求定积分，要详细过程，谢谢 关于定积分求面积，要详细过程，谢谢
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x的上下限为(0，-1/2)
当x=0时，t=1
当x=-1/2时，t=0
所以积分上下限变为(1，0)
x=(1-t)/2
dx=-1/2 dt
所以原式=
∫(1，0) t^99 (-1/2) dt
=(-1/2)∫(1,0) t^99 dt
=(-1/2)t^100/100 +C | (1，0)
=-1/200-0
=-1/200
绛旓細t鏄鍑芥暟(t鈭(1-t^2))鐨勭Н鍒涓0锛夛紝t^2鏄伓鍑芥暟(鈭(1-t^2)鐨勭Н鍒嗕负涓鍗婄Н鍒嗗尯闂绉垎鐨2鍊嶏級鍘熷紡=鈭(-1,1)(t+1)(鈭(1-t^2))dt =2鈭(0,1)鈭(1-t^2)dt (鐢ㄧН鍒嗗叕寮忔垨鑰呭埄鐢╰=siny)=2锛坱鈭(1-t^2)/2+(1/2)arcsint)|(0,1)=蟺/2 ...
绛旓細锛1锛4.5锛2锛夆垰2-1 瑙ｏ紙1锛夊師寮=½x²+x|[-1锛2]=½*4+2-锛½-1锛=4.5 锛2锛夊師寮=sinx|[0,蟺/4]+cosx[|[0,蟺/4]]=鈭2-0+锛0-1锛=鈭2-1
绛旓細浠2x+1=t x鐨涓婁笅闄愪负(0锛-1/2)褰搙=0鏃讹紝t=1 褰搙=-1/2鏃讹紝t=0 鎵浠绉垎涓婁笅闄愬彉涓(1锛0)x=(1-t)/2 dx=-1/2 dt 鎵浠ュ師寮= 鈭(1锛0) t^99 (-1/2) dt =(-1/2)鈭(1,0) t^99 dt =(-1/2)t^100/100 +C | (1锛0)=-1/200-0 =-1/200 ...
绛旓細鍒嗛儴绉垎 =(x-0.56)蠁(x)|(0.56,+鈭)-鈭(0.56,+鈭)蠁(x)dx =-鈭(0.56,+鈭)蠁(x)dx =鈭(-鈭,0.56)蠁(x)dx-鈭(-鈭,+鈭)蠁(x)dx =-1+桅(0.56)
绛旓細鍒嗘垚涓ら儴鍒 鍗冲墠闈㈠垎寮鐨绉垎鍔犱笂1鐨勭Н鍒 鍓嶉潰杩欎釜鏄剧劧鏄鍑芥暟 涓旂Н鍒嗛檺鍏充簬鍘熺偣瀵圭О 鎵浠瀹氱Н鍒涓0 鎵浠ュ師寮=鈭(-2锛2)1dx =x(-2锛2)=4
绛旓細鈭(x0->鏃犵┓) dx/x^2 棣栧厛鍒╃敤 鈭玠x/x^2 =-1/x +C =-[1/x]|(x0->鏃犵┓)浠ｅ叆瀹氱Н鍒嗙殑涓婁笅闄 =1/x0- lim(x0->鏃犵┓) 1/x 璁＄畻 lim(x0->鏃犵┓) 1/x =0 =1/x0 - 0 鍖栫畝 =1/x0 鎵浠ョ敱涓婅堪璁＄畻锛鍙緱 鈭(x0->鏃犵┓) dx/x^2 = 1/x0 ...
绛旓細鍘熷紡=鈭玔a~b]xf(x)路d[f(x)]=鈭玔a~b]x路d[1/2路f²(x)]=x/2路f²(x) |[a~b]-鈭玔a~b]1/2路f²(x)路dx =0-0-1/2 =-1/2
绛旓細鍏堟眰鍑轰笉瀹氱Н鍒嗭紝闇鐢ㄤ竾鑳芥崲鍏冩硶 浠 = tan(x/2)锛宒x = 2dz/(1 + z²)锛宻inx = 2z/(1 + z²)鈭 1/(2 + sinx)dx = 鈭 [2/(1 + z²)]/[2 + (2z)/(1 + z²)]dz = 鈭 1/[(1 + z²)+ z]dz = 鈭 1/[(z + 1/2)²+ 3...
绛旓細鈭玔-1,1](1+sinx)鈭(1-x^2)dx 瀵圭О鍖洪棿涓婂鍑芥暟鐨勭Н鍒涓0 =鈭玔-1,1]鈭(1-x^2)dx 浠 x = sint =鈭玔-蟺/2,蟺/2]costdsint =鈭玔-蟺/2,蟺/2]cos²tdt =鈭玔-蟺/2,蟺/2](1+cos2t)/2dt =1/4鈭玔-蟺/2,蟺/2](1+cos2t)d2t =1/4*(2t+sin2t)[-蟺/2,蟺/...
绛旓細鍘熷紡绛変簬鈭玿^4dx-鈭玿⁶dx =x⁵/5-x⁷/7+C銆傜劧鍚璁＄畻瀹氱Н鍒嗭紝绠椾笂浜屽垎涔嬩竴 寰楀埌锛1/5-1/7锛/2=(7-5)/70=2/70=1/35=0.028571銆傘傘備笉瀹氱Н鍒嗙粨鏋滀笉鍞竴姹傚楠岃瘉搴旇鑳藉鎻愰珮鍑戝井鍒嗙殑璁＄畻鑳藉姏鍏堝啓鍒棶鍞夈傘傛暟瀛楀笣鍥 GG娉涙互浣嗘槸鏄竴涓绠楀櫒缃戦〉銆傘
