Evaluate : ∫ 0 1 2 − x 2 dx .

= ∫ 2 − x 2 dx = sin − 1 2 x + c So ∫ 0 1 2 − x 2 dx = sin − 1 2 x 0 1 sin − 1 ( 2 1 ) + c − sin − 1 ( 0 ) − c = 4 π − 0 = 4 π

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDefinite Integration by SubstitutionMedium2 minPYQ_2009

MathematicsMediumsingle choice

Evaluate : $\int_{0}^{1} \frac{dx}{2 - x ^{2}}$ .

Options:

Answer:

Solution:

$= \int \frac{dx}{2 - x ^{2}} = sin^{- 1} \frac{x}{2} + c So \int_{0}^{1} \frac{dx}{2 - x ^{2}} = sin^{- 1} \frac{x}{2}_{0}^{1} sin^{- 1} (\frac{1}{2}) + c - sin^{- 1} (0) - c = \frac{π}{4} - 0 = \frac{π}{4}$

Stream:BITSATSubject:MathematicsTopic:Definite IntegrationSubtopic:Definite Integration by Substitution

⏱ 2mℹ️ Source: PYQ_2009

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