Question

数学微积分问题，求大神！！

a(t)=ln(1+2^t)，求该函数和t=1和t=2和x轴所成图形面积
哦...t轴，写的时候太着急写错了...就是求积分，好久不学有点忘了

Answer 1

x=a(t)=ln(1+2^t)
t=1 x=a(1)=ln3
t=2 x=a(2)=ln5
S=∫[1,2]ln(1+2^t)dt
1+2^t=u t=log<2>u-1=ln(u-1)/ln2 dt=(1/ln2)du/(u-1)
=∫[3,5]lnu (1/ln2)du/(u-1)
=(1/ln2)∫[3,5]lnudu/(u-1)

∫lnudu/(u-1)=∫lnudln(u-1)
=ln(u-1)lnu-∫ln(u-1)du/u
=ln(u-1)lnu-ln(u-1)+∫du/u-∫(ln(u-1)du/u^2
=ln(u-1)lnu-ln(u-1)+lnu-ln(u-1) /u +∫du/u^2-2∫ln(u-1)du/u^3
=ln(u-1)lnu-ln(u-1)+lnu-ln(u-1)/u-1/u-2ln(u-1)/u^2+2∫du/u^3-2*3∫ln(u-1)du/u^4
=ln(u-1)lnu-ln(u-1)+lnu-ln(u-1)/u-1/u-ln(u-1)/u^2-1/u^2-2*3ln(u-1)/u^3-2/u^3-2*3*4∫ln(u-1)du/u^5

S=(1/ln2)∫[3,5]lnudu/(u-1)
=(1/ln2)[ln(u-1)lnu-ln(u-1)+lnu-ln(u-1)/u-1/u-ln(u-1)/u^2-1/u^2-2*3ln(u-1)/u^3-2/u^3 |[3,5]
-2*3*4∫[3,5]ln(u-1)du/u^5 ]
=(1/ln2)(ln5ln4-ln4+ln5-ln4/5-1/5-ln4/5^2-1/5^2-6ln4/125-2/125
-ln3ln2+ln2-ln3+ln2/3-1/3-ln2/9-1/9-6ln2/27-2/27 ]
-24∫[3,5]ln(u-1)du/u^5 ]
只能取近似值

Answer 2

不知道X轴对应式中那个量？是t吗？ 如果是的话，就相当于求a（t）在（1，2）上的积分。

Answer 3

你这几个变量了？

Answer 4

555