Question

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

Answer 1

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

Explanation:

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`.