1/0 is infinity so it has a problem with that...furthermore the anti-derivative of 1/x is ln(x), which is also infinity (negative infinity) at 0. So the integral from 0 to 1 of 1/x doesn't exist, so this is why your calculator is giving an error.
On the other hand from 1 to 2, everything has a perfectly defined value:
ln(2) - ln(1) = ln(2) ~ 0.69314718056
Edit:
Mathematically: 0 ≤ x ≤ 1 is the same as 0 < x ≤ 1 (because leaving off a single REAL number amounts to leaving off a strip of area with 0 width, therefore the additional area by including x = 0 adds exactly 0)
I doubt any numerical integrator will integrate this properly because they will always assume the first: 0 ≤ x ≤ 1 (there's no way to specify it's the second... because it doesn't make sense to anyway). The only way you could get an approximation for this, is if you did a right-hand rule Riemann sum (because the value right to the right of 0 DOES have a valid value: i.e. 1/0.000 01 = 100,000). In this case the numerical integrator would just give you some random very large number (depending on how you initialized the integrator).
Any normal, numerical integrator will ALWAYS include the points x₀ and x₁ for an integral from x₀ ≤ x ≤ x₁. Therefore in this case, including x₀ = 0 means 1/0 = infinity. This infinity will likely trigger the integrator to say it's impossible to take this integral.
You are telling your calculator that 0 is IN the domain. Since you cannot divide by zero, this throws an error.
If you were to change it to 0 < x ≤ 1, I bet it would accept it, since x is not equal to zero, so you can perform all of the divisions needed properly.
Answers & Comments
1/0 is infinity so it has a problem with that...furthermore the anti-derivative of 1/x is ln(x), which is also infinity (negative infinity) at 0. So the integral from 0 to 1 of 1/x doesn't exist, so this is why your calculator is giving an error.
On the other hand from 1 to 2, everything has a perfectly defined value:
ln(2) - ln(1) = ln(2) ~ 0.69314718056
Edit:
Mathematically: 0 ≤ x ≤ 1 is the same as 0 < x ≤ 1 (because leaving off a single REAL number amounts to leaving off a strip of area with 0 width, therefore the additional area by including x = 0 adds exactly 0)
I doubt any numerical integrator will integrate this properly because they will always assume the first: 0 ≤ x ≤ 1 (there's no way to specify it's the second... because it doesn't make sense to anyway). The only way you could get an approximation for this, is if you did a right-hand rule Riemann sum (because the value right to the right of 0 DOES have a valid value: i.e. 1/0.000 01 = 100,000). In this case the numerical integrator would just give you some random very large number (depending on how you initialized the integrator).
Any normal, numerical integrator will ALWAYS include the points x₀ and x₁ for an integral from x₀ ≤ x ≤ x₁. Therefore in this case, including x₀ = 0 means 1/0 = infinity. This infinity will likely trigger the integrator to say it's impossible to take this integral.
You should actually write the domain as 0<x <= 1. 1/0 is nasty. No computer or calculator likes it much.
You are telling your calculator that 0 is IN the domain. Since you cannot divide by zero, this throws an error.
If you were to change it to 0 < x ≤ 1, I bet it would accept it, since x is not equal to zero, so you can perform all of the divisions needed properly.
The area under the graph?
As the graph approaches X=0, the y-value approaches the infinities. You can't find the area of something with an infinite length.
integral 1/x = ln x
0 ≤ x ≤ 1 = ln 1 - ln 0 = 0 - undefined
1 ≤ x ≤ 2 = ln2 - ln 1 = ln2 - 0 = ln 2