①一质点在直线上以速度v=3t2﹣2t﹣1(m/s)运动,从时刻t=0(s)到t=3(s)时质点运动的路程为15(m);
②若x∈(0,π),则sinx<x;
③若f′(x0)=0,则函数y=f(x)在x=x0取得极值;
④已知函数 f(x)=−x2+4x ,则 ∫02f(x)dx=π .
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计算题:
(1)(﹣a3)4•(﹣a)3
(2)8a(3a2﹣b)﹣a(5b+4a2)
(3)(2x+5y)(3x﹣2y)
(4)(﹣32x2yz3)•(﹣43xz3)•(13xy2z)