How do you evaluate the definite integral #int (2x^-1) dx# from #[1,3.5]#?

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Answer 1 · 2016-12-13T06:42:29

The answer is #2ln3.5=2.51#

We use

#intdx/x=lnx+C#

So, our integral

#int_1^3.5(2x^(-1))dx=int_1^3.5(2dx)/x#

#=[2ln∣x∣] _1^3.5#

#=2(ln3.5-ln1)#

#=2ln3.5-0#

#=2.51#