If I 1 = ∫ 0 1 1 - x 50 100 d x and I 2 = ∫ 0 1 1 - x 50 101 d x such that I 2 = α I 1 then α equals to :

Given α = I 2 I 1 = ∫ 0 1 1 - x 50 101 d x ∫ 0 1 1 - x 50 100 d x Now, I 2 = ∫ 0 1 1 - x 50 1 - x 50 100 d x I 2 = I 1 - ∫ 0 1 x . x 49 1 - x 50 100 d x I 2 = I 1 - - x 1 - x 50 101 5050 0 1 - ∫ 0 1 1 - x 50 101 5050 d x I 2 = I 1 - I 2 5050 ⇒ α = I 2 I 1 = 5050 5051

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Mathematics - Indefinite Integration Question with Solution | TestHub

MathematicsIndefinite IntegrationBy partMedium2 minPYQ_2020

MathematicsMediumsingle choice

If $I_{1} = \int_{0}^{1} {(1 - x^{50})}^{100} d x$ and $I_{2} = \int_{0}^{1} {(1 - x^{50})}^{101} d x$ such that $I_{2} = α I_{1}$ then $α$ equals to :

Options:

Answer:

Solution:

Given

$α = \frac{I_{2}}{I_{1}} = \frac{\int_{0}^{1} {(1 - x^{50})}^{101} d x}{\int_{0}^{1} {(1 - x^{50})}^{100} d x}$
Now,

$I_{2} = \int_{0}^{1} (1 - x^{50}) {(1 - x^{50})}^{100} d x$
$I_{2} = I_{1} - \int_{0}^{1} x_{.} x^{49} {(1 - x^{50})}^{100} d x$
$I_{2} = I_{1} - {[\frac{- x {(1 - x^{50})}^{101}}{5050}]}_{0}^{1} - \int_{0}^{1} \frac{{(1 - x^{50})}^{101}}{5050} d x$
$I_{2} = I_{1} - \frac{I_{2}}{5050}$
$\Rightarrow α = \frac{I_{2}}{I_{1}} = \frac{5050}{5051}$

Stream:JEESubject:MathematicsTopic:Indefinite IntegrationSubtopic:By part

⏱ 2mℹ️ Source: PYQ_2020

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