Question

If for \[n \geqslant 1\], \[{P_n} = \int\limits_1^e {{{(\log x)}^n}dx} \], then \[{P_{10}} - 90{P_8}\] is equal to
A. \[ - 9\]
B. \[10\]
C. \[ - 9e\]
D. \[10e\]

Answer 1

Hint: We solve this question by assuming the value of \[\log x\] as a variable and then changing the limits and the integration in terms of that variable. Solve the integration by using integration by parts. Then after we get the value of integration in terms of n, we substitute the values for n and solve for \[{P_{10}} - 90{P_8}\].
*In integration by parts we use the sequence of ILATE (Inverse trigonometric, Logarithms , algebraic, trigonometric , exponential) for choosing the value of \[u,v\] and then substitute in the integration formula of by parts integration.
* If we have \[\int {uvdx} \] then the value of integral is given by \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } } \] where \[u' = \dfrac{{du}}{{dx}}\] is differentiation of \[u\] with respect to \[x\].

Complete step-by-step answer:
We have \[{P_n} = \int\limits_1^e {{{(\log x)}^n}dx} \]
Let us assume \[\log x = t\]
Taking exponential function on both sides of the equation we get,
\[
   \Rightarrow {e^{\log x}} = {e^t} \\
   \Rightarrow x = {e^t} \\
 \] {Since, exponential and log function cancel each other}
Now differentiating both sides with respect to their respective variables we get
\[ \Rightarrow dx = {e^t}dt\]
Also, for the limits of integral
\[\because \log x = t\]
When \[x = 1 \Rightarrow \log 1 = 0\], So \[t = 0\]
When \[x = e \Rightarrow \log e = 1\], So \[t = 1\]
Now we will write the integral in terms of variable t as
\[
   \Rightarrow \int\limits_1^e {{{(\log x)}^n}dx} = \int\limits_0^1 {{t^n}.{e^t}dt} \\
   \Rightarrow {P_n} = \int\limits_0^1 {{t^n}.{e^t}dt} \\
 \]
Now we will solve the integration by using integration by parts.
Looking at the integral \[\int\limits_0^1 {{t^n}.{e^t}dt} \]
Using the ILATE sequence we can tell that the term comes first in the sequence is \[u\] and which comes later is \[v\] i.e. in this case we have the terms exponential and algebraic , first comes the algebraic term then comes the exponential term so \[u\] is \[{t^n}\] and \[v\]is \[{e^t}\]
\[u = {t^n},v = {e^t}\]
Now we calculate the differentiation of the term \[u\]
\[
  u' = \dfrac{{du}}{{dt}} \\
  u' = \dfrac{{d({t^n})}}{{dt}} \\
  u' = n{t^{n - 1}} \\
 \]
Now substitute the values in equation \[\int {uvdt = u\int {vdt - \int {u'\left( {\int {vdt} } \right)dt} } } \]
\[ \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = {t^n}\int\limits_0^1 {{e^t}dt} - \int\limits_0^1 {(n{t^{n - 1}})\left( {\int {{e^t}dt} } \right)dt} \] \[...(i)\]
Now we know \[\int {{e^t}dt = {e^t}} \]
Therefore substitute the value of\[\int {{e^t}dt = {e^t}} \] in equation \[(i)\]
\[ \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = \left[ {{t^n}{e^t}} \right]_0^1 - \int\limits_0^1 {(n{t^{n - 1}}).{e^t}dt} \]
Now we can solve and bring out constant terms from the integral sign.
\[ \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = \left[ {{1^n}{e^1} - {0^n}{e^0}} \right] - n\int\limits_0^1 {{t^{n - 1}}.{e^t}dt} \]
Since the value of any number having power zero gives the value one.
\[
   \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = \left[ {1e - 0} \right] - n\int\limits_0^1 {{t^{n - 1}}.{e^t}dt} \\
   \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = e - n\int\limits_0^1 {{t^{n - 1}}.{e^t}dt} \\
 \]
Since, we know \[{P_n} = \int\limits_0^1 {{t^n}.{e^t}dt} \], therefore, replacing n by n-1 we can write\[{P_{n - 1}} = \int\limits_0^1 {{t^{n - 1}}.{e^t}dt} \]. Substituting the value in integration
\[ \Rightarrow \int\limits_0^1 {{t^n}.{e^t}dt} = e - n{P_{n - 1}}\]
So we can write \[{P_n} = e - n{P_{n - 1}}\]
We also find the value of \[{P_{n - 1}} = e - (n - 1){P_{n - 2}}\]
Substitute the value of \[{P_{n - 1}}\] in the value of \[{P_n}\]
\[
   \Rightarrow {P_n} = e - n[e - (n - 1){P_{n - 2}}] \\
   \Rightarrow {P_n} = e - ne + n(n - 1){P_{n - 2}} \\
 \]
Now substitute the values of n as 10
\[
   \Rightarrow {P_{10}} = e - 10e + 10(10 - 1){P_{10 - 2}} \\
   \Rightarrow {P_{10}} = - 9e + 90{P_8} \\
 \]
Shift the value of \[90{P_8}\]to LHS
\[{P_{10}} - 90{P_8} = - 9e\]

So, the correct answer is “Option C”.

Note: Students are likely to make mistakes solving the second part of the integration again by using parts integration which is wrong, keep in mind we have to compare the integration to the value of \[{P_n}\] and then transform that part.