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JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Indefinite Integrals

Let

I_{n} = \int_{1}^{e} x^{19} (lo g ∣ x ∣)^{n} d x

, where

n \in N

. If (20)

l_{10} = α l_{9} + β l_{8}

, for natural numbers

α

and

β

then

α - β

equals to ______.

Answer: 1

Solution

Integrating by parts with

x^{19}

as the factor to be integrated,

I_{n} = [\frac{x ^{20}}{20} (lo g x)^{n}]_{1}^{e} - \frac{n}{20} \int_{1}^{e} x^{19} (lo g x)^{n - 1} d x = \frac{e ^{20}}{20} - \frac{n}{20} I_{n - 1}

so

20 I_{n} = e^{20} - n I_{n - 1}

for every

n \in N

. Taking

n = 10

and

n = 9

,

20 I_{10} = e^{20} - 10 I_{9}, e^{20} = 20 I_{9} + 9 I_{8}

Eliminating

e^{20}

between them,

20 I_{10} = 20 I_{9} + 9 I_{8} - 10 I_{9} = 10 I_{9} + 9 I_{8}

so

α = 10

,

β = 9

and

α - β = 1

.

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Integer answer

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