Please help me with this calc question?

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3 Answers
May 21, 2018

Have deleted answer due to an error.

May 21, 2018

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]

May 21, 2018

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