Question

Integral definida:
Calcule:
b. A área sob o gráfico f nos intervalos dados

f(x) = x^2-6x+5 em [0 ,10]
Resposta: 323/3

Answer 1

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Solução!

$A=\displaystyle\int_{0}^{10} ( x^{2} -6x+5)dx\\\\\\ A=\frac{ x^{3} }{3}- \frac{6 x^{2} }{2}+5x~\Bigg|_{0} ^{10}\\\\\\ A=\left(\frac{ x^{3} }{3}- \frac{6 x^{2} }{2}+5x\right )- \left(\frac{ x^{3} }{3}- \frac{6 x^{2} }{2}+5x\right )~~ \Bigg|_{0} ^{10}\\\\\\ A=\left(\frac{ (10)^{3} }{3}- \frac{6 (10)^{2} }{2}+5(10)\right )- \left(\frac{ (0)^{3} }{3}- \dfrac{6 (0)^{2} }{2}+5(0)\right )$


$A=\left(\dfrac{ 1000 }{3}- \dfrac{6 .100 }{2}+50\right )- \left(\dfrac{ 0 }{3}- \dfrac{6 .0 }{2}+5(0)\right )\\\\\\ A=\left(\dfrac{ 1000 }{3}- \dfrac{600 }{2}+50\right )- \left(\dfrac{ 0 }{3}- \dfrac{6 .0 }{2}+5(0)\right )\\\\\\ A=\left(\dfrac{ 1000 }{3}- 300+50\right )- \left(\dfrac{ 0 }{3}- \dfrac{6 .0 }{2}+5(0)\right )\\\\\\ A=\left(\dfrac{ 1000 }{3}- 900+150\right )- \left(0)\right )\\\\\\ A=\left(\dfrac{ 250}{3}\right )- \left(0)\right )\\\\\\ A=\left(\dfrac{ 250}{3}\right )$

$\boxed{Resposta:\displaystyle\int_{0}^{10} ( x^{2} -6x+5)dx\Rightarrow A=\dfrac{ 250}{3}}$



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