Question

aria cuprinsa intre graficul functiei f: x^3/1+x^4 si dreptele y=0, x=1, x=-1

Answer 1

[tex]\text{Avand in vedere ca e vorba despre o arie rezultatul nu poate fi 0.}\\ (\text{cu toate ca asa ar fi dat daca calculam direct.) }\\ \text{Fiind o functie impara,putem trage concluzia ca }\\ \displaystyle \int_{-1}^1 f(x)dx=2\int_0^1 f(x)dx\\ \text{Prin urmare :}\\ A=2\int _{0}^{1} \dfrac{x^3}{1+x^4}dx= \dfrac{1}{2} \int_0^1 \dfrac{4x^3}{1+x^4}dx=\dfrac{1}{2} \ln(1+x^4)|_0^1=calcule =\\ =\dfrac{1}{2}\ln 2=\boxed{\ln \sqrt 2} [/tex]