definite Integral: int[2, 8] [1/x] dx

[cmnalo]

find the area of the region under the graph of the function on the given interval

f(x) = 1/x dx on [2,8]

A= ∫dx/x on [2,8]

A= ∫d/dx lnx dx on [2,8]

I'm not sure if this is set up right or what to do for my next step.

Answer: 2ln2

 

[galactus]

The integral of 1/x is ln(x). Continue.

 

[cmnalo]

I may have taken a long route but it seems to still work out.

A= ln(8)-ln(2)

A= log of 8 to base e - log of 2 to base e

A= log 8/2 to base e

A= log 4 to base e = ln(4)

Why do I change to 2ln2? Please let me know if there is a faster way to do this.

 

[Unco]

A = ln(8) - ln(2) is perfect - the rest is just dressing.

Why do I change to 2ln2?

Perhaps to practise your logarithm algebra...