Question

Here's a neat problem.
\[I = \int \limits_2^3\frac{dx}{\sqrt{x^3 - 3x^2 + 5}}\]Evaluate \(\lfloor I + \sqrt 3 \rfloor\).

Answer 1

Apply any mechanical quadrature technique( any numerical method integral rule)

Answer 2

I don't know any cubic tricks and didn't look anything up so I substituted \(x-2=u\) to simplify it down to:

\[I = \int_0^1 \frac{du}{\sqrt{u^3+3u^2+1}} > \int_0^1 \frac{du}{(u+1)^{3/2}}\]
Then I substituted \(u+1=v\)
\[I > \int_1^2 v^{-3/2}dv = 1-\frac{1}{\sqrt{2}}\]

I'm thinking from here there might be some trick with this:
\[\int_0^1 \frac{du}{\sqrt{(u^3+3u^2+3u+1) - 3u}} \]

so that we can do it. Of course since I'm not dumb, I will simply tell you the answer geometrically since it's 100% rigorous and reliable.

\[\lfloor I + \sqrt{3} \rfloor= \left\lfloor \int_2^3 \frac{1}{\sqrt{3x^3-3x^2+5}} + \sqrt{3} dx \right\rfloor \]

https://www.desmos.com/calculator/c0hqcrbp6k

therefore the answer is 2 :P

I'd like to see "the real way" though.

Answer 3

Let \(f(x)=\dfrac{1}{\sqrt{x^3-3x^2+5}}\). For \(x\in[2,3]\) you can show that \(f'(x)\le0\), so \(f(x)\) is decreasing.

From that we have the fact that \(\dfrac{1}{\sqrt{x^3-3x^2+5}}\le1\) for \(x\in[2,3]\), since \(f(2)=1\) and \(f(3)=\dfrac{1}{\sqrt5}\). This means
\[\int_2^3f(x)\,\mathrm dx\le\int_2^3\,\mathrm dx=1\implies\lfloor I\rfloor=0\implies\lfloor I+\sqrt3\rfloor=\lfloor\sqrt3\rfloor=2\]

Answer 4

Actually, that last implication isn't true, and \(\lfloor\sqrt3\rfloor\neq2\)...

Answer 5

Hmm, I suppose you have by the mean value theorem that there's some \(2\le c\le3\) such that
\[f(c)=\frac{1}{3-2}\int_2^3f(x)\,\mathrm dx=I\]which means \(f(3)<I<f(2)\), or \(0.4472\ldots<I<1\).

From there, you have \(2.1792\ldots<I+\sqrt3<2.7320\ldots\) (assuming using a numerical approximation for \(\sqrt3\) is considered valid) and so \(\lfloor I+\sqrt3\rfloor=2\).

Answer 6

Yup, that last post is the solution.