Question

求定积分x^2根号下4-x^2(上限为2,下限为0)

Answer 1

令x=sint,t∈(-∏/2,∏/2)
∵∫x^2•√(4-x^2)dx
=∫ (2sint)^2•√(4-(2sint)^2)d(2sint)
=∫ 4(sint)^2•2cost•2costdt
=∫16(sint)^2•(cost)^2dt
=∫16(sintcost)^2dt
=∫16(1/2•sin2t)^2dt
=∫ 16•1/4•(sin2t)^2dt
=∫ 4•(1-cos4t)/2 dt
=∫ 2dt - ∫ 2cos4tdt
=∫2dt - 1/4∫2cos4td4t
=2x-1/2•sin4t+C
∴∫(上限为2,下限为0)x^2•√(4-x^2)dx
=(2x-1/2•sin4t)|(上限为2,下限为0)
=4-1/2 sin8