不利用Newton-leibniz公式,而从Riemann积分的定义出发,得出:integral from n=a to b(dx/x′)=1/r-1[(1/a^(r-1))-(1/b^(r-1))](a>0,r为正整数,且r≥2)integral from n=a to b(x′dx)=b^(r+1)-a^(r+1)/r+1(a>0,r为正整数)About th...不利用Newton-leibniz公式,而从Riemann积分的定义出发,得出:integral from n=a to b(dx/x′)=1/r-1[(1/a^(r-1))-(1/b^(r-1))](a>0,r为正整数,且r≥2)integral from n=a to b(x′dx)=b^(r+1)-a^(r+1)/r+1(a>0,r为正整数)About the Research of Two Integral展开更多
文摘不利用Newton-leibniz公式,而从Riemann积分的定义出发,得出:integral from n=a to b(dx/x′)=1/r-1[(1/a^(r-1))-(1/b^(r-1))](a>0,r为正整数,且r≥2)integral from n=a to b(x′dx)=b^(r+1)-a^(r+1)/r+1(a>0,r为正整数)About the Research of Two Integral
