用分部积分法求定积分 用分部积分法求定积分
\u7528\u5206\u90e8\u79ef\u5206\u6cd5\u6c42\u4e0b\u5217\u5b9a\u79ef\u5206(1)\uff1a
\u222b(0\u2192\u03c0) xsinx dx
= \u222b(0\u2192\u03c0) x d(- cosx)
= - xcosx\uff1a[0\u2192\u03c0] + \u222b(0\u2192\u03c0) cosx dx
= - \u03c0(- 1) + sinx\uff1a[0\u2192\u03c0]
= \u03c0
(2)\uff1a
\u222b(0\u21921) xe^x dx
= \u222b(0\u21921) x d(e^x)
= xe^x\uff1a[0\u21921] - \u222b(0\u21921) e^x dx
= e - e^x\uff1a(0\u21921)
= e - (e - 1)
= 1
(3)\uff1a
\u222b(1\u2192e) x(x - 1)lnx dx
= \u222b(1\u2192e) (x^2 - x)lnx dx
= \u222b(1\u2192e) lnx d(x^3/3 - x^2/2)
= (x^3/3 - x^2/2)lnx\uff1a(1\u2192e) - \u222b(1\u2192e) (x^3/3 - x^2/2)(1/x) dx
= (1/3)e^3 - (1/2)e^2 - \u222b(1\u2192e) (x^2/3 - x/2) dx
= (1/3)e^3 - (1/2)e^2 - (x^3/9 - x^2/4)\uff1a(1\u2192e)
= (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)\uff1a
\u222b(0\u21921) x^2e^(2x) dx
= (1/2)\u222b(0\u21921) x^2 d(e^(2x))
= (1/2)x^2e^(2x)\uff1a(0\u21921) - (1/2)\u222b(0\u21921) 2xe^(2x) dx
= (1/2)e^2 - (1/2)\u222b(0\u21921) x d(e^(2x))
= (1/2)e^2 - (1/2)xe^(2x)\uff1a(0\u21921) + (1/2)\u222b(0\u21921) e^(2x) dx
= (1/2)e^2 - (1/2)e^2 + (1/4)e^(2x)\uff1a(0\u21921)
= (1/4)(e^2 - 1)
\u5982\u56fe
∫ xlnx dx = ∫ lnx d(x²/2)
= (1/2)x²lnx - (1/2)∫ x² d(lnx)
= (1/2)x²lnx - (1/2)∫ x dx
= (1/2)x²lnx - (1/2)(x²/2) + C
= (1/2)x²lnx - (1/4)x² + C
代入上下限(e,1)
=(1/2)e²-(1/4)e²-[0-(1/4)]
=e²/4+1/4
绛旓細濡傚浘锛
绛旓細浠=1鍗冲彲锛岃鎯呭鍥炬墍绀
绛旓細鍒嗛儴绉垎娉鏄敱寰垎鐨勪箻娉曞畾鍒欏拰寰Н鍒嗗熀鏈畾鐞嗘帹瀵艰屾潵鐨勩傚叾鍩烘湰鎬濊矾鏄皢涓嶆槗姹傚緱缁撴灉鐨勭Н鍒嗗舰寮忚浆鍖栦负绛変环鐨勪絾鏄撲簬姹傚嚭缁撴灉鐨勭Н鍒嗗舰寮忋傚浜庨偅浜涚敱涓や釜涓嶅悓鍑芥暟缁勬垚鐨勮绉嚱鏁颁笉渚夸簬杩涜鎹㈠厓鐨勭粍鍚堝垎鎴愪袱閮ㄥ垎杩涜绉垎锛屽叾鍘熺悊鏄嚱鏁板洓鍒欒繍绠楃殑姹傚娉曞垯閫嗙敤銆瀹氱Н鍒鍐呬笌涓嶅畾绉垎鐨勫垎閮ㄧН鍒嗘硶涓鏍凤紝鍙緱...
绛旓細鍏紡濡備笅锛氱浉鍏充粙缁嶏細鍒嗛儴绉垎娉锛堝鏂囧悕锛欼ntegration by parts锛夋槸寰Н鍒嗗涓殑涓绫婚噸瑕佺殑銆佸熀鏈殑璁＄畻绉垎鐨勬柟娉曘傚畠鏄敱寰垎鐨勪箻娉曟硶鍒欏拰寰Н鍒嗗熀鏈畾鐞嗘帹瀵艰屾潵鐨勩傚叾涓昏鍘熺悊鏄皢涓嶆槗鐩存帴姹傜粨鏋滅殑绉垎褰㈠紡锛岃浆鍖栦负绛変环鐨勬槗姹傚嚭缁撴灉鐨勭Н鍒嗗舰寮忋瀹氱Н鍒锛堝鏂囧悕锛歞efinite integral锛夋槸绉垎鐨勪竴绉嶏紝鏄...
绛旓細(1)锛氣埆(0鈫捪) xsinx dx = 鈭(0鈫捪) x d(- cosx)= - xcosx锛歔0鈫捪] + 鈭(0鈫捪) cosx dx = - 蟺(- 1) + sinx锛歔0鈫捪]= 蟺 (2)锛氣埆(0鈫1) xe^x dx = 鈭(0鈫1) x d(e^x)= xe^x锛歔0鈫1] - 鈭(0鈫1) e^x dx = e - e^x锛(0鈫1)= e - ...
绛旓細瑙佷笅鍥撅細
绛旓細1銆佷粈涔堟椂鍊欒鐢ㄥ垎閮ㄧН鍒嗘硶銆2銆佷粈涔堟椂鍊欑敤瀹氱Н鍒鐨勫垎閮ㄧН鍒嗘硶銆3銆佷粈涔堟儏鍐典笅鐢ㄥ垎閮ㄧН鍒嗘硶銆4銆佸垎閮ㄧН鍒嗘硶鐨勯鐩1.鎸囨暟鍨嬪拰骞傚嚱鏁扮粨鍚堢殑锛屽鏁板嚱鏁板拰骞傚嚱鏁扮粨鍚堢殑锛屽弽涓夎鍑芥暟鍜屽箓鍑芥暟缁撳悎鐨勮繖涓夌鏄瘮杈冨吀鍨嬬殑鐢ㄥ垎閮ㄧН鍒嗘硶绠楃殑銆2.寰Н鍒嗕腑鐨勪竴绫荤Н鍒嗗姙娉曪細瀵逛簬鐢变袱涓笉鍚屽嚱鏁扮粍鎴愮殑琚Н鍑芥暟锛屼笉...
绛旓細鐢ㄥ垎閮ㄧН鍒嗘硶姹傚畾绉垎 鎴戞潵绛 1涓洖绛 #鐑# 璇ヤ笉璇ヨ瀛╁瓙寰堟棭瀛︿範浜烘儏涓栨晠?鐧惧害缃戝弸af34c30f5 2015-04-02 路 TA鑾峰緱瓒呰繃4.3涓囦釜璧 鐭ラ亾澶ф湁鍙负绛斾富 鍥炵瓟閲:1.8涓 閲囩撼鐜:65% 甯姪鐨勪汉:5079涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鏈洖绛旂敱鎻愰棶鑰呮帹鑽 宸茶禐杩 宸茶俯杩< 浣犲杩欎釜...
绛旓細,鈭(e,1)xlnxdx =1/2鈭(e,1)lnxdx²=1/2*x²lnx(e,1)-1/2鈭(e,1)x²dlnx =1/2*x²lnx(e,1)-1/2鈭(e,1)x²*1/xdx =1/2*x²lnx(e,1)-1/2鈭(e,1)xdx =[1/2*x²lnx-x²/4](e,1)=e²/2-e²/4+...
绛旓細銆愮瓟妗堛戯細瀹氱Н鍒涓庝笉瀹氱Н鍒嗙殑鍒嗛儴绉垎娉鍦ㄤ娇鐢ㄦ潯浠朵笂鏄竴鏍风殑锛屽湪缁撹涓婏紝鍓嶈呭彲浠ュコ鍙ｄ笅闈㈠叕寮忛偅鏍凤紝鍋氬嚭涓閮ㄥ垎涔嬪悗灏卞皢涓娿佷笅闄愪唬鍏ワ細鐒跺悗璁＄畻鍙宠竟绗簩椤圭Н鍒嗭紟
