Question

Please help me with this calc question?

Answer 1

Have deleted answer due to an error.

Answer 2

`color(blue)[int_a^(2a)1/x*dx=int_(3a)^(6a)1/x*dx=0.693]`

Explanation:

The first integral

`int_a^(2a)1/x*dx=[lnx]_a^(2a)=ln2a-lna=ln2+lna-lna=ln2=0.693`

The second integral

`int_(3a)^(6a)1/x*dx=[lnx]_(3a)^(6a)=ln6a-ln3a`

`=ln6+lna-ln3-lna=ln6-ln3=1.792-1.099=0.693`

Note that the two integrals have same value

`color(blue)[int_a^(2a)1/x*dx=int_(3a)^(6a)1/x*dx=0.693]`

note that

`color(red)[ln(x*y)=lnx+lny]`

Answer 3

The two integrals are identical.

Explanation:

We seek the greater of:

`I_1 = int_a^(2a) \ 1/x \ dx` or `I_2 = int_(3a)^(6a) \ 1/x \ dx`

Where `a gt 0`. Consider the general case:

`I(alpha,beta) = int_(alpha)^(beta) \ 1/x \ dx`

Where `alpha,beta gt 0`, Which is a standard integral, so we can directly integrate:

`I(alpha,beta) = [ \ ln |x| \ ]_(alpha)^(beta)`

And as `alpha,beta gt 0`, we can additionally use the properties of logarithms to write:

`I(alpha,beta) = ln beta - ln alpha = ln (beta/alpha)`

This integral is in fact used to define the Napier logarithm and the unexpected ratio is an alternative proof of the logarithm of a product property! Given this result we now conclude that

`I_1 = ln((2a)/a) = ln 2`
`I_2 = ln ((6a)/(3a)) =ln 2`

Making the two integrals identical