几道简单的高数题1.设f(x)=∫1到x(Int/1+t)dt,(x>0),求f(x)+f(1/x)2.设光滑曲线y=f(x)过原点,且当x>0时f(x)>0,对应于[0,x]一段曲线的弧长为e^x-1,求f(x).

Question

几道简单的高数题
1.设f(x)=∫1到x (Int/1+t)dt ,(x>0),求f(x)+f(1/x)
2.设光滑曲线y=f(x)过原点,且当x>0时f(x)>0,对应于[0,x]一段曲线的弧长为e^x-1,求f(x).

Answer 1

1.f(x)+f(1/x)=∫(1,x)[lnt/(1+t)]dt+∫(1,1/x)[lnt/(1+t)]dt
=∫(1,x)[lnt/(1+t)]dt+∫(1,x)[ln(1/t)/(1+1/t)]d(1/t) (第二个积分用1/t代换t)
=∫(1,x)[lnt/(1+t)]dt+∫(1,x)[-lnt/(1+1/t)](-1/t²)dt
=∫(1,x)[lnt/(1+t)]dt+∫(1,x)[lnt/(t(1+t))]dt
=∫(1,x)[lnt(1/(1+t)+1/(t(1+t))]dt
=∫(1,x)(lnt/t)dt
=∫(1,x)lntd(lnt)
=[ln²t/2]│(1,x)
=ln²x/2
2.∵对应于[0,x]一段曲线的弧长为e^x-1
∴e^x-1=∫(0,x)√[1+(f'(x))²]dx
==>e^x=√[1+(f'(x))²] (两端求导数)
==>e^(2x)=1+(f'(x))²
==>(f'(x))²=1-e^(2x)
==>f'(x)=√[1-e^(2x)] (∵当x>0时f(x)>0,∴f'(x)>0)
==>f(x)=∫[1-e^(2x)]dx
=√[e^(2x)-1]+arcsec(e^x)+C (C是积分常数,
∵曲线y=f(x)过原点,即f(0)=0
∴有C=0
故f(x)=√[e^(2x)-1]+arcsec(e^x).