Calculate the area under the graph of f of x equals the negative of the natural logarithm of x above the x-axis on the interval (0, 1]

Question

a) infinite
b) 2
c) 1
d) 0

Answer 1

Calculate the area under the graph of f(x)=-ln(x) above the x-axis on the interval (0, 1].

Answer 2

hmmm.
∫[0,1] -lnx dx = x(1-lnx) [0,1]
OK. Since the value at x=1 is 1(1-0) = 1, we just need to evaluate
lim(x->0) x(1-lnx) = lim(x->0) -x*lnx
using l'Hospital's rule
lim(x->0) -x*lnx
lim(x->0) -lnx / (1/x) = lim(x->0) -(1/x) / (-1/x^2) = x -> 0

So the area is just = 1-0 = 1

Answer 3

In your repertoire of basic integrals you should have:
∫ lnx dx = xlnx - x

so for your case
area = ∫ -lnx dx from 0 to 1
= x - xlnx from 0 to 1
= 1 - 1ln1 - (0 - 0(-∞))

the problem will be in the 0(-∞) which turns out to be 0

Here is a short video where the problem is ∫ lnx dx from 0 to 1
which is simply the opposite of yours.
www.youtube.com/watch?v=5RhbrPAlx2E

If you really want to be intimidated watch beginning at 7:50 . Looks like the kid is in about grade 6.