lim x → 1 ∫ 0 ( x - 1 ) 2 t cos t 2 d t ( x - 1 ) sin ( x - 1 )

Applying L'Hospital's Rule, for numerator we have to apply Newton Leibnitz integral rule lim x → 1 2 ( x - 1 ) × ( x - 1 ) 2 cos x - 1 4 - 0 ( x - 1 ) cos ( x - 1 ) + sin ( x - 1 ) 0 0 Divide x - 1 = lim x → 1 2 ( x - 1 ) 2 cos ( x - 1 ) 4 cos ( x - 1 ) + sin ( x - 1 ) ( x - 1 ) = 0 1 + 1 = 0

Mathematics - Definite Integration Question with Solution | TestHub

MathematicsDefinite IntegrationDerivatives (Newton- Leibnitz)Medium2 minPYQ_2020

MathematicsMediumsingle choice

$\lim_{x \to 1} (\frac{\int_{0}^{(x - 1)^{2}} t \cos t^{2} d t}{(x - 1) \sin (x - 1)})$

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Answer:

Solution:

Applying L'Hospital's Rule, for numerator we have to apply Newton Leibnitz integral rule

$\lim_{x \to 1} (\frac{2 (x - 1) \times (x - 1)^{2} \cos {(x - 1)}^{4} - 0}{(x - 1) \cos (x - 1) + \sin (x - 1)}) (\frac{0}{0})$

Divide $(x - 1)$

$= \lim_{x \to 1} \frac{2 (x - 1)^{2} \cos (x - 1)^{4}}{\cos (x - 1) + \frac{\sin (x - 1)}{(x - 1)}}$

$= \frac{0}{1 + 1}$

$= 0$

Stream:JEESubject:MathematicsTopic:Definite IntegrationSubtopic:Derivatives (Newton- Leibnitz)

⏱ 2mℹ️ Source: PYQ_2020

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