How do you integrate #int 6^x-2^xdx# from #[1,e]#?

Question

1640 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-19T18:01:09

The answer is #=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129#

Let #u=6^x#

Then, #lnu=xln6#

#u=e^(xln6)#

Let #v=2^x#

#lnv=xln2#

#v=e^(xln2)#

Therefore

#int_1 ^e(6^x-2^x)dx#

#=int_1 ^e(e^(xln6)-e^(xln2))dx#

#= [e^(xln6)/ln6-e^(xln2)/ln2 ]_1^e #

#= [6^x/ln6-2^x/ln2 ]_1 ^e #

#=(6^e/ln6-2^e/ln2)-(6/ln6-2/ln2)#

#=6^e/ln6-2 ^e/ln2-6/ln6+2/ln2=62.8129#