JEE Mains · Maths · STD 12 - 7.2 definite integral
If for \(n \geq 1\) , \({P_n} = \int\limits_1^e {{{\left( {\log \,x} \right)}^n}\,dx} \) , then \(P_{10} - 90P_8\) is equal to
- A \(-9\)
- B \(10\,e\)
- C \(-9\,e\)
- D \(10\)
Answer & Solution
Correct Answer
(C) \(-9\,e\)
Step-by-step Solution
Detailed explanation
\({P_n} = \int\limits_1^e {{{\left( {\log x} \right)}^n}dx} \) put \(\log x=t\) then \(x=e^{t}\) and \(d x=e^{t} d t\) A bo, when \(x=1,\) then \(t=\log 1=0\) and when \(x=e,\) then \(t=\log _{e} e=1\) \(\therefore {P_n} = \int\limits_0^1 {{t^n}.{e^t}dt} \)…
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