Integration

[jlbowles]

Let R be the shaded region by the graph of y = lnx and the line y = x -2, as shown above.

 

a) find the area of R

[Klaynos]

What does integration do?

[5614]

I'm assuming you're familiar with integration, but just not how to do this specific problem.

 

Firstly if you can picture the graphs it will help:

 

You need to work out the points where the two graphs intercept. That will give you the upper and lower bounds. Set: y=ln[x]=x-2 and solve for x. But err, I can't see how to solve that myself.

 

ln[x] = x-2

e^ln[x] = x = e^x-2 = e^xe^-2

xe^-x = e^-2

but then what?

 

if I use my calculator to do it for me then I get: x=0.1586 ; x=3.146.

 

Then you want to integrate for one function, and then subtract the integral of the other. So you start off with the entire area under the higher graph (the ln[x] one) and then subtract the area under the lower graph (the x-2 one) to leave just the area between them.

 

[edit] oh, and to integrate ln[x] what you do is the integral, by parts, of 1*ln[x]. So you integrate the 1 and differentiate the ln[x], as per the "by parts" method of integration.

 

[edit2] also you're gonna have to be careful about where the function crosses the y-axis, else you might get 0 area (where "positive" area cancels with "negative" area). Do you know what I mean? Like the integral of (x-2) from 0 to 2 is -2, and the integral from 2 to 3 is 1/2. The total area is 2+½, not ½-2, because area has to be positive - hopefully you know what I mean! So maybe if you do the problem in 2 parts, firstly for the positive y values, and then for negative.

 

[edit3] ok, I did the question and got a correct looking answer. If you want the solution I'll post it, but if you want to do it yourself (it's good practice, seriously!) then use the two intercept points I gave you, because I still can't see how to get them without a calculator, and do what I said in [edit2], treating +y and -y as two seperate areas, because under the y-axis will give you a "negative" area.

[Country Boy]

I'm assuming you're familiar with integration, but just not how to do this specific problem.

 

Firstly if you can picture the graphs it will help:

 

You need to work out the points where the two graphs intercept. That will give you the upper and lower bounds. Set: y=ln[x]=x-2 and solve for x. But err, I can't see how to solve that myself.

 

ln[x] = x-2

e^ln[x] = x = e^x-2 = e^xe^-2

xe^-x = e^-2

but then what?

 

if I use my calculator to do it for me then I get: x=0.1586 ; x=3.146.

 

Then you want to integrate for one function, and then subtract the integral of the other. So you start off with the entire area under the higher graph (the ln[x] one) and then subtract the area under the lower graph (the x-2 one) to leave just the area between them.

I would recommend, instead, subtracting the two functions and integrating ln(x)- (x-2)= ln(x)- x+ 2. That way you don't have to worry about where the functions are positive or negative. Within the limits of integration ln(x)- x+ 2 is always positive.

 

[edit] oh, and to integrate ln[x] what you do is the integral, by parts, of 1*ln[x]. So you integrate the 1 and differentiate the ln[x], as per the "by parts" method of integration.

 

[edit2] also you're gonna have to be careful about where the function crosses the y-axis, else you might get 0 area (where "positive" area cancels with "negative" area). Do you know what I mean? Like the integral of (x-2) from 0 to 2 is -2, and the integral from 2 to 3 is 1/2. The total area is 2+½, not ½-2, because area has to be positive - hopefully you know what I mean! So maybe if you do the problem in 2 parts, firstly for the positive y values, and then for negative.

 

[edit3] ok, I did the question and got a correct looking answer. If you want the solution I'll post it, but if you want to do it yourself (it's good practice, seriously!) then use the two intercept points I gave you, because I still can't see how to get them without a calculator, and do what I said in [edit2], treating +y and -y as two seperate areas, because under the y-axis will give you a "negative" area.

[5614]

Is it possible to solve ln[x]=x-2 directly?

[Country Boy]

Not in terms of elementary functions. It might be possible to solve in terms of the "Lambert w-function" which is defined as the inverse function to f(x)= xe^x.