國立政治大學112學年度轉學生招生考試
科目:微積分(二)系所別:應用數學系二年級
解答:(a) limx→0∫3x+22ln(t−1)dtx2=limx→0ddx∫3x+22ln(t−1)dtddxx2=limx→03ln(3x+1)2x=limx→0ddx(3ln(3x+1))ddx(2x)=limx→093x+12=92(b) u=cosx⇒du=−sinxdx⇒∫sinx1+cos2xdx=∫−11+u2du=−tan−1u+C=−tan−1cosx+C(c) {u=arctan(x)dv=4dx⇒{du=dx1+x2v=4x⇒∫4arctan(x)dx=4xarctan(x)−∫4x1+x2dx=4xarctan(x)−2ln(1+x2)+C⇒∫104arctan(x)dx=[4xarctan(x)−2ln(1+x2)]|10=π−2ln2(d) {u=csc(x)dv=csc2(x)dx⇒{du=−csc(x)cot(x)dxv=−cotx⇒I=∫4sin3(x)dx=4∫csc3(x)dx=−4csc(x)cot(x)−4∫csc(x)cot2(x)dx=−4csc(x)cot(x)−4∫csc(x)(csc2(x)−1)dx=−4csc(x)cot(x)−I+4∫csc(x)dx⇒2I=−4csc(x)cot(x)+4∫1sinxdxI=−4csc(x)cot(x)+2∫1sin(x/2)cos(x/2)dx⇒I=−2csc(x)cot(x)+∫tan2(x/2)tan2(x/2)sin(x/2)cos(x/2)dx=−2csc(x)cot(x)+∫sec2(x/2)tan(x/2)dx=−2csc(x)cot(x)+2ln(|tanx2|)+C解答:(a) n√3+np<n√np=1n(p/2−1)By p-series test, if p2−1>1(p>4),then ∞∑n=11np/2−1 will converge.And by comparison-test, the series ∞∑n=1n√3+np will converge(b) limn→∞n∑i=13π4nsec2(iπ4n)=∫103π4sec2πx4dx=3[tanπx4]|10=3
解答:dydx=√1−y21+x2⇒∫1√1−y2dy=∫11+x2dx⇒arcsiny=arctanx+C⇒y=sin(arctanx+C)⇒y(0)=sinC=0⇒C=0⇒y=sin(arctanx)⇒y=x√1+x2
解答:{x=ρsinϕcosθy=ρsinϕsinθz=ρcosϕ⇒z=ρcosϕ=√x2+y2=ρsinϕ⇒ϕ=π4⇒V=∫π/20∫π/2π/4∫√120ρ2sinϕdρdϕdθ=8√3∫π/20∫π/2π/4sinϕdϕdθ=4√6∫π/201dθ=2√6π
解答:r(t)=eti+(1+t)j⇒r′(t)=eti+j⇒∫CF⋅dr=∫10((et(1+t)2+2(1+t))i+(e2t(1+t)+2et+2)j)⋅(eti+j)dt=∫10(e2t(1+t)(2+t)+2et(t+2)+2)dt=[12e2t(t+1)2+2et(t+1)+2t]|10=2e2+4e−12
解答:r(t)=(2cost)i+(2sint)j⇒C:x2+y2=4⇒∫C−y3dx+x3dy=∬
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解題僅供參考,轉學考歷年試題及詳解
第六題最後那個雙重積分.D表示的是以原點為圓心半徑為2的圓的區域,但不能直接把(x^2+y^2)看作4,要當作r^2下去做積分,答案應是24*pi.
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