A calculus problem by Nishanth Subramanian

As Sambhrant Sachan has done,

$-f(x) = x^3-9x^2+24x-20$

$\begin{aligned} -f(x) & = (x-2)^3-3x^2+12x-12 \\ & = (x-2)^3-3(x-2)^2 \\ & = (x-2)^2(x-2-3) \\ -f(x) & = (x-2)^2 (x-5) \end{aligned}$

Let $I = \displaystyle \int_{2}^{5} (x-2)^{17}(x-5)^{78}(-x^3+9x^2-24x+20) dx$

$-I = \displaystyle \int_{2}^{5} (x-2)^{19}(x-5)^{79} dx$

$I = \displaystyle \int_{2}^{5} (x-2)^{19}(5-x)^{79} dx$

Now, consider the beta function $I_2$ (20,80) $=$ $\int_{0}^{1} (x)^{19}(1-x)^{79} dx$

Substitute in $I_2$ (20,80)
$x=\frac {y-2}{5-2}$ $\Rightarrow$ $dx=\frac {dy}{3}$

$I_2$ (20,80) $3^{99}$ $=$ $\displaystyle \int_{2}^{5} (y-2)^{19}(5-y)^{79} dy$

Now, evaluating the beta function $I_2$ (20,80) , we get $I_2$ (20,80) $=$ $\frac {19! 79!}{99!}$

Therefore the required integral is equal to $\frac{3^{99}}{99!} \times 19! \times 79!$

Pack some food , some water and a mind full of integration , because this solution will be a hell of a ride for you ...

Let start with the polynomial given to us $-f(x) = x^3-9x^2+24x-20$

$\begin{aligned} -f(x) & = (x-2)^3-3x^2+12x-12 \\ & = (x-2)^3-3(x-2)^2 \\ & = (x-2)^2(x-2-3) \\ -f(x) & = (x-2)^2 (x-5) \end{aligned}$

Let $I = \displaystyle \int_{2}^{5} (x-2)^{17}(x-5)^{78}(-x^3+9x^2-24x+20) dx$

$-I = \displaystyle \int_{2}^{5} (x-2)^{19}(x-5)^{79} dx$

Let's take a general case of : $I_{\{n,m\}} = \displaystyle \int \underbrace{(x-a)^{n}}_{1^{st} f^n} \underbrace{(x-b)^{m}}_{2^{nd} f^n} dx$

$\begin{aligned} I_{\{n,m\}} & = (x-a)^n \displaystyle \int (x-b)^{m} dx - \displaystyle \int n(x-a)^{n-1} \left( \displaystyle\int (x-b)^{m} dx\right) dx \\ & = (x-a)^n \dfrac{(x-b)^{m+1}}{m+1} - \dfrac{n}{m+1} \displaystyle \int (x-a)^{n-1}(x-b)^{m+1}dx \\ I_{\{n,m\}} & = (x-a)^n \dfrac{(x-b)^{m+1}}{m+1} - \dfrac{n}{m+1} \times I_{\{n-1,m+1\}} \\ I_{\{n-1,m+1\}} & = (x-a)^{n-1} \dfrac{(x-b)^{m+2}}{m+2} - \dfrac{n-1}{m+2} \times I_{\{n-2,m+2\}} \hspace{10mm} \times \small \left( \dfrac{-(n)}{m+1} \right) \\ I_{\{n-2,m+2\}} & = (x-a)^{n-2} \dfrac{(x-b)^{m+3}}{m+3} - \dfrac{n-2}{m+3} \times I_{\{n-3,m+3\}} \hspace{10mm} \times \small \left( \dfrac{-(n)}{m+1} \cdot \dfrac{-(n-1)}{m+2} \right) \\ I_{\{n-3,m+4\}} & = (x-a)^{n-3} \dfrac{(x-b)^{m+4}}{m+4} - \dfrac{n-3}{m+4} \times I_{\{n-4,m+4\}} \hspace{10mm} \times \small \left( \dfrac{-(n)}{m+1} \cdot \dfrac{-(n-1)}{m+2} \cdot \dfrac{-(n-2)}{m+2} \right) \\ \vdots \\ I_{\{0,m+n\}} & = (x-a)^0 \dfrac{(x-b)^{m+n+1}}{m+n+1} - 0 \hspace{10mm} \times \small \left( \dfrac{-(n)}{m+1} \cdot \dfrac{-(n-1)}{m+2} \cdot \dfrac{-(n-2)}{m+2} \cdots \dfrac{-(1)}{m+n} \right) \end{aligned}$

Add All the equations . We get :

$\begin{aligned} I_{\{n,m\}} & = (x-a)^n \dfrac{(x-b)^{m+1}}{m+1} + (x-a)^{n-1} \dfrac{(x-b)^{m+2}}{m+2} \times \dfrac{-(n)}{m+1} + (x-a)^{n-2} \dfrac{(x-b)^{m+3}}{m+3} \times \dfrac{-(n)}{m+1} \cdot \dfrac{-(n-1)}{m+2} \\ & \cdots + (x-a)^0 \dfrac{(x-b)^{m+n+1}}{m+n+1} \times \dfrac{-(n)}{m+1} \cdot \dfrac{-(n-1)}{m+2} \cdot \dfrac{-(n-2)}{m+2} \cdots \dfrac{-(1)}{m+n} \\ & = \displaystyle \sum_{r=0}^{n} (-1)^{n-r} \dfrac{n!}{r!} \cdot \dfrac{m!}{(m+n-r)!} \cdot \dfrac{(x-a)^{r}(x-b)^{m+n+1-r}}{m+n+1-r} \end{aligned}$

$\boxed{ I_{\{n,m\}} = \dfrac{(x-b)^{m+1} \cdot n! \cdot m!}{(n+m+1)!} \displaystyle \sum_{r=0}^{n} (-1)^{n-r}\dbinom{m+n+1}{r} (x-b)^{n-r}(x-a)^{r}}$

$a= +2 \hspace{2.5mm} ,\hspace{2.5mm} b=+5\hspace{2.5mm} , \hspace{2.5mm} n=19 \hspace{2.5mm} , \hspace{2.5mm} m = 79$

$I_{\{19,79\}} = \dfrac{(x-5)^{80} \cdot 19! \cdot 79!}{99!} \displaystyle \sum_{r=0}^{19} (-1)^{19-r}\dbinom{99}{r} (x-5)^{19-r}(x-2)^{r}$

Putting the limits

$\begin{aligned} I_{\{19,79\}} & = 0 - \dfrac{(3)^{80} \cdot 19! \cdot 79!}{99!} \times \dbinom{99}{0} \times -1 \times -(3)^{19} \\ & = -\dfrac{3^{99}}{99!} \times 19! \times 79! \\ -I & = -\dfrac{3^{99}}{99!} \times 19! \times 79! \\ I & = \dfrac{3^{99}}{99!} \times 19! \times 79! \end{aligned}$

Answer : $3+99+19+79 = \boxed{200}$

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Special Case : Lower limit $= a$ , upper limit $= b$

$\begin{aligned} I_{\{n,m\}} & = 0 - \dfrac{(a-b)^{m+1} \cdot n! \cdot m!}{(n+m+1)!} \times \dbinom{n+m+1}{0} \times (-1)^{n} \times (a-b)^{n} \\ & = (-1)^{n+1}\dfrac{(a-b)^{n+m+1} \cdot n! \cdot m!}{(n+m+1)!} \end{aligned}$