∫（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
绛旓細鈭玔0,a] 鈭歛^2 -x^2dx 棣栧厛绠椾笉瀹氱Н鍒 鈭垰锛坅^2 -x^2锛塪x 璁緓=asinu,dx=acosudu,鈭氾紙a^2 -x^2锛=acosu 鈭垰锛坅^2 -x^2锛塪x =鈭玜cosuacosudu =鈭玜^2cos^2udu =鈭玜^2(1+cos2u)/2du =a^2/4鈭(1+cos2u)d(2u)=a^2/4[2u+sin2u]
绛旓細=鈭玔0,a]-鈭(a^2-x^2)^5dx+鈭玔0,a]a^2鈭(x^2-a^2)^3dx x=asinu x=a,u=蟺/2 ,x=0,u=0 =鈭玔0,蟺/2]-a^6(cosu)^6du+a^4鈭玔0,蟺/2]cosu^4du =(-a^6/8)鈭玔0,蟺/2](1+cos2u)^3du+(a^4/4)鈭玔0,蟺/2](1+cos2u)^2du =(-a^6/8)鈭玔0,蟺...
绛旓細浠x/a=sint,鍒檇x=acostdt,x=0鏃,t=0锛泋=a鏃,t=蟺/2.鏁呭師寮=(0,蟺/2)a⁴鈭玸in²tcos²tdt=(0,蟺/2)(a⁴/4)鈭玸in²(2t)dt=(0,蟺/2)(a⁴/8)鈭玸in²(2t)d(2t)=(0,蟺/2)(a⁴/16)鈭玔(1-cos4t)/2]d(4t)=(0,蟺/...
绛旓細璁x=r cost锛寉=r sint 鍒0<t<2蟺,0<r<a 鍘熷紡=鈭玔0,2蟺]dt 鈭玔0,a] 鈭(a²-r²) rdr =2蟺 鈭玔0,a] 鈭(a²-r²) rdr =-蟺 鈭玔0,a] 鈭(a²-r²) d(a²-r²)=-蟺 2/3 (a²-r²)^(3/2) |[0,a]=...
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝璇︽儏濡傚浘鎵绀
绛旓細璁緓=asinu锛鈭(a^2-x^2)=acosu锛宒x=acosudu 鍘熷紡=鈭 a^2(sinu)^2*acosu*acosu du =a^4鈭 (sinu*cosu)^2du =a^4/4鈭 (sin2u)^2du =a^4/8鈭 (1-cos4u)du =a^4/8*(u-(1/4)sin4u)+C =a^4/8*(u-(1/2)sin2ucos2u)+C =a^4/8*(u-2sinu*cosu*((cosu)...
绛旓細鍥炵瓟锛氳 x=atant 鍒 dx=asec^2(t)dt 鍘熷紡=1/|a|鈭玸ec^2(t)dt/(tantsect) =-1/|a|鈭玸ectdt =-1/|a| *ln(sec t + tan t) +C =-1/|a|*ln{鈭歔1+(x/a)^2]+x/a}+C
绛旓細瑕佽缁嗚繃绋,绛旀鏄1/2... 8 2018-02-09 鈭玿鈭2/鈭(a鈭2-x鈭2)dx (a>0) 30 2017-01-09 鈭(涓婇檺a 涓嬮檺0)1/鈭(a^2-x^2)=? (a>0) 1 2016-09-24 姹備笉瀹氱Н鍒嗏埆1/((a鈭2+x鈭2)鈭(b鈭2+x鈭2)) 1 2017-12-18 鐢ㄥ嚑浣曟剰涔夋眰鈭0鍒癮(a^2-x^2)^1/2dx 3 2019-07-19...
绛旓細鈭埆鈭(a²-x²-y²)d蟽 =鈭埆 r鈭(a²-r²) drd胃 =鈭玔0鈫2蟺]d胃鈭玔0鈫抋] r鈭(a²-r²) dr =2蟺鈭玔0鈫抋] r鈭(a²-r²) dr =蟺鈭玔0鈫抋] 鈭(a²-r²) d(r²)=-(2蟺/3)(a²-r²...
绛旓細棣栧厛绉垎鏇查潰鍏充簬xoz锛寉oz骞抽潰閮芥槸瀵圭О鐨勶紝鑰岃绉嚱鏁帮紙x锛媦锛夊垎鍒槸鍏充簬x锛寉鐨勫鍑芥暟锛屾墍浠モ埆鈭紙x锛媦锛夛紳0锛屽師绉垎锛濃埆鈭珃ds锛岃岋紙z锛x锛锛2锛嬶紙z锛噛锛夛季2锛1锛x锛2锛弞锛2锛媦锛2锛弞锛2锛1锛a锛2锛弞锛2锛屾墍浠ョН鍒嗭紳鈭埆azdxdy锛弞锛a鈭鈭玠xdy锛澫a锛3 姹傛洸闈锛漻y锛廰琚煴闈...
