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

Feb 7, 2016

$\frac{74}{3} + \ln \left(\frac{27}{15} ^ 15\right) \approx 6.00421$

Explanation:

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

${\int}_{2}^{4} {x}^{2} - x \ln \left({x}^{2} - 1\right) \mathrm{dx}$

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:

$\int x \ln \left({x}^{2} - 1\right) \mathrm{dx}$

We could try the substitution: $u = {x}^{2} - 1$.
We then have that: $\mathrm{du} = 2 x \mathrm{dx}$

We can then apply the substitution to obtain:

$\int \frac{1}{2} \ln \left(u\right) \mathrm{du} = \frac{1}{2} \int \ln \left(u\right)$

Now to find the integral of $\ln \left(u\right)$ 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 $\ln \left(u\right)$ by $1$ and then take$1$ as our second function:

$= \frac{1}{2} \int 1 \cdot \ln \left(u\right) \mathrm{du}$ Set:

$s = \ln \left(u\right) , s ' = \frac{1}{u}$
$t ' = 1 , t = u$

So now using integration by parts gives us:

$\frac{1}{2} u \ln \left(u\right) - \frac{1}{2} \int 1 \mathrm{du} = \frac{1}{2} \left(u \ln u - u\right) + C$

Now reverse the substitution of $u$ gives us:

$\frac{1}{2} \left(\left({x}^{2} - 1\right) \ln \left({x}^{2} - 1\right) - \left({x}^{2} - 1\right)\right) + C = \frac{1}{2} \left({x}^{2} - 1\right) \left(\ln \left({x}^{2} - 1\right) - 1\right) + C$

So we have effectively shown that:

$\int x \ln \left({x}^{2} - 1\right) \mathrm{dx} = \frac{1}{2} \left({x}^{2} - 1\right) \left(\ln \left({x}^{2} - 1\right) - 1\right) + C$

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

${\int}_{2}^{4} {x}^{2} - x \ln \left({x}^{2} - 1\right) \mathrm{dx}$

Using our integral that we just we found out:

${\left[{x}^{3} / 3 - \frac{1}{2} \left({x}^{2} - 1\right) \left(\ln \left({x}^{2} - 1\right) - 1\right)\right]}_{2}^{4}$

Now putting the limits in:

$\left\{{\left(4\right)}^{3} / 3 - \frac{1}{2} \left({\left(4\right)}^{2} - 1\right) \left(\ln \left({\left(4\right)}^{2} - 1\right) - 1\right)\right\} - \left\{{\left(2\right)}^{3} / 3 - \frac{1}{2} \left({\left(2\right)}^{2} - 1\right) \left(\ln \left({\left(2\right)}^{2} - 1\right) - 1\right)\right\}$

$= \frac{64}{3} - \frac{15}{2} \ln \left(15\right) + \frac{15}{2} - \frac{8}{3} + \frac{3}{2} \ln \left(3\right) - \frac{3}{2}$

Simplifying:

$= \frac{56}{3} + 6 + \frac{3}{2} \ln \left(3\right) - \frac{15}{2} L o g \left(15\right)$

$\frac{74}{3} + \ln \left(\frac{27}{15} ^ 15\right)$

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