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