111年一般警察人員考試
考 試 別:一般警察人員考試
等 別:三等考試
類科組別:消防警察人員
科 目:微積分
解答:f′(x)=3x2−1⇒f(x)=x3−x+C,又圖形通過(1,2),即f(1)=2⇒1−1+C=2⇒C=2⇒y=x3−x+2
解答:令f(x)=x−√x2+1⇒f′(x)=1−x√x2+1因此y=tan−1f(x)⇒y′=f′(x)1+(f(x))2=1−x√x2+12(x2+1)−2x√x2+1=√x2+1−x√x2+12x2+2−2x√x2+1=√x2+1−x(2x2+2)√x2+1−2x(x2+1)⇒y′=√x2+1−x(2x2+2)√x2+1−2x(x2+1)
解答:先求limn→∞3nln(1+3n)=limn→∞ln(1+3n)13n=limn→∞−3n(n+3)−13n2=limn→∞9n2n(n+3)=9因此=limn→∞(1+3n)3n=limn→∞e3nln(1+3/n)=e9
解答:∫∞2x+3x3−x2+x−1dx=∫∞2x+3(x2+1)(x−1)dx=∫∞22x−1−2x+1x2+1dx=∫∞22x−1−2xx2+1−1x2+1dx=[2ln(x−1)−ln(x2+1)−tan−1x]|∞2=[ln(x−1)2x2+1−tan−1x]|∞2=(0−π2)−(ln15−tan−12)=ln5+tan−12−π2
解答:∫0−∞1x2−3x+2dx=∫0−∞1(x−2)(x−1)dx=∫0−∞1x−2−1x−1dx=[ln(x−2)−ln(x−1)]|0−∞=[lnx−2x−1]|0−∞=ln2
解答:
第一象限心臟線面積=∫π/2012⋅(2(1+cosθ))2dθ=∫π/202+4cosθ+2cos2θdθ=∫π/203+4cosθ+cos2θdθ=[3θ+4sinθ+12sin2θ]|π/20=4+32π第一象限圓面積=14⋅22π=π欲求之面積=2(4+32π−π)=8+π
解答:質心坐標(ˉx,ˉy),其中ˉx=∫π/40x(cosx−sinx)dx∫π/40cosx−sinxdx,ˉy=∫π/40(cosx−sinx)2dx∫π/40cosx−sinxdx而{∫π/40cosx−sinxdx=[sinx+cosx]|π/40=√2−1∫π/40x(cosx−sinx)dx=[(x−1)sinx+(x+1)cosx]|π/40=√24π−1∫π/40(cosx−sinx)2dx=∫π/401−sin(2x)dx=[x+12cos(2x)]|π/40=π4−12⇒{ˉx=√2π/4−1√2−1=2+√24π−(√2+1)ˉy=π/4−1/2√2−1=(√2+1)(π−2)4⇒質心坐標=(2+√24π−(√2+1),(√2+1)(π−2)4)================ END ===============
解答:質心坐標(ˉx,ˉy),其中ˉx=∫π/40x(cosx−sinx)dx∫π/40cosx−sinxdx,ˉy=∫π/40(cosx−sinx)2dx∫π/40cosx−sinxdx而{∫π/40cosx−sinxdx=[sinx+cosx]|π/40=√2−1∫π/40x(cosx−sinx)dx=[(x−1)sinx+(x+1)cosx]|π/40=√24π−1∫π/40(cosx−sinx)2dx=∫π/401−sin(2x)dx=[x+12cos(2x)]|π/40=π4−12⇒{ˉx=√2π/4−1√2−1=2+√24π−(√2+1)ˉy=π/4−1/2√2−1=(√2+1)(π−2)4⇒質心坐標=(2+√24π−(√2+1),(√2+1)(π−2)4)================ END ===============
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