高数题求解答，这个积分的详细计算过程 高数题，定积分求体积，这个答案怎么算出来的？
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\u672c\u9898\u9002\u7528\u89e3\u6790\u6cd5,\u662f\u6309\u7167\u529b\u5b66\u79ef\u5206\u7684\u65b9\u5f0f\u63a8\u51fa\u8ba1\u7b97\u516c\u5f0f\u7684,\u5177\u4f53\u8fc7\u7a0b\u4f60\u4e0d\u5fc5\u6df1\u7a76
\u9700\u8981\u638c\u63e1\u68af\u5f62\u7684\u51e0\u4e2a\u5e38\u7528\u53c2\u6570,\u3010\u4f8b\u5982,\u68af\u5f62\u4e0a\u5bbda,\u4e0b\u5bbdb,\u9ad8\u4e3ah,\u9762\u79ef\u4e3aA,\u5f62\u5fc3\u4e3aC\u3011
\u5f62\u5fc3\u5750\u6807yc=(h/3)[(a+2b)/(a+b)] \u60ef\u6027\u77e9Ic=(h³/36)[(a²+4ab+b²)/(a+b)]
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\u603b\u538b\u529bP=pcA pc\u4e3a\u5f62\u5fc3\u5904\u7684\u76f8\u5bf9\u538b\u5f3a
\u7531\u6df1\u5ea6\u4e0e\u5bbd\u5ea6\u7684\u5173\u7cfb\u5f0f\u4e3aL=\ufe630.2h+10
\u4e8e\u662f\u53d6\u5176\u4e2d\u4e00dh\u5176\u6240\u53d7\u538b\u529b\u4e3adF=\u03c1ghLdh
\u5728\u5bf9\u4e24\u8fb9\u8fdb\u884c\u79ef\u5206\u53ef\u5f97F\u22481.47\u00d710\u22277N
绛旓細涓嶅畾绉垎缁撴灉涓嶅敮涓姹傚楠岃瘉搴旇鑳藉鎻愰珮鍑戝井鍒嗙殑璁＄畻鑳藉姏銆
绛旓細=lim[x-->鈭瀅[(arctan x)^2鈭(1+1/x^2)]=蟺^2/4 (3)鐢ㄥ垎閮ㄧН鍒嗘硶锛氬師寮=鈭0锛2蟺銆憍^2dsinx=x^2sinx|[0,2蟺]-2鈭玔0,2蟺]xsinxdx =2xcosx|[0,2蟺]-2鈭玔0,2蟺]cosxdx =4蟺-2sinx|[0,2蟺]=4蟺 (4)鍒╃敤瀹绉垎鐨鎬ц川锛屽垎鎴愪袱娈典笂鐨勫畾绉垎鐨勫拰 鍘熷紡=-鈭玔1...
绛旓細濡備笂鍥炬墍绀恒
绛旓細瑙ｏ細鍘熷紡=-鈭(0,蟺)x^2d(cosx)+(1/3)鈭(0,蟺)(sinx)^2d(cosx)锛岃屸埆x^2d(cosx)=(x^2)cosx-2鈭玿cosxdx=(x^2)cosx-2[xsinx-鈭玸inxdx]=(x^2)cosx-2xsinx-2cosx+C1锛屸埆(sinx)^2cosxdx=鈭玔1-(cosx)^2]d(cosx)=cosx-(1/3)(cosx)^3+C2锛屸埓鍘熷紡=[-(x^2)cosx+2x...
绛旓細濡傛灉浣犵敤鑷繁涓婅堪鐨瑙ｇ瓟鏂规硶锛屽緱鍒扮殑搴旇鏄涓婂浘鐨勭Н鍒嗭紝璇ョН鍒嗗緢闅剧Н鍑烘潵銆傝屼綘鍐欑殑鏂规硶鏄繃绋嬫槸閿欒鐨勶紝-1/2t涓璽鏄绉垎鐨鍙橀噺锛屼笉鑳界洿鎺ユ彁鍑烘潵鐨勩傛墍浠ラ噰鐢ㄤ笁瑙掑嚱鏁版硶姣旇緝绠鍗曘傝棰樺彲鍏堟眰涓嶅畾绉垎锛岀劧鍚庡啀浠ｅ叆鏁板煎嵆鍙傝棰樺彲鐢▁锛漵int杩涜鏇挎崲锛屾渶缁堝緱鍒1/2脳x+1/4脳sin2x+C锛屾浛鎹负t...
绛旓細(1)let u=x^(1/6)du = (1/6)x^(-5/6) dx dx = 6u^5 du 鈭 dx/[鈭歺.(1+ x^(1/3)) ]=鈭 6u^5 du/[ u^3.(1+ u^2) ]=6鈭 u^2 /(1+ u^2) du =6鈭 [1 - 1/(1+ u^2)] du =6(u -arctanu ) + C =6{ x^(1/6) -arctan[x^(1/6)]...
绛旓細鈭4/[ 4(sinx)^2.(cosx)^2] dx =鈭4/(sin2x)^2 dx =鈭4(csc2x)^2 dx =2鈭(csc2x)^2 d(2x)=-2cot2x + C
绛旓細鍒嗛儴绉垎鍗冲彲
绛旓細鎬濊矾锛氱洿鎺ュ垎閮绉垎 杩囩▼锛氬弬鑰冧笅鍥
绛旓細1銆亂^2=x涓巟=2y鐨勪氦鐐癸細(0,0)銆(4,2)V=鈭(0,2)蟺(4y^2-y^4)dy =蟺(4/3y^3-1/5/y^5)|(0,2)=蟺[4/3(2^3-0^3)-1/5(2^5-0^5)]=蟺(32/3-32/5)=32蟺(5-3)/15 =64蟺/15 2銆佲埆(0,蟺/6)cos^(-3)2胃sin2胃d胃 =-1/2鈭(0,蟺/6)cos^(-3)2胃...
