Question

Calculo 1.
Calcular Integral!!!

Answer 1

Olá, segue a resolução:

$\int\limits^2_1 {(3x^2+6x+1)} \, dx$

$=[\frac{3x^3}{3}+\frac{6x^2}{2}+x]|\limits^{2}_{1}$

$=[x^3+3x^2+x]|\limits^{2}_{1}$

$=[2^3+3(2)^2+2]-[1^3+3(1)^2+1]$

$=[8+12+2]-[1+3+1]$

$=22-5$

$=17$