Find the area under the graph of f(x) = e-ln(x) on the interval [1, 2].

Answer 1

You do realize that
f(x) = e^(-lnx) , can be changed to
f(x) = x^-1 = 1/x

so area = ∫(1/x) dx from 1 to 2
= [lnx] from 1 to 2
= ln2 - ln1
= ln2 or appr .6931...

proof:
http://www.wolframalpha.com/input/?i=%E2%88%AB(e%5E(-lnx))+dx+from+1+to+2

Answer 2

huh? e^(-lnx) = 1/e^(lnx) = 1/x

You can probably handle that, right?

Answer 3

no

Answer 4

shut up Steve