Let f : R $\to$ R be a continuous function. Then $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ is equal to :
Solution
$$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}f({\sec ^2}x){\sec ^3}x.{{\sin x} \over x}$$<br><br>= $${\pi \over 4}f(2).{\left( {\sqrt 2 } \right)^3}.{1 \over {\sqrt 2 }} \times {4 \over \pi }$$<br><br>= 2f (2)
About this question
Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals
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