A) -9
B) 10e
C) -9 e
D) 10
Correct Answer: C
Solution :
\[{{P}_{n}}=\int\limits_{1}^{e}{{{(logx)}^{n}}dx}\] put log x = t then \[x={{e}^{t}}\] and \[dx={{e}^{t}}dt\] Also, when x = 1, then t = log 1 = 0 and when x = e, then t = loge e = 1 \[\therefore \]\[{{P}_{n}}=\int\limits_{0}^{1}{{{t}^{n}}.{{e}^{t}}dt}\] \[\therefore \]\[{{P}_{10}}=\int\limits_{0}^{1}{{{t}^{10}}{{e}^{t}}dt}\]and\[{{P}_{8}}=\int\limits_{0}^{1}{{{t}^{8}}{{e}^{t}}dt}\] Now, \[{{P}_{10}}-90{{P}_{8}}=\int\limits_{0}^{1}{\underset{I}{\mathop{{{t}^{10}}}}\,\underset{II}{\mathop{{{e}^{t}}}}\,dt}-90\int\limits_{0}^{1}{{{t}^{8}}{{e}^{t}}dt}\] \[{{P}_{10}}-90{{P}_{8}}\] \[{{P}_{10}}-90{{P}_{8}}=\left[ {{t}^{10}}{{e}^{t}} \right]_{0}^{1}-10\int_{0}^{1}{{{t}^{9}}{{e}^{t}}dt}-90\int_{0}^{1}{{{t}^{8}}{{e}^{t}}dt}\] \[{{P}_{10}}-90{{P}_{8}}=\] \[e-10\left[ {{t}^{9}}\int_{0}^{1}{{{e}^{t}}dt}-\int\limits_{0}^{1}{\frac{d}{dt}({{t}^{9}})\int_{{}}^{{}}{{{e}^{t}}dt}} \right]-90\int\limits_{0}^{1}{{{t}^{8}}{{e}^{t}}}dt\] \[{{P}_{10}}-90{{P}_{8}}=e-10\left[ e-9\int_{0}^{1}{{{e}^{8}}{{e}^{t}}dt} \right]-90\int_{0}^{1}{{{t}^{8}}{{e}^{t}}dt}\] \[{{P}_{10}}-90{{P}_{8}}=e-10e+90\int_{{}}^{{}}{{{t}^{8}}\cancel{{{e}^{t}}}}dt-90\int_{0}^{1}{{{t}^{8}}\cancel{{{e}^{t}}}}dt\] \[\therefore \]\[{{P}_{10}}-90{{P}_{8}}=-9e\]
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