Matemática, perguntado por LuanaBeatriz321, 1 ano atrás

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.

Soluções para a tarefa

Respondido por Usuário anônimo
2
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}}

Anexos:

LuanaBeatriz321: Muito Obrigado !
Usuário anônimo: Dê nada!
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