

Answer 1

原式
=∫[0,a]√(a²-a²+2ax-x²)dx
=∫[0,a]√[a²-(x-a)²]dx
令x-a=asint，dx=acostdt
x=0时，t=-π/2；x=a时，t=0.
故原式=∫[-π/2,0]a²cos²tdt
=a²/2 ∫[-π/2,0](1+cos2t)dt
=a²/2 (t+½ sin2t)|[-π/2,0]
=a²/2 [(0+0)-(-π/2+0)]
=πa²/4

Answer 2

有用望采纳！！