Question
Question: If for \[n \geqslant 1\], \[{P_n} = \int\limits_1^e {{{(\log x)}^n}dx} \], then \[{P_{10}} - 90{P_8}...
If for n⩾1, Pn=1∫e(logx)ndx, then P10−90P8 is equal to
A. −9
B. 10
C. −9e
D. 10e
Solution
We solve this question by assuming the value of logx 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 P10−90P8.
*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 ∫uvdx then the value of integral is given by ∫uvdx=u∫vdx−∫u′(∫vdx)dx where u′=dxdu is differentiation of u with respect to x.
Complete step-by-step answer:
We have Pn=1∫e(logx)ndx
Let us assume logx=t
Taking exponential function on both sides of the equation we get,
\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} \\
u' = \dfrac{{du}}{{dt}} \\
u' = \dfrac{{d({t^n})}}{{dt}} \\
u' = n{t^{n - 1}} \\
\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} \\
\Rightarrow {P_n} = e - n[e - (n - 1){P_{n - 2}}] \\
\Rightarrow {P_n} = e - ne + n(n - 1){P_{n - 2}} \\
\Rightarrow {P_{10}} = e - 10e + 10(10 - 1){P_{10 - 2}} \\
\Rightarrow {P_{10}} = - 9e + 90{P_8} \\