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 -1 = -2. but 1/x^2 is a positive function and so its integral over [-1,1] should be positive. what is wrong here?

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

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p