Question

O mână de ajutor, vă rog ..?

Answer 1

$f(x) = \dfrac{2x+3}{x^2+3x+3} \\ \\ \int\limits^2_1 {(x^2+3x+3)\cdot f(x)} \, dx = \int\limits^2_1 {(x^2+3x+3)\cdot \dfrac{2x+3}{x^2+3x+3} } \, dx= \\ \\ = \int\limits^2_1 {(2x+3)} \, dx = \Big(2\cdot\dfrac{x^2}{2}+ 3x\Big)\Big|_1^2= \Big(x^2+ 3x\Big)\Big|_1^2 = \\ \\ =2^2+3\cdot 2-(1^2+3\cdot 1) = 4+6-1-3 = 6$

Answer 2

dac efectuezi calculul SIMPLIFICAND cu x²+3x+3>0,  vezi ca sub integrala ramane doar (2x+3)dx
deci
∫(2x+3)dx, de la 1 la 2 . este
x²+3x de la 1 la 2=2²+3*2-(1²+3*1)=4+6-(1+3)=4+6-4=6