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JEE MainMathematicsDefinite Integration

Let f(x) = _ n=0 ^ ( - ^2 x 2 )^n . If I = _ - /2 ^ /2 f(x) 1 + 2024^ ^3 x dx , then the value of 24 I^2 ^2 is

Correct answer

4

Step-by-step solution

First, we sum the infinite geometric series for f(x) . The common ratio is r = - ^2 x 2 , which satisfies |r| f(x) = 1 1 - (- ^2 x 2 ) = 1 1 + ^2 x 2 = 2 2 + ^2 x Now, substitute f(x) into the integral: I = _ - /2 ^ /2 2 (2 + ^2 x)(1 + 2024^ ^3 x ) dx Using the property _ -a ^a g(x) dx = ₀^a (g(x) + g(-x)) dx , we observe that ^3(-x) = - ^3 x and ^2(-x) = ^2 x . g(x) + g(-x) = 2 2 + ^2 x ( 1 1 + 2024^ ^3 x + 1 1 + 2024^ - ^3 x ) Since 1 1 + a^y + 1 1 + a^ -y = 1 , we get: g(x) + g(-x) = 2 2 + ^2 x So, I = ₀^ /2 2 2

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