Question:
If \(I_n = \int _1^e(log_ex)^n\,dx\) where \(n\in Z^+\) and \(I_m+mI_{2026} = e\), then \(m =\)
If \(I_n = \int _1^e(log_ex)^n\,dx\) where \(n\in Z^+\) and \(I_m+mI_{2026} = e\), then \(m =\)
Show Hint
Integrate by parts to get \(I_n=e-nI_{n-1}\).
Updated On: Oct 1, 2026
- \(2024\)
- \(2025\)
- \(2026\)
- \(2027\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
\(I_n=\int_1^e(\log x)^n\,dx\). Use integration by parts with \(u=(\log x)^n\) and \(dv=dx\).
Step 2: Key Formula or Approach
\[ I_n=\left[x(\log x)^n\right]_1^e-n\int_1^e(\log x)^{n-1}dx=e-nI_{n-1} \]
Step 3: Detailed Explanation
So \(I_n+nI_{n-1}=e\).
Put \(n=2027\): \(I_{2027}+2027\,I_{2026}=e\).
Compare with \(I_m+mI_{2026}=e\). Matching terms gives \(m=2027\).
Final Answer:
The value of \(m\) is 2027, option (D). \[ \boxed{2027\ \text{(D)}} \]
\(I_n=\int_1^e(\log x)^n\,dx\). Use integration by parts with \(u=(\log x)^n\) and \(dv=dx\).
Step 2: Key Formula or Approach
\[ I_n=\left[x(\log x)^n\right]_1^e-n\int_1^e(\log x)^{n-1}dx=e-nI_{n-1} \]
Step 3: Detailed Explanation
So \(I_n+nI_{n-1}=e\).
Put \(n=2027\): \(I_{2027}+2027\,I_{2026}=e\).
Compare with \(I_m+mI_{2026}=e\). Matching terms gives \(m=2027\).
Final Answer:
The value of \(m\) is 2027, option (D). \[ \boxed{2027\ \text{(D)}} \]
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