∫(0到a)√a^2-x^2dx 有题目图 ∫[0到a] (x^2)根号(a^2-x^2) dx
\u5b9a\u79ef\u5206\u222b[0\uff0ca] \u221aa^2 -x^2dx(a>0)\u6c42\u8be6\u7ec6\u8fc7\u7a0b\u222b[0\uff0ca] \u221aa^2 -x^2dx
\u9996\u5148\u7b97\u4e0d\u5b9a\u79ef\u5206
\u222b\u221a\uff08a^2 -x^2\uff09dx
\u8bbex=asinu,dx=acosudu,\u221a\uff08a^2 -x^2\uff09=acosu
\u222b\u221a\uff08a^2 -x^2\uff09dx
=\u222bacosuacosudu
=\u222ba^2cos^2udu
=\u222ba^2(1+cos2u)/2du
=a^2/4\u222b(1+cos2u)d(2u)
=a^2/4[2u+sin2u]
\u4f9b\u53c2\u8003\uff1a
方法2:用几何意义,定积分的结果是一个曲边梯形的面积。
y=√(a^2-x^2)是个上半圆周,[0,a]范围就是1/4个圆,因此本题结果就是1/4圆面积,
结果为πa^2/4
本结果作为楼上答案的补充,请采纳楼上答案。
令x = a sinθ,dx = a cosθ dθ
∫ √(a² - x²) dx,x∈[0,a] => θ∈[0,π/2]
= ∫ a²cos²θ dθ
= (a²/2)∫ (1 + cos2θ) dθ
= (a²/2)(θ + (1/2)sin2θ)
= (a²/2)(π/2)
= πa²/4
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