Find the value $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ Let $I = \int\limits_2^3 {\frac{{dx}}{{\sqrt {{x^3} - 3{x^2} + 5} }}} $ find the value of $\left[ {I + \sqrt 3 } \right] $ {where [.] represent greatest integral function}
Let $T(x)={{x^3} - 3{x^2} + 5}$, $T'\left( x \right) = {x^3} - 3{x^2} + 5 = 3{x^2} - 6x = 3x\left( {x - 2} \right) > 0,x \in \left( {2,3} \right)$
$T(x)$ is increasing for $x\in(2,3)$
Not able to proceed further
 A: Built around $x=\frac 52$ the series expansion of the integrand is
$$2 \sqrt{\frac{2}{15}}-2 \sqrt{\frac{2}{15}} \left(x-\frac{5}{2}\right)+\frac{1}{5}
   \sqrt{\frac{6}{5}} \left(x-\frac{5}{2}\right)^2+\frac{1}{3} \sqrt{\frac{10}{3}}
   \left(x-\frac{5}{2}\right)^3-\frac{111}{50} \sqrt{\frac{3}{10}}
   \left(x-\frac{5}{2}\right)^4+O\left(\left(x-\frac{5}{2}\right)^5\right)$$
Integrate (some terms will disappear because of the symmetry) and you have a very good approximation.
A: Well, I have an idea to compute that integral. Let,
$$I(t)=\int_{2}^{3}\sqrt{x^3-3x^2+5t} dx$$
Notice that $I'(1)$ gives the required integral but a constant term in front.
Now we are only required to evaluate $I(t)$.
Note that for all $t\in\mathbb{N}$ and $x\in [2,3]$, $5t>x^{3}-3x^{2} \implies \frac{x^3-3x^2}{5t} <1$.
$$I(t)=\sqrt{5t}\int_{2}^{3}\sqrt{1+\frac{x^3-3x^2}{5t}} dx$$
Since, $\frac{x^3-3x^2}{5t} <1$ we can use fractional binomial theorem
Now at last you will be left with a summation, whose bounds are easily predictable by adjusting the sum to another well known convergent sums. This easily enables you to calculate your required floor value.
