用定积分的分部积分法做下题 归纳一下定积分的换元积分和分部积分法的一般解题步骤？
\u7528\u5206\u90e8\u79ef\u5206\u6cd5\u6c42\u8be5\u5b9a\u79ef\u5206\uff0c\u4e0b\u56fe\u7b2c\u516b\u9898\uff0c\u975e\u5e38\u611f\u8c22\uff011\u3001\u6362\u5143\u6cd5\uff0c\u4e5f\u5c31\u662f\u53d8\u91cf\u4ee3\u6362\u6cd5 substitution\uff0c
\u8ddf\u5206\u90e8\u79ef\u5206\u6cd5 inegral by parts\uff0c\u8fd9\u4e24\u79cd\u65b9\u6cd5
\u65e2\u9002\u7528\u4e8e\u5b9a\u79ef\u5206 definite integral\uff0c\u4e5f\u9002\u7528\u4e8e
\u4e0d\u5b9a\u79ef\u5206 indefinite integral\u3002
.
2\u3001\u6709\u5f88\u591a\u65b9\u6cd5\uff0c\u5bf9\u4e8e\u4e0d\u5b9a\u79ef\u5206\u4e0d\u80fd\u9002\u7528\uff0c\u4f46
\u662f\u9002\u7528\u4e8e\u5b9a\u79ef\u5206\u3002\u4f8b\u5982\uff0c\u8fd0\u7528\u7559\u6570\u8ba1\u7b97\u79ef\u5206\u5c31
\u53ea\u80fd\u9002\u7528\u4e8e\u5b9a\u79ef\u5206\uff1b\u5bf9\u4e8e\u6b63\u6001\u5206\u5e03\u51fd\u6570\u7684\u79ef\u5206\uff0c
\u5fc5\u987b\u8981\u4f7f\u7528\u6781\u5750\u6807\u4e0b\u7684\u5e7f\u4e49\u79ef\u5206\uff0c\u4e5f\u5c31\u662f\u5b9a\u79ef
\u5206\uff0c\u624d\u80fd\u79ef\u51fa\u6765\u3002
.
3\u3001\u5bf9\u5bf9\u4e8e\u4e0d\u5b9a\u79ef\u5206\u8ddf\u5b9a\u79ef\u5206\uff0c\u7b2c\u4e09\u79cd\u5171\u540c\u4f7f
\u7528\u7684\u65b9\u6cd5\u662f\u6709\u7406\u5206\u5f0f\u7684\u5206\u89e3\u6cd5 partial fraction\u3002
.
=∫[0,π/2]e^(2x)d(sinx)
=e^(2x)sinx |[0,π/2] - 2∫[0,π/2]e^(2x)sinxdx
=e^π -0 +2∫[0,π/2]e^(2x)d(cosx)
=e^π + 2e^(2x)cosx |[0,π/2] - 4∫[0,π/2]e^(2x)cosxdx
=e^π + 0 - 2 -4∫[0,π/2]e^(2x)cosxdx
故∫[0,π/2]e^(2x)cosxdx=(e^π -2)/5
(1)：
∫(0→π) xsinx dx
= ∫(0→π) x d(- cosx)
= - xcosx：[0→π] + ∫(0→π) cosx dx
= - π(- 1) + sinx：[0→π]
= π
(2)：
∫(0→1) xe^x dx
= ∫(0→1) x d(e^x)
= xe^x：[0→1] - ∫(0→1) e^x dx
= e - e^x：(0→1)
= e - (e - 1)
= 1
(3)：
∫(1→e) x(x - 1)lnx dx
= ∫(1→e) (x^2 - x)lnx dx
= ∫(1→e) lnx d(x^3/3 - x^2/2)
= (x^3/3 - x^2/2)lnx：(1→e) - ∫(1→e) (x^3/3 - x^2/2)(1/x) dx
= (1/3)e^3 - (1/2)e^2 - ∫(1→e) (x^2/3 - x/2) dx
= (1/3)e^3 - (1/2)e^2 - (x^3/9 - x^2/4)：(1→e)
= (1/3)e^3 - (1/2)e^2 - [(e^3/9 - e^2/4) - (1/9 - 1/4)]
= (2/9)e^3 - (1/4)e^2 - 5/36
(4)：
∫(0→1) x^2e^(2x) dx
= (1/2)∫(0→1) x^2 d(e^(2x))
= (1/2)x^2e^(2x)：(0→1) - (1/2)∫(0→1) 2xe^(2x) dx
= (1/2)e^2 - (1/2)∫(0→1) x d(e^(2x))
= (1/2)e^2 - (1/2)xe^(2x)：(0→1) + (1/2)∫(0→1) e^(2x) dx
= (1/2)e^2 - (1/2)e^2 + (1/4)e^(2x)：(0→1)
= (1/4)(e^2 - 1)
绛旓細濡備笅锛氭敞鎰忥細瀹氱Н鍒嗙殑姝ｅ紡鍚嶇О鏄粠鏇肩Н鍒嗐傜敤榛庢浖鑷繁鐨勮瘽鏉ヨ锛屽氨鏄妸鐩磋鍧愭爣绯讳笂鐨勫嚱鏁扮殑鍥捐薄鐢ㄥ钩琛屼簬y杞寸殑鐩寸嚎鎶婂叾鍒嗗壊鎴愭棤鏁颁釜鐭╁舰锛岀劧鍚庢妸鏌愪釜鍖洪棿[a,b]涓婄殑鐭╁舰绱姞璧锋潵锛屾墍寰楀埌鐨勫氨鏄繖涓嚱鏁扮殑鍥捐薄鍦ㄥ尯闂碵a,b]鐨勯潰绉傚疄闄呬笂锛屽畾绉垎鐨勪笂涓嬮檺灏辨槸鍖洪棿鐨勪袱涓鐐筧,b銆
绛旓細鈭玿f(x)dx=xF(x)-鈭獸(x)dx=xF(x)-G(x)+C 瑙ｉ杩囩▼濡備笅锛氳嫢宸茬煡f(x)鐨勫師鍑芥暟涓篎(x),F(x)鐨勫師鍑芥暟涓篏(x),鍒欏彲鐢鍒嗛儴绉垎娉姹傦細鈭玿f(x)dx=xF(x)-鈭獸(x)dx=xF(x)-G(x)+C
绛旓細浣犺浣忎竴涓『搴忥紝鍙嶅骞備笁鎸囷紝鍙嶏細鍙嶄笁瑙掑嚱鏁帮紝瀵癸細瀵规暟鍑芥暟锛屽箓锛氬箓鍑芥暟锛屼笁锛氫笁瑙掑嚱鏁帮紝鎸囷細鎸囨暟鍑芥暟銆傛寜鐓ц繖涓『搴忥紝鍙绗﹀悎杩欎釜椤哄簭鐨勶紝鐣欏湪鍓嶉潰銆傛瘮濡傝鏈锛歽³鏄箓鍑芥暟锛宔^(-y²)鏄寚鏁板嚱鏁帮紝鎸夌収杩欎釜椤哄簭鏉ワ紝搴旇骞傚嚱鏁扮暀鍦ㄥ墠闈紝鎸囨暟鍑芥暟鏀惧埌鍚庨潰鐨刣y閲屻
绛旓細cosx) dx = 蟺²/4 + 鈭<0,蟺>arctan(cosx) dx 浠 t=cosx 锛寈=arccost 鈭<0,蟺>arctan(cosx) dx = 鈭 <1,-1> arctant / -鈭(1-t²) dt 濂囧嚱鏁帮紝绉垎鍖洪棿瀵圭О锛屽叾鍊间负0 鏁呭師绉垎= 蟺²/4 鍙︽硶锛歨ttp://zhidao.baidu.com/question/177015643.html ...
绛旓細鈭玸inx.e^x dx =鈭玸inx de^x =sinx.e^x -鈭玞osx.e^x dx =sinx.e^x -鈭玞osx de^x =sinx.e^x -cosx.e^x -鈭玸inx e^x dx 2鈭玸inx.e^x dx=sinx.e^x -cosx.e^x 鈭玸inx.e^x dx=(1/2)[sinx.e^x -cosx.e^x] +C ...
绛旓細浠=1鍗冲彲锛岃鎯呭鍥炬墍绀
绛旓細浣犵煡閬撴槸鐢ㄥ垎閮ㄧН鍒鏉ュ仛鐨勶紝閭ｄ箞+鍙峰彸杈规槸杩欐牱:鎴戜滑鏄妸鈭玔x-1,2]e^(y^2)dy浣滀负鍏充簬x鐨勫嚱鏁拌繘琛屽垎閮ㄧН鍒,鎵浠ュ師寮=浣犵殑寮忓瓙+宸﹁竟-鈭玔1,3]x*[鈭玔x-1,2]e^(y^2)dy]'dx 鍗宠绉嚱鏁颁负x*[鈭玔x-1,2]e^(y^2)dy鐨勫鏁癩鎶婇偅涓Н鍒嗗啓鎴愪竴鑸殑锛屽彉閲忓湪涓婏紝灏辨槸涓婁笅闄愪簰鎹紝鍑烘潵涓涓...
绛旓細鈭 (0,2) f(x-1)dx= (0,1) [x/(2-x)]dx+ (1,2) (2x-2)dx= (0,1) {[2/(2-x)]-[(2-x)/(2-x)]}dx+ x^2-2x涓(1,2)=-2ln(2-x) 涓(0,1)+(4-4-1+2)=2ln2+1銆備互涓婄己灏绉垎绗﹀彿锛屾槸鐢变簬鐧惧害鐭ラ亾鐩墠鏈夌己闄凤紝鏃犳硶杈撳叆涔熸棤娉曞鍒剁矘璐寸Н鍒嗙...
绛旓細璁＄畻杩囩▼濡備笅锛氱敱寰垎鐨勪箻娉曟硶鍒欏拰寰绉垎鍩烘湰瀹氱悊鎺ㄥ鑰屾潵鐨勩傚畠鐨勪富瑕佸師鐞嗘槸灏嗕笉鏄撶洿鎺ユ眰缁撴灉鐨勭Н鍒嗗舰寮忥紝杞寲涓虹瓑浠风殑鏄撴眰鍑虹粨鏋滅殑绉垎褰㈠紡鐨勩
绛旓細濡備笅
