Let $$\mathop {\lim }\limits_{ \in \to 0 + } \int\limits_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1,\left( {0 < x < {\pi \over 4}} \right)} $$. Then a and b are given by

Let $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$. Then $$f'\left( {{\pi \over 4}} \right)$$ equals

If I is the greatest of $${I_1} = \int\limits_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} $$, $${I_2} = \int\limits_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} $$, $${I_3} = \int\limits_0^1 {{e^{ - {x^2}}}dx} $$, $${I_4} = \int\limits_0^1 {{e^{ - {x^2}/2}}dx} $$, then