第1个回答 2018-05-19
case 1: x<1
∫(0->x) f(t) dt
=∫(0->x) t dt
= (1/2)x^2
case 2: x≥1
∫(0->x) f(t) dt
=∫(0->1) t dt +∫(1->x) (2t+1) dt
=1/2 + [t^2 +t]|(1->x)
=1/2 + (x^2 +x) -(1 +1)
=x^2 +x - 3/2本回答被网友采纳
∫(0->x) f(t) dt
=∫(0->x) t dt
= (1/2)x^2
case 2: x≥1
∫(0->x) f(t) dt
=∫(0->1) t dt +∫(1->x) (2t+1) dt
=1/2 + [t^2 +t]|(1->x)
=1/2 + (x^2 +x) -(1 +1)
=x^2 +x - 3/2本回答被网友采纳
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