If f(x) = -1/x, then f'(x) = 1/x^2. Theorem seems to suggest that the integral from -1 to 1 of 1/x^2 dx would equal f(1) - f(-1)
1 answer:
Answer:
The flaw is at x=0 where the function is not defined
Step-by-step explanation:
For any function we have
The integral can be evaluated only if the function f(x) is defined in the interval [a,b] in which this integral is evaluated
In our our case function f(x) is not defined at x=0 in the interval [-1,+1] thus this results in the flaw that we have obtained in the reasoning
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