how would you do this improper integral
1/(x-1)
from 0 to 2
this is improper at one, so I split it up into two integrals
ln(x-1) from 0-1 and
ln(x-1) from 1-2
I then did for the first one the (lim t->1(-) of ln(t-1))-(ln(0-1))
and then the same thing for the second part
I didn't know if this was right though, or what the answer would be
Did you read my previous answer?
yes, but I wasn't sure of the final answer from your answer.
I left the final calculation up to you. If you define the integral as the sum of two parts that each approach within a distance a of x=1, and let a gapproach zero, than the answer will be zero because the two parts will always cancel, not matter how small a is.
To solve the improper integral 1/(x-1) from 0 to 2, you correctly identified that the integral is improper at x = 1. To handle this, you split the integral into two parts—one from 0 to 1, and another from 1 to 2.
Let's first focus on the first part of the integral from 0 to 1. We can write it as:
∫[0 to 1] (1/(x-1)) dx
Now, you correctly observed that the limit as x approaches 1 from the left side (lim x->1(-)) should be used to evaluate the function at x = 1. So, the first part of the integral becomes:
lim t->1(-) ∫[0 to t] (1/(x-1)) dx
To evaluate this, we can integrate the function 1/(x-1) with respect to x, which gives us the natural logarithm ln(x-1). Therefore, the first part can be written as:
lim t->1(-) [ln(x-1)] evaluated from 0 to t
= lim t->1(-) [ln(t-1) - ln(0-1)]
Now, let's focus on the second part of the integral from 1 to 2. We can write it as:
∫[1 to 2] (1/(x-1)) dx
Similar to the first part, the limit as x approaches 1 from the right side (lim x->1(+)) should be used to evaluate the function at x = 1. So, the second part of the integral becomes:
lim t->1(+) ∫[t to 2] (1/(x-1)) dx
Again, integrating the function 1/(x-1) with respect to x gives us ln(x-1). Therefore, the second part can be written as:
lim t->1(+) [ln(x-1)] evaluated from t to 2
= lim t->1(+) [ln(2-1) - ln(t-1)]
Now, to find the value of the improper integral, you need to evaluate both limits:
∫[0 to 2] (1/(x-1)) dx = lim t->1(-) [ln(t-1) - ln(0-1)] + lim t->1(+) [ln(2-1) - ln(t-1)]
= [ln(1-1) - ln(0-1)] + [ln(2-1) - ln(1-1)]
= [ln(0) - ln(-1)] + [ln(1) - ln(0)]
However, ln(0) is undefined, and ln(-1) is also undefined. So, the integral is divergent. In other words, it does not have a finite value.
Therefore, the improper integral of 1/(x-1) from 0 to 2 is undefined.