Question

设f(x)=∫(上限x^2,下限x)[xt+x^2*e^(-t^2)]dt,求f'(x)

Answer 1

f(x) = ∫(x->x²) [xt + x²e^(-t²)] dt

= x∫(x->x²) t dt + x²∫(x->x²) e^(-t²) dt

f'(x) = x(x²*2x - x*1) + x²[e^(-x⁴)*2x - e^(-x²)*1]

= x(2x³-x) + x²[2xe^(-x⁴) - e^(-x²)]

= 2x⁴ - x² + 2x³e^(-x⁴) - x²e^(-x²)