求定积分∫(π~0)xsinx dx xsinx在π到0的定积分
\u6c42\u5b9a\u79ef\u5206\u222b(-\u03c0,\u03c0) xsinx dx(0,\u03c0/2) \u222b xsinx dx
=(0,\u03c0/2) \u222b -x dcosx
= -xcosx | (0,\u03c0/2) + (0,\u03c0/2) \u222bcosxdx
= 0 + sinx | (0,\u03c0/2)
= 1
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=-xcosx+∫cosxdx
=-xcosx+sinx+C
计算结果就是原式=-π*cosπ+sinπ+0-sin0=π
绛旓細=-xcosx+sinx+C 璁＄畻缁撴灉灏辨槸鍘熷紡=-蟺*cos蟺+sin蟺+0-sin0=蟺
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绛旓細绠鍗璁＄畻涓涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細濡傛灉浣犲墠闈㈡噦锛屾垜鐩存帴缁欎綘璇存渶鍚庝竴姝 閭ｄ釜琛ㄧず鐨勬槸cos蟺 - cos0 = -1 - 1 = -2
绛旓細瑙ｉ杩囩▼濡備笅锛
