Question

A figura representada o grafico dada a função:
f(x)= x^{2} - 2x.
A area compreendida entre o grafico de uma função e o eixo dos x é dada por
\int\limits^b_a f({x}) \, dx
Assim a area sombreada da figura é igual a:
y= x^{ 2} - 2 x

alternativas:

a) 12
b) \frac{4}{3}
c) 4
d) \frac{2}{3}

058c63283524fffe4315aaf123a98264.pdf

Answer 1

Boa tarde Pirata!

Solução!

$\displaystyle \int_{0}^{2} } (x^{2} -2x)dx\\\\\\ \displaystyle \int_{0}^{2} } \frac{(x^{2+1}) }{2+1} - \frac{(2x^{1+1}) }{1+1} )dx\\\\\\ I= \frac{x^{3} }{3}- x^{2} \Bigg|_{0}^{2} }= \left ( \frac{ x^{3}}{3}-2 \right )-\left ( \frac{ x^{3} }{3}- 2}\right )\\\\\\\\\ I= \frac{x^{3} }{3}- x^{2} \Bigg|_{0}^{2} }= \left ( \frac{ 2^{3}}{3}- 2\right )-\left ( \frac{ 0^{3} }{3}-2 \right )$



$I= \frac{x^{3} }{3}- x^{2} \Bigg|_{0}^{2} }= \left ( \dfrac{ 8}{3}-2 \right )-\left ( \dfrac{ 0 }{3}-2 \right )\\\\\\\\\ I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 8-6}{3} \right )-\left ( \dfrac{ 0 -6}{3} \right )\\\\\\\\\ I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 2}{3} \right )+\left ( \dfrac{ 0+6}{3} \right )\\\\\\\\\ I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 2}{3} \right )+\left ( \dfrac{ 6}{3} \right )$



$I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 2}{3} \right )+\left ( 2 \right )\\\\\\\\\ I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 2+2}{3} \right )\\\\\\ I= \dfrac{ x^{3} }{3}- x^{2} \Bigg|_{0}^{2} =\left ( \dfrac{ 4}{3} \right )$



$Resposta:\displaystyle \int_{0}^{2} } (x^{2} -2x)dx\Rightarrow Area= \frac{4}{3}~~Alternativa~~B}}$

Boa tarde!
Bons estudos!