2025年新版定積分題庫(kù)及答案1.計(jì)算定積分∫?^π(x2+3sinx)dx解：原積分=∫?^πx2dx+3∫?^πsinxdx=[x3/3]?^π+3[-cosx]?^π=(π3/30)+3[(-cosπ)(-cos0)]=π3/3+3[(1)(-1)]=π3/3+62.計(jì)算∫?^4(√x+1/√x)dx解：原積分=∫?^4x^(1/2)dx+∫?^4x^(-1/2)dx=[(2/3)x^(3/2)]?^4+[2x^(1/2)]?^4=(2/3)(81)+2(21)=14/3+2=20/33.利用換元法計(jì)算∫?^1x√(1x2)dx解：令u=1x2，則du=-2xdx，當(dāng)x=0時(shí)u=1，x=1時(shí)u=0。原積分=∫?^0√u(-du/2)=(1/2)∫?^1u^(1/2)du=(1/2)(2/3)u^(3/2)|?^1=1/34.計(jì)算∫?^π/2cos3xsinxdx解：令t=cosx，則dt=-sinxdx，當(dāng)x=0時(shí)t=1，x=π/2時(shí)t=0。原積分=∫?^0t3(-dt)=∫?^1t3dt=[t?/4]?^1=1/45.計(jì)算∫??^1(x?3x3+2x)dx解：被積函數(shù)為奇函數(shù)（x?、-3x3、2x均為奇函數(shù)），在對(duì)稱區(qū)間[-1,1]上積分值為0，故原積分=06.計(jì)算∫?π/2^π/2cos?xdx解：被積函數(shù)為偶函數(shù)，原積分=2∫?^π/2cos?xdx。利用遞推公式∫?^π/2cos?xdx=(n-1)!!/n!!π/2（n為偶數(shù)），n=4時(shí)，(3!!/4!!)π/2=(31)/(42)π/2=3π/16，故原積分=2(3π/16)=3π/87.用分部積分法計(jì)算∫?^1x2e^xdx解：設(shè)u=x2，dv=e^xdx，則du=2xdx，v=e^x。原積分=x2e^x|?^12∫?^1xe^xdx=(e0)2[xe^x|?^1∫?^1e^xdx]=e2[(e0)(e^x|?^1)]=e2[e(e1)]=e2(1)=e28.計(jì)算∫?^exlnxdx解：設(shè)u=lnx，dv=xdx，則du=(1/x)dx，v=x2/2。原積分=(x2/2)lnx|?^e∫?^e(x2/2)(1/x)dx=(e2/210)(1/2)∫?^exdx=e2/2(1/2)(x2/2)|?^e=e2/2(e2/41/4)=e2/4+1/49.計(jì)算∫?^πxsinxdx解：設(shè)u=x，dv=sinxdx，則du=dx，v=-cosx。原積分=-xcosx|?^π+∫?^πcosxdx=-π(-1)+0+sinx|?^π=π+0=π10.求變限積分函數(shù)F(x)=∫?^xt√(1+t2)dt的導(dǎo)數(shù)F’(x)解：根據(jù)微積分基本定理，F(xiàn)’(x)=x√(1+x2)11.求lim?→0[∫??(1cost)dt]/x3解：分子分母均趨于0，用洛必達(dá)法則，分子導(dǎo)數(shù)=1cosx，分母導(dǎo)數(shù)=3x2，原式=lim?→0(1cosx)/(3x2)。再用等價(jià)無(wú)窮小1cosx~x2/2，故原式=lim?→0(x2/2)/(3x2)=1/612.計(jì)算∫?^1dx/(1+√x)解：令t=√x，則x=t2，dx=2tdt，當(dāng)x=0時(shí)t=0，x=1時(shí)t=1。原積分=∫?^12t/(1+t)dt=2∫?^1[(t+1)1]/(1+t)dt=2∫?^1[11/(1+t)]dt=2[tln(1+t)]?^1=2[(1ln2)0]=22ln213.計(jì)算∫?^2|x1|dx解：分段積分，x∈[0,1]時(shí)|x1|=1x；x∈[1,2]時(shí)|x1|=x1。原積分=∫?^1(1x)dx+∫?^2(x1)dx=[xx2/2]?^1+[x2/2x]?^2=(11/2)+[(22)(1/21)]=1/2+(0(-1/2))=1/2+1/2=114.計(jì)算反常積分∫?^+∞dx/(x√x)解：原積分=∫?^+∞x^(-3/2)dx=lim?→+∞[-2x^(-1/2)]?^a=lim?→+∞(-2/√a+2/√1)=0+2=2（收斂）15.計(jì)算∫?^1lnxdx（瑕積分）解：x=0為瑕點(diǎn)，原積分=lim?→0?∫?^1lnxdx=lim?→0?[xlnxx]?^1=lim?→0?[(01)(tlntt)]。其中tlnt→0（當(dāng)t→0?時(shí)），故原式=(-1)(00)=-116.計(jì)算∫?^π/4tan2xdx解：tan2x=sec2x1，原積分=∫?^π/4(sec2x1)dx=[tanxx]?^π/4=(1π/4)(00)=1π/417.計(jì)算∫??2(x3cosx+√(4x2))dx解：x3cosx為奇函數(shù)（x3奇，cosx偶，乘積奇），在[-2,2]上積分=0；√(4x2)為上半圓，積分=半圓面積=π22/2=2π。故原積分=0+2π=2π18.求∫?^2f(x)dx，其中f(x)=｛x2,0≤x≤1；2x,1<x≤2｝解：分段積分，原積分=∫?^1x2dx+∫?^2(2x)dx=[x3/3]?^1+[2xx2/2]?^2=(1/3)+[(42)(21/2)]=1/3+(23/2)=1/3+1/2=5/619.計(jì)算∫?^ln2√(e?1)dx解：令t=√(e?1)，則t2=e?1，e?=t2+1，x=ln(t2+1)，dx=2t/(t2+1)dt。當(dāng)x=0時(shí)t=0，x=ln2時(shí)t=1。原積分=∫?^1t(2t)/(t2+1)dt=2∫?^1(t2)/(t2+1)dt=2∫?^1[11/(t2+1)]dt=2[tarctant]?^1=2[(1π/4)0]=2π/2