Let f:[−π2,π2]→R be a differentiable function such that f(0)=12, If the limx→0x∫x0f(t)dtex2−1=α,then 8α2 is equal to :

The given limit is of the form 0/0, so we can apply L'Hopital's Rule.

L'Hopital's Rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

So, we differentiate the numerator and the denominator with respect to x.

The derivative of the numerator, using the Fundamental Theorem of Calculus, is x*f(x).

The derivative of the denominator, using the chain rule, is 2x*(e^(x^2)).

So, the limit becomes lim (x→0) [xf(x)] / [2xe^(x^2)].

We can simplify this to lim (x→0) [f(x)] / [2*e^(x^2)].

As x approaches 0, e^(x^2) approaches 1, so the limit simplifies to f(0) / 2.

We know that f(0) = 12, so the limit is 12 / 2 = 6.

Therefore, α = 6.

Finally, we find 8α^2 = 8*6^2 = 288.