这两道高数题 怎么做?
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令lnx=t,则x=e^t
x∈[1:2],t∈[0,ln2]
∫[1:2]lnxdx
=∫[0:ln2]td(e^t)
=∫[0:ln2]t·e^t dt
=(t-1)·e^t|[0:ln2]
=(ln2-1)·e^(ln2) -(0-1)·e^0
=(ln2-1)·e²+1
∫[1:2](lnx)²dx
=∫[0:ln2]t²d(e^t)
=∫[0:ln2]t²·e^t dt
=(t²-2t+2)e^t|[0:ln2]
=[(ln2)²-2ln2+2]·e^(ln2) -(0-0+2)e^0
=[(ln2)²-2ln2+2]·e²-2
[(ln2)²-2ln2+2]·e²-2-[(ln2-1)·e²+1]
=[(ln2)²-3ln2+1]·e²-3
<0
∫[1:2]lnxdx>∫[1:2](lnx)²dx
本题应该有简便方法比较大小,不过为了说明情况,还是把两个定积分都求出来了。
f(1,y/x)
=[1²·(y/x)]/[1²+(y/x)²]
=(y/x)/(1+ y²/x²)
=xy/(x²+y²)
x∈[1:2],t∈[0,ln2]
∫[1:2]lnxdx
=∫[0:ln2]td(e^t)
=∫[0:ln2]t·e^t dt
=(t-1)·e^t|[0:ln2]
=(ln2-1)·e^(ln2) -(0-1)·e^0
=(ln2-1)·e²+1
∫[1:2](lnx)²dx
=∫[0:ln2]t²d(e^t)
=∫[0:ln2]t²·e^t dt
=(t²-2t+2)e^t|[0:ln2]
=[(ln2)²-2ln2+2]·e^(ln2) -(0-0+2)e^0
=[(ln2)²-2ln2+2]·e²-2
[(ln2)²-2ln2+2]·e²-2-[(ln2-1)·e²+1]
=[(ln2)²-3ln2+1]·e²-3
<0
∫[1:2]lnxdx>∫[1:2](lnx)²dx
本题应该有简便方法比较大小,不过为了说明情况,还是把两个定积分都求出来了。
f(1,y/x)
=[1²·(y/x)]/[1²+(y/x)²]
=(y/x)/(1+ y²/x²)
=xy/(x²+y²)
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