Question

Integrala de la 0 la 1 din x la puterea 2020+ x la puterea 2019 + x la puterea 2 + x totul supra x+1​

Answer 1

Răspuns:

Explicație pas cu pas:

Transformăm funcția de sub integrală...

$Fie f(x)=\dfrac{x^{2020}+x^{2019}+x^{2}+x}{x+1}=\dfrac{x^{2019}(x+1)+x(x+1)}{x+1}=\dfrac{(x+1)(x^{2019}+x)}{x+1}.~~\\Deci~f(x)=x^{2019} +x.\\\int\limits^1_0 {f(x)} \, dx=\int\limits^1_0 {(x^{2019} +x)} \, dx=(\dfrac{x^{2020}}{2020}+\dfrac{x^{2}}{2})|_{0}^{1}=(\dfrac{1}{2020}+\dfrac{1}{2})-(\dfrac{0}{2020}+\dfrac{0}{2})=\dfrac{1011}{2020}$