Question

A área da região limitada pelo gráfico de y =x² -3x+2, o eixo das abcissas e as retas verticais x=0 e x=2 é:
a) 10/3
b) 8/3
c) 5/3
d) 2/3
e) Nenhuma das anteriores

Answer 1

Boa noite!

Solução!


$A=\displaystyle \int_{0}^{2} x^{2} -3x+2dx\\\\\\\\\\ A= \dfrac{ x^{3} }{3}- \frac{3 x^{2} }{2} +2x~\bigg|_{0} ^{2} \\\\\\\\\\ A= \left(\dfrac{ x^{3} }{3}- \frac{3 x^{2} }{2} +2x\right )-\left( \dfrac{ x^{3} }{3}- \frac{3 x^{2} }{2}+2x \right )\\\\\\\\ A= \left(\dfrac{ (2)^{3} }{3}- \frac{3 (2)^{2} }{2} +2(2)\right )-\left( \dfrac{ (0)^{3} }{3}- \frac{3 (0)^{2} }{2}+2(0) \right )$



$A= \left(\dfrac{ 8 }{3}- \dfrac{3 .4 }{2} +4\right )-(0+0+0) \right )\\\\\\\\\\ A= \left(\dfrac{ 8 }{3}- \dfrac{12}{2} +4\right )-\left( 0+0+0 \right )\\\\\\\\\\ A= \left(\dfrac{ 8 }{3}- 6 +4\right )-\left( 0+0+0 \right )\\\\\\\\\\ A= \left(\dfrac{ 8-18 +12}{3}\right )-(0)\\\\\\\\\\ A= \left(\dfrac{ 20-18 }{3}\right )\\\\\\\ A= \dfrac{2}{3}~u.a$




$\boxed{Resposta:A= \frac{2}{3}~~\boxed{Alternativa~~D}}$


Boa noite!
Bons estudos!