Definite Integration Question 332
Question: If $ I _{m}=\int_1^{x}{{{(\log x)}^{m}}dx} $ satisfies the relation $ I _{m}=k-l{I _{m-1}}, $ then
Options:
A) $ k=e $
B) $ l=m $
C) $ k=\frac{1}{e} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I _{m}=\int_1^{x}{{{(\log x)}^{m}}dx}=({{(\log x)}^{m}}.x)_1^{x}-\int_1^{x}{m{{(\log x)}^{m-1}}.\frac{1}{x}xdx} $
$ ={{(\log x)}^{m}}.x-m{I _{m-1}} $
$ \therefore $ $ I _{m}=k-l{I _{m-1}}\Rightarrow k-l{I _{m-1}}=x{{(\log x)}^{m}}-m{I _{m-1}} $
therefore $ k=x{{(\log x)}^{m}},l=m $ .
BETA
