Question

Let ${\mathrm{I}}_{\mathrm{n}}={\int }_1^{\mathrm{e}}{\mathrm{x}}^{19}{\left(\log \right|\mathrm{x}\left|\right)}^{\mathrm{n}}\mathrm{dx}$, where $\mathrm{n}\in \mathrm{N}$. If $20\left({\mathrm{I}}_{10}\right)={\mathrm{\alpha I}}_9+{\mathrm{\beta I}}_8$, for natural numbers $\mathrm{\alpha }\mathrm{and}\mathrm{\beta }$, then $\mathrm{\alpha }–\mathrm{\beta }$ equal to ………………… .

Answer 1

Step 1: Given data

Given, ${\mathrm{I}}_{\mathrm{n}}={\int }_1^{\mathrm{e}}{\mathrm{x}}^{19}{\left(\log \right|\mathrm{x}\left|\right)}^{\mathrm{n}}\mathrm{dx}$.

Also, $20\left({\mathrm{I}}_{10}\right)={\mathrm{\alpha I}}_9+{\mathrm{\beta I}}_8\left(\mathrm{P}\right)$

Step 2: Finding the value of $\mathrm{\alpha }–\mathrm{\beta }$

Using ILATE, we can choose logarithmic function as first function and algebraic as second function. Now according to ILATE rule:

$$\begin{gathered} \int \mathrm{u}.\mathrm{v}d\mathrm{x}=\mathrm{u}\int \mathrm{v}d\mathrm{x}-\int [\frac{d\mathrm{u}}{d\mathrm{x}}\int \mathrm{v}d\mathrm{x}]d\mathrm{x} \\ \Rightarrow {\mathrm{I}}_{\mathrm{n}}=[\log {\left(\mathrm{x}\right)}^{\mathrm{n}}{\int }_1^{\mathrm{e}}{\mathrm{x}}^{19}d\mathrm{x}]-{\int }_1^{\mathrm{e}}[\mathrm{nlog}{\left(\mathrm{x}\right)}^{\mathrm{n}-1}\frac{1}{\mathrm{x}}\int {\mathrm{x}}^{19}]d\mathrm{x} \\ \Rightarrow {\mathrm{I}}_{\mathrm{n}}=\left[\log {\left(\mathrm{e}\right)}^{\mathrm{n}}\frac{{\mathrm{e}}^{20}}{20}-\log {\left(1\right)}^{\mathrm{n}}\frac{1}{20}\right]-{\int }_1^{\mathrm{e}}\mathrm{nlog}{\left(\mathrm{x}\right)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{20}}{20\mathrm{x}}d\mathrm{x} \\ \Rightarrow {\mathrm{I}}_{\mathrm{n}}=\frac{{\mathrm{e}}^{20}}{20}-\frac{\mathrm{n}}{20}{\int }_1^{\mathrm{e}}{\mathrm{x}}^{19}\log {\left(\mathrm{x}\right)}^{\mathrm{n}-1}d\mathrm{x} \\ \Rightarrow {\mathrm{I}}_{\mathrm{n}}=\frac{{\mathrm{e}}^{20}}{20}-\frac{\mathrm{n}}{20}{\mathrm{I}}_{\mathrm{n}-1} \\ \Rightarrow 20{\mathrm{I}}_{\mathrm{n}}={\mathrm{e}}^{20}-{\mathrm{nI}}_{\mathrm{n}-1}\left(1\right) \end{gathered}$$

Now put $\mathrm{n}=10$ in equation $\left(1\right)$ we get,

$20{\mathrm{I}}_{10}={\mathrm{e}}^{20}-10{\mathrm{I}}_9\left(\mathrm{a}\right)$

Put $\mathrm{n}=9$in equation $\left(1\right)$ we get,

$20{\mathrm{I}}_9={\mathrm{e}}^{20}-9{\mathrm{I}}_{10}\left(\mathrm{b}\right)$

Subtract equation $\left(\mathrm{b}\right)$ from equation $\left(\mathrm{a}\right)$ we get,

$20{\mathrm{I}}_{10}=10{\mathrm{I}}_9+9{\mathrm{I}}_8\left(2\right)$

Now comparing equations $\left(\mathrm{P}\right)$ and $\left(2\right)$ we get:

$$\begin{gathered} \mathrm{\alpha }=10,\mathrm{\beta }=9 \\ \therefore \mathrm{\alpha }-\mathrm{\beta }=10-9 \\ \mathrm{\alpha }-\mathrm{\beta }=1 \end{gathered}$$

Therefore, $\mathrm{\alpha }-\mathrm{\beta }$is equal to $1$.