2025年成人高考專(zhuān)升本《高等數(shù)學(xué)（一）》積分模擬專(zhuān)項(xiàng)卷考試時(shí)間：______分鐘總分：______分姓名：______一、選擇題：本題共5小題，每小題4分，共20分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1.函數(shù)f(x)=∫<0xE2><0x82><0x9B>^xtsin(t^2)dt的導(dǎo)數(shù)f'(x)等于).(A)xsin(x^2)(B)sin(x^2)(C)2x^2cos(x^4)(D)x^2cos(x^4)2.若∫<0xE2><0x82><0x9B>^1(2ax+b)dx=3x^2+2x在區(qū)間[0,1]上成立，則常數(shù)a,b的值為).(A)a=3,b=2(B)a=3,b=0(C)a=1,b=1(D)a=0,b=23.下列不定積分計(jì)算正確的是).(A)∫xcos2xdx=(1/2)xsin2x+C(B)∫e<0xE2><0x82><0x9B>xdx=e<0xE2><0x82><0x9B>+C(C)∫(1/x)dx=ln|x|+C(D)∫√(1+x^2)dx=(1/3)(1+x^2)^(3/2)+C4.函數(shù)F(x)=∫<0xE2><0x82><0x9B>^xsin(t+1)dt的單調(diào)遞減區(qū)間是).(A)(-∞,-1)(B)(-1,+∞)(C)(-∞,0)(D)(0,+∞)5.若反常積分∫<0xE2><0x82><0x9B>^1dx/(xlnx)收斂，則實(shí)數(shù)a的取值范圍是).(A)a>1(B)a<1(C)a≥1(D)a≤1二、填空題：本題共4小題，每小題4分，共16分。6.∫(x^2-1)/(x+1)dx=________.7.利用定積分的幾何意義，計(jì)算∫<0xE2><0x82><0x9B>^2|x-1|dx=________.8.若f'(x)=sinx+1，且f(0)=2，則f(π/2)=________.9.計(jì)算定積分I=∫<0xE2><0x82><0x9B>^π(cosx+1)dx=________.三、解答題：本題共5小題，共64分。解答應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟。10.(本題滿(mǎn)分10分)計(jì)算不定積分∫xlnxdx.11.(本題滿(mǎn)分12分)計(jì)算不定積分∫x^2sec^2(x^3)dx.12.(本題滿(mǎn)分12分)計(jì)算定積分∫<0xE2><0x82><0x9B>^1xe^xdx.13.(本題滿(mǎn)分12分)計(jì)算定積分∫<0xE2><0x82><0x9B>^π/2xsin2xdx.14.(本題滿(mǎn)分16分)計(jì)算定積分∫<0xE2><0x82><0x9B>^1(x^2+1)/(x^2+x)dx.要求：對(duì)被積函數(shù)進(jìn)行適當(dāng)?shù)淖冃魏螅謩e在不同區(qū)間內(nèi)計(jì)算積分。試卷答案一、選擇題1.(A)2.(A)3.(C)4.(D)5.(B)二、填空題6.x^2/2-x+C7.28.π-19.π+2三、解答題10.解：令u=lnx，dv=xdx，則du=1/xdx，v=x^2/2.由分部積分法公式∫udv=uv-∫vdu，得∫xlnxdx=(lnx)*(x^2/2)-∫(x^2/2)*(1/x)dx=(x^2/2)lnx-∫(x/2)dx=(x^2/2)lnx-(1/2)*(x^2/2)+C=(x^2/2)lnx-(x^2/4)+C.11.解：令u=x^3，則du=3x^2dx，即x^2dx=(1/3)du.當(dāng)x=0時(shí)，u=0；當(dāng)x=1時(shí)，u=1.∫x^2sec^2(x^3)dx=∫sec^2(u)*(1/3)du=(1/3)∫sec^2udu=(1/3)tanu+C=(1/3)tan(x^3)+C.（此處為不定積分結(jié)果，若題目要求定積分，需帶入積分限）若求定積分∫<0xE2><0x82><0x9B>^1x^2sec^2(x^3)dx，則=(1/3)∫<0xE2><0x82><0x9B>^1sec^2udu=(1/3)[tanu]0^1=(1/3)[tan1-tan0]=(1/3)tan1.12.解：令u=x，dv=e^xdx，則du=dx，v=e^x.由分部積分法公式∫udv=uv-∫vdu，得∫xe^xdx=xe^x-∫e^xdx=xe^x-e^x+C=e^x(x-1)+C.若求定積分∫<0xE2><0x82><0x9B>^1xe^xdx，則=[e^x(x-1)]0^1=[e^1*(1-1)]-[e^0*(0-1)]=0-(-1)=1.13.解：令u=x，dv=sin2xdx，則du=dx，v=-cos2x/2.由分部積分法公式∫udv=uv-∫vdu，得∫<0xE2><0x82><0x9B>^π/2xsin2xdx=[-xcos2x/2]0^π/2+∫<0xE2><0x82><0x9B>^π/2cos2x/2dx=(-π/4*cosπ)-(0*cos0)+(1/2)∫<0xE2><0x82><0x9B>^π/2cos2xdx=(π/4)+(1/2)*(1/2)∫<0xE2><0x82><0x9B>^π/2cos2xd(2x)=(π/4)+(1/4)[sin2x]0^π/2=(π/4)+(1/4)[sinπ-sin0]=(π/4)+(1/4)*0=π/4.14.解：對(duì)被積函數(shù)進(jìn)行多項(xiàng)式除法，得(x^2+1)/(x^2+x)=1-x/(x^2+x).∫<0xE2><0x82><0x9B>^1(x^2+1)/(x^2+x)dx=∫<0xE2><0x82><0x9B>^11dx-∫<0xE2><0x82><0x9B>^1x/(x^2+x)dx=∫<0xE2><0x82><0x9B>^11dx-∫<0xE2><0x82><0x9B>^1x/(x(x+1))dx=∫<0xE2><0x82><0x9B>^11dx-∫<0xE2><0x82><0x9B>^1(1/x)dx-∫<0xE2><0x82><0x9B>^1(-1/(x+1))dx=∫<0xE2><0x82><0x9B>^11dx-[ln|x|]0^1+[ln|x+1|]0^1=[x]0^1-(ln1-ln0)+(ln2-ln1)=(1-0)-(0-(-∞))+(ln2-ln1)