高数问题求解
y = x^2/4,√y = x/2,dy = (2x/4)dx = xdx/2;
ds 是曲线中一小段的长度;由勾股定理,
(ds)^2 = (dx)^2 + (dy)^2 = (dx)^2 + (xdx/2)^2 = (dx)^2[ 1 + x^2/4 ];
ds = √[ 1 + x^2/4 ]dx;
积分式为 ∫(0,2)(x/2)√[ 1 + x^2/4 ]dx = ∫(0,2)√[ 1 + x^2/4 ]d(1+x^2/4)
= [ 1/(1+1/2) ][ (1 + x^2/4)^(3/2)](0,2)
= (2/3)[ 2^(3/2) - 1] = 4√2/3 - 2/3 。
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