Question

大一微积分下考试题，“定积分”求解答如图

Answer 1

1.[0,4]∫1/(2+√x)dx
设 y=√x x=y² dx=2ydy
[0,4]∫1/(2+√x)dx
=[0,2]∫2ydy/(2+y)
=[0,2]∫[2(2+y)-4]/(2+y) dy
=[0,2]∫2-[4/(2+y)]dy
=[0,2]2y-4ln(y+2)+C
=(4-4ln6)-(-4ln2)
=4-4ln6+4ln2
=4+4ln(1/3)
=4ln(e/3)

[0,3]∫1/√(9-x²)dx
设 x=3sint -π/2<t<π/2
dx=3costdt
[0,3]∫1/√(9-x²)dx
=[0,3]∫3costdt/(3cost)
=[0,3]∫dt
=[0,3]arcsin(x/3)
=arcsin1-arcsin0
=π/2

Answer 2

未被采纳，隐去答案。