What is the net area between #f(x) = x^2-xln(x^2-1) # and the x-axis over #x in [2, 4 ]#?

Question

Paisley Johnson · Accepted Answer

#74/3+ln(27/15^15)~~6.00421#

To find the area we must integrate the function then apply the limits to the integral:

#int_2^4x^2-xln(x^2-1)dx#

Integrating the first term is straightforward but I suspect people may have trouble integrating the second term so let's look in more detail how we do that:

#intxln(x^2-1)dx#

We could try the substitution: #u=x^2-1#. We then have that: #du=2x dx#

We can then apply the substitution to obtain:

#int1/2ln(u)du=1/2intln(u)#

Now to find the integral of #ln(u)# you can either read the information from a table of standard integrals or use integration by parts. To use integration by parts: although it is only one function the trick is to multiply #ln(u)# by #1# and then take#1# as our second function:

#=1/2int1*ln(u)du# Set:

#s = ln(u), s' =1/u# #t'=1, t=u#

So now using integration by parts gives us:

#1/2u ln(u)-1/2int1 du=1/2(u lnu - u)+C#

Now reverse the substitution of #u# gives us:

#1/2((x^2-1) ln (x^2-1) -(x^2-1))+C=1/2(x^2-1)(ln(x^2-1)-1)+C#

So we have effectively shown that:

#intxln(x^2-1)dx =1/2(x^2-1)(ln(x^2-1)-1)+C #

We can now proceed with the original question stated, that is:

#int_2^4x^2-xln(x^2-1)dx#

Using our integral that we just we found out:

#[x^3/3-1/2(x^2-1)(ln(x^2-1)-1)]_2^4#

Now putting the limits in:

#{(4)^3/3-1/2((4)^2-1)(ln((4)^2-1)-1)}-{(2)^3/3-1/2((2)^2-1)(ln((2)^2-1)-1)}#

#=64/3-15/2ln(15)+15/2-8/3+3/2ln(3)-3/2#

Simplifying:

#=56/3+6+3/2ln(3) - 15/2Log(15)#

#74/3+ln(27/15^15)#

You can either leave this exact value or use a calculator to approximate the solution to #6.00421#.