Question

Se considera f: R->R , f(x)=x^4+x+1 . Aratati ca integrala de la 1 la e din ( f(x) - x^4 - 1)lnx dx = (e^2+1)/4

Answer 1

f(x) = x⁴ + x + 1

f(x) - x⁴ -1 = x⁴ + x + 1 - x⁴ -1 = x

Integrala devine :

$I=\int^e _1 xlnx\ dx$

Folosim inegrarea prin părți:

[tex]\it \int fg' = fg - \int f'g \\\;\\ f=lnx \Rightarrow f'=\dfrac{1}{x} \\\;\\ g'=x \Rightarrow g=\dfrac{x^2}{2}[/tex]

[tex]\it I = \dfrac{x^2}{2}lnx \Big{|}^e _1 - \int^e _1 \dfrac{1}{x}\cdot \dfrac{x^2}{2} \ dx = \dfrac{x^2lnx}{2} \Big{|}^e _1 - \int^e _1 \dfrac{x}{2}\ dx = \\\;\\ \\\;\\ = \dfrac{x^2lnx}{2} \Big{|}^e _1 - \dfrac{x^2}{4}\Big{|}^e _1 = \dfrac{e^2\cdot1}{2} -\dfrac{1\cdot 0}{2} -\dfrac{e^2}{4} +\dfrac{1}{4} = \dfrac{e^2}{2} -\dfrac{e^4}{4} +\dfrac{1}{4} = [/tex]


[tex]\it = \dfrac{e^2}{4} +\dfrac{1}{4} = \dfrac{e^2+1}{4}. [/tex]