the actual problem is integral(2 to +infinity) (1/x^2) dx
and after integrating and plugging in the bounds, i got
-1/(infinity sign)+ 1/2
does that converge to 1/2, Ln2, 1, 2 or is the answer nonexistent?
what i was thinking that everytime the bottom number increases seeing the number in infinity, the answer decreases so adding the 1/2 will make a difference but i don't know what it will converge to.
1/infinity = 0, so the answer is 1/2
You evaluated the integral correctly
To evaluate the integral ∫(2 to +∞) (1/x^2) dx, you first need to find the antiderivative of the function 1/x^2.
The antiderivative of 1/x^2 can be found by using the power rule for integration. The power rule states that the antiderivative of x^n is (1/(n+1))x^(n+1), where n is not equal to -1.
Applying the power rule to 1/x^2, you get the antiderivative as -1/x.
Now, you can find the definite integral by plugging in the upper and lower bounds into the antiderivative formula.
The definite integral of (1/x^2) from 2 to +∞ is [-1/x] evaluated from 2 to +∞.
Evaluating the antiderivative at the upper bound (+∞), you get -1/(+∞), which simplifies to 0.
Evaluating the antiderivative at the lower bound (2), you get -1/2.
Therefore, the value of the definite integral is (0 - (-1/2)), which is equal to 1/2.
Hence, the answer to the question is 1/2.