求x2乘sinx在0到1上的定积分 若f(x)在0到1上连续,那么f(x^2)sinx从-1到1...
\u600e\u6837\u6c42x\u4e0esinx\u7684\u5546\u57280\u52301\u4e0a\u5b9a\u79ef\u5206\u7684\u8fd1\u4f3c\u503c\uff1f\u62c9\u683c\u6717\u65e5\u63d2\u503c\u6cd5\u3002
http://www.hudong.com/wiki/%E6%8B%89%E6%A0%BC%E6%9C%97%E6%97%A5%E6%8F%92%E5%80%BC%E5%A4%9A%E9%A1%B9%E5%BC%8F%E9%80%BC%E8%BF%91
\u4ee4t=2x-1\uff0c\u5219x=(t+1)/2
\u4ee3\u5165\uff0c\u5f97f(t)=[(t+1)^2]/4
\u6240\u4ee5f(x)=[(x+1)^2]/4
=-∫(0->1) x^2.dcosx
=-[x^2.cosx]|(0->1) + 2∫(0->1) x.cosx dx
=-cos1 + 2∫(0->1) x.dsinx
=-cos1 + 2[x.dsinx]|(0->1) -2∫(0->1) sinx dx
=-cos1 + 2sin1 +2[cosx]|(0->1)
=-cos1 + 2sin1 +2cos1 -2
=2sinx + cos1 -2
∫(0->1) x^2.sinx dx
=-∫(0->1) x^2.dcosx
=-[x^2.cosx]|(0->1) + 2∫(0->1) x.cosx dx
=-cos1 + 2∫(0->1) x.dsinx
=-cos1 + 2[x.dsinx]|(0->1) -2∫(0->1) sinx dx
=-cos1 + 2sin1 +2[cosx]|(0->1)
=-cos1 + 2sin1 +2cos1 -2
=2sinx + cos1 -2
绛旓細=-鈭(0->1) x^2.dcosx =-[x^2.cosx]|(0->1) + 2鈭(0->1) x.cosx dx =-cos1 + 2鈭(0->1) x.dsinx =-cos1 + 2[x.dsinx]|(0->1) -2鈭(0->1) sinx dx =-cos1 + 2sin1 +2[cosx]|(0->1)=-cos1 + 2sin1 +2cos1 -2 =2sinx + cos1 -2 ...
绛旓細[(x鐨2娆℃柟)涔樹互sinx]dx,涓婇檺鏄2鍒嗕箣pai,涓嬮檺鏄0 =(-x²cosx+2xsinx+2cosx)|(0,蟺/2)=蟺-锛2cos0锛=蟺-2,4,姹傚畾绉垎:[(x鐨2娆℃柟)涔樹互sinx]dx,涓婇檺鏄2鍒嗕箣pai,涓嬮檺鏄0?(acontent)
绛旓細鍙皢sinx灞曞紑锛堝甫鐨簹璇轰綑椤圭殑椹厠鍔虫灄鍏紡锛塻inx=x-x^3/3!+x^5/5!-x^7/7!+...+x^97/97!+o(x^98)f(x)=x^2sinx=...+x^99/97!+...鏍规嵁娉板嫆瀹氱悊,f(x)=...+f(99)(0)/99!+...鏁協(99)(0)/99!=1/97!f(99)(0)=99*98=9702 ...
绛旓細鍙皢sinx灞曞紑锛堝甫鐨簹璇轰綑椤圭殑椹厠鍔虫灄鍏紡锛塻inx=x-x^3/3!+x^5/5!-x^7/7!+...+x^97/97!+o(x^98)f(x)=x^2sinx=...+x^99/97!+...鏍规嵁娉板嫆瀹氱悊锛宖(x)=...+f(99)(0)/99!+...鏁協(99)(0)/99!=1/97!f(99)(0)=99*98=9702 ...
绛旓細濡傚浘鎵绀猴細
绛旓細=x³/6-(1/4)鈭玿²d(sin2x)銆傝岋紝鈭玿²d(sin2x)=x²sin2x-鈭2xsin2xdx =x²sin2x+xcos2x-鈭玞os2xdx =x²sin2x+xcos2x-(1/2)sin2x+C锛屸埓鈭(0,蟺)x²sin²xdx =[x³/6-(1/4)(x²sin2x+xcos2x)+(1/8)sin2x]...
绛旓細鍏蜂綋鍥炵瓟濡傚浘锛涓涓嚱鏁帮紝鍙互瀛樺湪涓嶅畾绉垎锛岃屼笉瀛樺湪瀹氱Н鍒嗭紱涔熷彲浠ュ瓨鍦ㄥ畾绉垎锛岃屼笉瀛樺湪涓嶅畾绉垎銆備竴涓繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜屼笉瀹氱Н鍒嗭紱鑻ュ彧鏈夋湁闄愪釜闂存柇鐐癸紝鍒欏畾绉垎瀛樺湪锛涜嫢鏈夎烦璺冮棿鏂偣锛屽垯鍘熷嚱鏁颁竴瀹氫笉瀛樺湪锛屽嵆涓嶅畾绉垎涓瀹氫笉瀛樺湪銆
绛旓細=鈭(0-蟺)蟺 sinx dx-I 2I=蟺鈭(0-蟺)sinx dx 榛庢浖绉垎 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锛籥锛宐锛戒笂鐨勭煩褰㈢疮鍔犺捣鏉ワ紝鎵寰楀埌鐨勫氨鏄繖涓嚱鏁扮殑鍥捐薄鍦ㄥ尯闂达蓟a锛宐锛界殑闈㈢Н銆傚疄闄呬笂锛...
绛旓細鎹㈠厓鍗冲彲
绛旓細(-n蟺cosn蟺+(n蟺-蟺)cos(n蟺-蟺))=(-1)^(n-1)(-n蟺(-1)^n+(n蟺-蟺)(-1)^(n-1))=(2n-1)蟺 浜庢槸锛氣埆(0,n蟺)x|sinx|dx =鈭(0,蟺)xsinxdx-鈭(蟺,2蟺)xsinxdx+...+(-1)^(n-1)鈭(n蟺-蟺,n蟺)xsinxdx =蟺+3蟺+...+(2n-1)蟺 =n²蟺 ...
