Let f(x)=1/x and g(x)=1/x^2 Find the area A of the region between the graphs of f and g on the interval [1/2, 16].

Question

I set the two equations together and got x=1 and then I set the integral from 1/2 to 16 and solved and I got 1.528... but I am wrong. How would I solve this equation?

Answer 1

The tricky thing about this problem is that on the interval [0.5,1], the graph of g(x) = 1/x² is above the graph of f(x) = 1/x, whereas on the interval [1,16], the it's the opposite, with f(x) being above g(x). (plug the functions into a graphing calculator and it's easier to see).

This means that you can't just evaluate the integral of one function minus the other from 0.5 to 16 to get the area. You have to split the integral into two parts. The first integral is going to be the integral from 0.5 to 1 of g(x) - f(x) dx (because g(x) is the upper function on that interval), and the second integral is the integral from 1 to 16 of f(x) - g(x) dx (because f(x) is the upper function on that interval).

Evaluating this with the functions given, I believe the answer should be that the area between the two curves is 2.142.

Answer 2

The curves cross at x=1, so you really need to calculate two integrals: one from [1/2, 1] and one from [1,16].

So you should have:

A = ∫ 1/x^2 - 1/x dx from [1/2,1] + ∫1/x - 1/x^2 dx from [1,16]

= - (1/x + ln(x)) from [1/2,1] + (ln(x) + 1/x) from [1,16]

= -1 - ln(1) + 2 + ln(1/2) + ln(16) +1/16 - ln(1) - 1

= 1/16 - ln(2) + 4ln(2)

= 3ln(2) + 1/16 ~ 2.14

Lot of trickiness with the signs etc. so I may have made some mistakes simplifying but the idea of splitting it into two integrals is correct.

Answer 3

Question:

Let f(x)=1/x and g(x)=1/x^2 Find the area A of the region between the graphs of f and g on the interval [1/2, 16].

I set the two equations together and got x=1 and then I set the integral from 1/2 to 16 and solved and I got 1.528... but I am wrong. How would I solve this equation?

Solution:

Given:

f(x)=1/x

g(x)=1/x^2

Interval [1/2, 16]

Find the crossover point:

f(x) = g(x)

1/x = 1/x^2

x = x^2

x^2 - x = 0

x(x-1) = 0

Roots: 0, 1

g(x) is the upper bound between [0, 1]

f(x) is the upper bound between [1, 16]

A = INT [g(x) - f(x)] dx + INT [f(x) - g(x)] dx with the above bounds

= INT [1/x^2 - 1/x] dx + INT [1/x - 1/x^2] dx with the above bounds

= INT [x^-2 - 1/x] dx from 1/2 to 1

+ INT [1/x - x^-2] dx from 1 to 16

= [-x^-1 - ln(x)] from 1/2 to 1

+ [ln(x) - -x^-1] from 1 to 16

= [-1/x - ln(x)] from 1/2 to 1

+ [ln(x) + 1/x] from 1 to 16

= [-1/1 - ln(1)] - [-1/(1/2) - ln(1/2)]

+ [ln(16) + 1/16] - [ln(1) + 1/1]

= [-1 - 0] - [-2 + ln(2)] + [ln(16) + 1/16] - [0 + 1]

= -1 + 2 - ln(2) + ln(16) + 1/16 - 1

= - ln(2) + ln(16) + 1/16

= - ln(2) + ln(2^4) + 1/16

= - ln(2) + 4*ln(2) + 1/16

= 3*ln(2) + 1/16

ANSWER: A = 3*ln(2) + 1/16