Let f(x)= 1/x and g(x) = 1/(x^2). Find the area of the region between the graphs of f and g on the interval?

Question

[1/2,16]. Give both exact(symbolic) and approx. (numeric) answers.

Answer 1

okay you have your limits so...
we call it fun calculus, for fundamental theroem 1.

This is an ABC problem

Integral of (top - bottom)/(b-a)
a=1\2 b=16

Graph it. whichever graph is on the top goes first, graph on bottom 2nd.
with me?

so in this case it'll be
integral (1\x ) - (1\x^2) dx
let dx in
call this "integral.." |

|(1\x dx) - (1\x^2 dx)

get the integral. why does that look so hard to me.. hmm.
anyway, once u have the integral, plug in the bounds. upper bound first.
subtract.
tada!

or u could put that into the calculator somehow..
2nd>calc>7>enter the bounds>enter
Theres your answer

this is hurting my brain bc i want to help but..

Answer 2

You should have a graphic calculator for this problem. First of all, draw both of the graphics using calculator then find the upper function and the lower function between the interval.

the intersection of the graphic will be (1,1), then for X ∈ (1/2 , 1) the 1/(x^2) will be the upper function and 1/x will be the lower function.For X ∈(1,16), the 1/x will be the upper function and 1/(x^2) will be the lower function.

Next, use this formula to count the area between the graphs:

Area = ∫ (upper function-lower function) and integrate from the lower interval and upper interval.

So ...

Area = ∫ (1/(x^2) - 1/x) dx evaluate from 1/2 to 1 + ∫ (1/x - 1/x^2) dx evaluate from 1 to 16

The answer will be

Area = 3ln(2) + 1/16
= 2.14194

Answer 3

= integ .5 to 16[1/x^2] - integ .5 to 16[1/x]
= 1/x .5 to 16 - lnx .5 to 16
= 1/16-1/.5 -(ln16 -ln.5)
= 1/16 -2 +ln.5-ln16 = -1 15/16 + ln (16/.5)
= approx = 1.532