Question

求5 6两道题的解

Answer 1

5.

令x-1=sint，则x=1+sint

x：0→2，则t：-π/2→π/2

∫[0：2]x√(2x-x²)dx

=∫[-π/2：π/2](1+sint)√(1-sin²t)d(1+sint)

=∫[-π/2：π/2](1+sint)·cos²tdt

=∫[-π/2：π/2]cos²tdt+∫[-π/2：π/2]sint·cos²tdt

=∫[0：π/2](1+cos2t)dt +0

=(t+½sin2t)|[0：π/2]

=π/2+½sinπ-0-½sin0

=π/2

6.

令√x=u，则x=u²

x：0→t²，则u：0→t

∫[0：t²](sin√x/t³)dx

=(1/t³)∫[0：t]sinud(u²)

=(2/t³)∫[0：t]usinudu

=(-2/t³)∫[0：t]ud(cosu)

=(-2/t³)ucosu|[0：t]+(2/t³)∫[0：t]cosudu

=(-2/t³)(tcost-0)+(2/t³)sinu|[0：t]

=-2cost/t²+(2/t³)(sint-0)

=2(sint-tcost)/t³

lim ∫[0：t²](sin√x/t³)dx

t→0

=lim 2(sint-tcost)/t³

t→0

=lim 2(cost-cost+tsint)/(3t²)

t→0

=lim 2t²/(3t²)

t→0

=⅔