Let, y=x→0limex2−1x∫0xf(t)dt
⇒y=x→0lim(x2ex2−1)×x2x∫0xf(t)dt
We know that, x→0limx2ex2−1=1
⇒y=x→0limx∫0xf(t)dt
Applying L-hospital's rule and Newton Leibnitz Theorem we get,
⇒y=x→0lim1f(x)
⇒y=f(0)
It is given that, f(0)=21
⇒y=21
⇒α=21
⇒8α2=2