JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(I_{n}=\int_{1}^{e} x^{19}(\log |x|)^{n} d x,\) where \(n \in N\). If \((20) I _{10}=\alpha I _{9}+\beta I _{8},\) for natural numbers \(\alpha\) and \(\beta\), then \(\alpha-\beta\) equal to ..... .
- A \(2\)
- B \(1\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
\(I _{ n }=\int_{1}^{ e } x ^{19}(\log | x |)^{ n } d x\) \(I _{ n }=\left|(\log | x |)^{19} \frac{ x ^{20}}{20}\right|_{1}^{e}-\int n (\log | x |)^{ n -1} \cdot \frac{1}{ x } \cdot \frac{ x ^{20}}{20} dx\) \(20 I _{ n }= e ^{20}- nI _{ n -1}\)…
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