Let I 1 = ∫ 1 2 1 + x 2 1 dx and I 2 = ∫ 1 2 x 1 d x , then

If I 1 = ∫ 1 2 1 + x 2 dx , I 2 = ∫ 1 2 x dx I 1 = ℓ n ( 1 + 2 2 + 5 ) , I 2 = ℓ n 2 ⇒ I 1 < I 2

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationProperties Involving Inequalities (Approximation)Hard2 minPYQ_2011

MathematicsHardsingle choice

Let $I_{1} = \int_{1}^{2} \frac{1}{1 + x ^{2}} dx$ and $I_{2} = \int_{1}^{2} \frac{1}{x} d x$ , then

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Answer:

Solution:

If $I_{1} = \int_{1}^{2} \frac{dx}{1 + x ^{2}}, I_{2} = \int_{1}^{2} \frac{dx}{x}$ $I_{1} = ℓ n (\frac{2 + 5}{1 + 2}), I_{2} = ℓ n 2 \Rightarrow I_{1} < I_{2}$

Stream:BITSATSubject:MathematicsTopic:Definite IntegrationSubtopic:Properties Involving Inequalities (Approximation)

⏱ 2mℹ️ Source: PYQ_2011

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