中央電大《微積分初步》模擬試題

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁眉頁腳。本資料屬于網(wǎng)絡(luò)整理，如有侵權(quán)，請聯(lián)系刪除，謝謝！微積分初步模擬試題一41f(x)．ln(x)x22x3y=．x11(x)x1(xf在．12(x)dxcos2xcf(x)f．．42x()0y324yxxBx2xf(x)limf(x)〔.x01A．201Ay(x2(2,2)〕ACCindsxxedxx．．20011dxx．dx．x11B1yy〕１xcy1x2c；y2；2yec；yce1xxDlimx23x2⒈．x21x13x2(xx2)(xxx2x11limx2limlim解12x2x1x1x13xlncosxy⒉設(shè)y.23131x21x2解y(sin)xtanx2cosx2exdx5exexd(5ex)5exdxx25ec解x5exdxx20xdxxxdxx1解2220200034mhShx4x2S(x)x4x,22x()2Sxxx2()0令Sxx2，xh1S(2)２微積分初步模擬試題二4f(xx2x2f(x)．x212xsink,x0(x)x0fx在=k．x02f(x)x1(在．1211()fx，將x0代入上式得(0)f22x1⒋152)dx3xx．14()sin0y3y．24x1y232xx．x2．x1C．x2x0且x2x1．且Df(x)x1limf(x)〔.xx1．2A．01C(〕yx4x72ACC３1x1．．xx2311x11dxxdx．．x11A〕yyx；ye；xyyyx；yxBlimx26x8⒈．xx22x16x8x2(x2)(x4)(x2)(x(x4)(x23limx2limlim解x2x1x2x2cos5xlnxy.⒉設(shè)y33lnx2(cos5)(ln)sin5)3ln(ln)5sin5解yx3xxx2xxxxsinxdx.xsinxdx=2sindxx2cosxc解x⒌xdx20xdxxsinxsindxx1解222020220062.5hx2xyh62.5，hx262.5yx4x4xx222x2x2yxx252令y0x5x5hx5，h2.5４微積分初步期末模擬試題三41f(x)．4x2(2,2)4x2k．x02⒊f(x)xf(x)=．1x2111())()()fxx，fxxxx2sinxdx．cosxc提示：sinxdxcosxc()sinxe3y4．xy4yxxAB02xxx0.k=〔f(x)kx0112．0B()0fxfx()５A(chǔ)．f(x)af(x)dx〔〕-a00f(x)dx．a(chǎn)2f(x)dxf(x)dxA．．D．0-a-a01yy〕；xyxC；y1CyeCx1；yCe12x2Blimx23x2．x24x23x2(xx2)(x2)(x2)x11limx2limlim解4x24x2x2x2x2sin5xcosxy.y35cos53cos(sin)5cos532x解yx2xxxxx)2dxxx)22dx=2))x)Cxx解233xx