Question

Derivada e Integral !!!

1) Dada a função $y= f(x) = x^{3}-6 x^{2}$
faça o gráfico.

2) Calcule a área entre as curvas $y= f(x) = 1- x^{2}$ e $y= g(x) = x^{2} -1$

3) Calcule a área entre as curvas $y= f(x) = x^{2} -1$ e $y= g(x) = 3x+3.$

Answer 1

Boa noite Luana!

Solução!

Gráfico da função esta anexo.


$y= x^{3}-6 x^{2}$


$Exercicio~~2\\\\\ f(x)=1- x^{2} \\\\\ g(x)= x^{2} -1\\\\\ Primeiro~~vamos~~encontrar~~o~~intervalo~~de~~integrac\~ao!\\\\\ g(x)=f(x)\\\\\ x^{2}-1=1-x^{2} \\\\\\ x^{2} + x^{2} =1+1\\\\\ 2 x^{2} =2\\\\\\ x^{2} = \dfrac{2}{2}\\\\\ x^{2} =1\\\\\\ x= \sqrt{1}\\\\\ x=\pm1~~logo~~[-1,1]$


$A=\displaystyle \int_{-1} ^{1} (f(x)-g(x))dx\\\\\\ A=\displaystyle \int_{-1} ^{1} (1- x^{2})-(x^{2} -1)dx\\\\\\ A=\displaystyle \int_{-1} ^{1} (1- x^{2}-x^{2} +1)dx\\\\\\ A=\displaystyle \int_{-1} ^{1} (2- 2x^{2})dx$



$A= (2- 2x^{2})\Bigg|_{-1}^{1} \\\\\\ A= \left(2x- \dfrac{2x^{3} }{3} \right )\Bigg|_{-1}^{1} \\\\\\\ A= \left(2x- \dfrac{2x^{3} }{3} \right )-\left(2x- \dfrac{2x^{3} }{3} \right )\Bigg|_{-1}^{1} \\\\\\\ A= \left(2(1)- \dfrac{2(1)^{3} }{3} \right )-\left(2(-1)- \dfrac{2(-1)^{3} }{3} \right )\Bigg|_{-1}^{1} \\\\\\\ A= \left(2- \dfrac{2 }{3} \right )-\left(-2+ \dfrac{2 }{3} \right )\Bigg|_{-1}^{1}$

$A= \left( \dfrac{6-2 }{3} \right )-\left( \dfrac{-6+2 }{3} \right )\\\\\\ A= \left( \dfrac{6-2 }{3} \right )+\left( \dfrac{6-2 }{3} \right )\\\\\\ A= \left( \dfrac{4 }{3} \right )+\left( \dfrac{4 }{3} \right )\\\\\\ A= \left( \dfrac{4 }{3} + \dfrac{4 }{3} \right )\\\\\\ A= \dfrac{8}{3}\\\\\\ \boxed{Resposta:A= \dfrac{8}{3}u^{2} }$



$Exercicio~~3\\\\\\ f(x)= x^{2} -1\\\\\ g(x)=3x+3\\\\\\\ Calculo ~~do~~ intevalo ~~de ~~integrac\~ao!\\\\\\ f(x)=g(x)\\\\\\\ x^{2} -1=3x+3\\\\\ x^{2} -1-3x-3=0\\\\\ x^{2} -3x-4=0$


$x= \dfrac{3\pm \sqrt{(-3)^{2}-4.1.-4 } }{2.1}\\\\\\\ x= \dfrac{3\pm \sqrt{9+16 } }{2}\\\\\\\ x= \dfrac{3\pm \sqrt{25 } }{2}\\\\\\\ x= \dfrac{3\pm 5}{2}\\\\\\\ x_{1} = \dfrac{3+5}{2} =4\\\\\\ x_{2} = \dfrac{3-5}{2} =-1~~intervalo~~~~~~[-1,4]$


$A=\displaystyle\int_{-1}^{4} (f(x)-g(x))dx \\\\\\\\\ A=\displaystyle\int_{-1}^{4} ((3x+3)-( x^{2} -1))dx\\\\\\\\\ A=\displaystyle\int_{-1}^{4} (3x+3- x^{2} +1)dx\\\\\\\\\ A=\displaystyle\int_{-1}^{4} (- x^{2}+3x+4)dx$


$A= \left ( \dfrac{-x^{3} }{3} + \dfrac{3 x^{2} }{2} +4x\right )\Bigg| _{-1}^{4} \\\\\\\\\ A= \left ( \dfrac{-x^{3} }{3} + \dfrac{3 x^{2} }{2} +4x\right )- \left ( \dfrac{-x^{3} }{3} + \dfrac{3 x^{2} }{2} +4x\right ) \Bigg| _{-1}^{4} \\\\\\\\\ A= \left ( \dfrac{-(4)^{3} }{3} + \dfrac{3 (4)^{2} }{2} +4(4)\right )- \left ( \dfrac{-(-1)^{3} }{3} + \dfrac{3 (-1)^{2} }{2} +4(-1)\right ) \Bigg| _{-1}^{4}$

$A= \left ( \dfrac{-64 }{3} + \dfrac{3 (16) }{2} +16\right )- \left ( \dfrac{1 }{3} + \dfrac{3 (1)}{2} -4\right )\\\\\\\ A= \left ( \dfrac{-64 }{3} + \dfrac{48 }{2} +16\right )- \left ( \dfrac{1 }{3} + \dfrac{3}{2} -4\right )\\\\\\\ A= \left ( \dfrac{-64 }{3} + 24 +16\right )- \left ( \dfrac{1 }{3} + \dfrac{3}{2} -4\right )\\\\\\\ A= \left ( \dfrac{-64+72+48 }{3} \right )- \left ( \dfrac{2+9-24 }{6} + \right )$

$A= \left ( \dfrac{-64+72+48 }{3} \right )+ \left ( \dfrac{-11+24 }{6} + \right )\\\\\\\\ A= \left ( \dfrac{56 }{3} \right )+ \left ( \dfrac{13 }{6} \right )$



$A= \left ( \dfrac{112+13}{6} \right )\\\\\\ A= \dfrac{125}{6}~~ou~~20.83 \\\\\\\\\\\ \boxed{Resposta:A= \dfrac{125}{6}u^{2}~~ou~~20.83u^{2}}$