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Pareto's Law of Income states that the number of people with incomes between $ x = a $ and $ x = b $ is $ N = \int_a^b Ax^{-k}\ dx $ where $ A $ and $ k $ are constants with $ A > 0 $ and $ k > 1 $. The average income of these people is

$$ \bar{x} = \frac{1}{N} \int_a^b Ax^{1 - k}\ dx $$

Calculate $ \bar{x} $.

$\overline{x}=\frac{(1-k)\left(b^{2-k}-a^{2-k}\right)}{(2-k)\left(b^{1-k}-a^{1-k}\right)}$

Applications of Integration

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Campbell University

Oregon State University

University of Michigan - Ann Arbor

Okay, So the problem asks us using Perrotta's love income to find the average income of people between to specific incomes. So first part is that we're going to evaluate the number of people capital end, which was given to be the integral from A to B of capital a x to the minus k d. X. So evaluating this we get a is a constant. And then when you add one to the power of X and divide, you get one minus K and the denominator and then X to the one minus k. Hey, hey! Which gives us a Times B to the one minus K minus a to the one minus K over one minus K. Now for explore, it was given as one over end times the into girl from A to B of a ext the one minus K. D X. Now valuing this, we have won over end times a over to minus k times X the two months kay taken from a to B. So for one over end, we're just going to take the value and we have and flip the numerator and denominator. So we have one minus k divided by a beat, the one minus K. Maya's a the one minus Okay times. This remains a minus two K. And then finally our last term B to the two minus K minus a Jew minus k. These A's will cancel, leaving us with the final result of explore you one minus k times B to the two minus K minus A. To this you minus K divided by to minus K trying to be too, the one minus K minus a to the one minus K. And that right there is the answer.