定积分[0到x](t-1)(t-2)^2 dt,求指教
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∫[0到x](t-1)(t-2)^2 dt
=∫[0到x](t-1)(t-1-1)^2 dt
=∫[0到x](t-1)^3-2(t-1)^2+(t-1) d(t-1)
=1/4(t-1)^4-2/3 (t-1)^3+1/2 (t-1)^2|(0,x)
=1/4(x-1)^4-2/3 (x-1)^3+1/2 (x-1)^2-(1/4+2/3+1/2)
=1/4(x-1)^4-2/3 (x-1)^3+1/2 (x-1)^2-17/12
=∫[0到x](t-1)(t-1-1)^2 dt
=∫[0到x](t-1)^3-2(t-1)^2+(t-1) d(t-1)
=1/4(t-1)^4-2/3 (t-1)^3+1/2 (t-1)^2|(0,x)
=1/4(x-1)^4-2/3 (x-1)^3+1/2 (x-1)^2-(1/4+2/3+1/2)
=1/4(x-1)^4-2/3 (x-1)^3+1/2 (x-1)^2-17/12
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