∫(0,x)tf(x-t)dt=-∫(x,0)(x-u)f(u)du=∫(0,x) (x-u)f(u)du
=x∫(0,x)f(u)du-∫(0,x)uf(u)du
Derivada de sustitución: 3x 2 = xf (x)+∫ (0,x) f (u) du-xf (x).
Entonces: ∫ (0, x) f (u) du = 3x 2.
∫(0,1)f(2x-3)dx =(1/2)∫(0,1)f(2x-3)d(2x-3)=(1/2) ∫(-3,-1)f(u)d(u)
=(1/2)(∫(0,-1)f(u)d(u)-∫(0, -3)f(u)d(u))
=(1/2)(3-27)=-12
2.ln(1+f(x) /sinx) es equivalente a f(x)/sinx, y 2 x-1 es equivalente a xln2.
entonces:3 = lim[f(x)/sinx]/(XL N2)=(1/LN2)LIMF(x)/xsin x)=(1/LN2)LIMF(x)/ x ^ 2)
Límite=3ln2