intergration of e and 1/x

[Bryn]

I've got this question i'm somewhat flumexed with.

 

[math]\int_{e}^{e^2} \left(\frac{5}{x} + e\right) dx[/math]

 

What've i've tried is as follows

 

[math]\int_{e}^{e^2} (5x^{-1} + e) dx = [5ln|x| +e]_{e}^{e^2}[/math]

 

[math] = (5ln(e^2) + e) - (5ln(e) + e)[/math]

 

[math] = (5(2) + e) - (5(1) + e)[/math]

 

[math] = (10 + e) - (5 + e)[/math]

 

[math] = 10 + e - 5 - e[/math]

 

[math] = 5 [/math]

 

which evan i can see from just looking at the graph is wrong, but for the life of my i can't see why. I'm sure i've got everything right after the intergration, but i'm almost just as sure that've i've intergrated it correctly. Any help much appreciated.

[bloodhound]

it seems u forgot to integrate e with respect to x

 

[math]\int_e^{e^2}(\frac{5}{x}+e)dx=[5lnx+ex]_e^{e^2}[/math]

[Bryn]

why does e intergrate to ex? [math]e^x[/math] intergrates to [math]e^x[/math] and as [math]e = e^1[/math] surely the intergral of e is also e?

[bloodhound]

oh, no no no. its absolutely not like that. e is just a constant, just like any other constants, eg pi. etc

 

so when u integrate a constant function, ur just basically finding the area of the rectangle the function makes between the two bounds of the definite integral. for example:

[math]\int_1^{2}3dx=[3x]_1^{2}=6-3=3[/math]

which is the same thing as finding the area of the rectagle with height 3 and length (2-1)=1.

 

 

while e is not a function of anything. e^x is a function. its quite a nice function as the value of its derivative at a point is equal to the value of the function at that point.

 

i am not really good at explaining stuff. maybe someone will help you better

[Dave]

That's basically it. e^x involves a variable x, whereas e to the power of a constant is just that - a constant.

[Bryn]

ah ok i get.