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]
a) infinite
b) 2
c) 1
d) 0
Calculate the area under the graph of f(x)=-ln(x) above the x-axis on the interval (0, 1].
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
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.
To calculate the area under the graph of the function f(x) = -ln(x) above the x-axis on the interval (0, 1], we can use integration.
First, we need to find the antiderivative (or integral) of f(x) = -ln(x).
The antiderivative of -ln(x) can be found using integration rules. We have:
∫ -ln(x) dx = -x ln(x) + x + C
Next, we can find the definite integral of f(x) = -ln(x) on the interval (0, 1] by evaluating the antiderivative at the endpoints and subtracting:
∫[0, 1] -ln(x) dx = (-1)(1) ln(1) + (1) - ((-1)(0) ln(0) + (0))
= (-1)(0) + 1 - ((-1)(-∞) + (0))
= 1 - (-∞)
= 1 + ∞
The result is 1 + ∞, which is infinite. Therefore, the area under the graph of f(x) = -ln(x) above the x-axis on the interval (0, 1] is infinite.
So, the correct answer is:
a) infinite