Question

Ache a área total entre a curva y=1-x2 e o intervalo [0, 1]. Assinale a alternativa correta.
a´= --2/3
b= -2
c= 2/3
d= 150

Answer 1

✅ Após resolver os cálculos, concluímos que a área sobe a curva compreendida no referido intervalo de integração é:

 $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S = \int\limits_{0}^{1}(-x^{2} + 1)\,dx = \frac{2}{3}\,u.\,a.\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

              $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf Alternativa\:C\:\:\:}}\end{gathered} }$

Sejam os dados:

                      $\Large\begin{cases}\tt y = 1 - x^{2}\\\tt I = [0,\,1]\end{cases}$

Organizando a função, temos:

                    $\Large\begin{cases}\tt f(x) = - x^{2} + 1\\\tt I = [0,\,1]\end{cases}$

Para calcular a área "S" entre a função polinomial e o intervalo referido devemos calcular a integral definida no referido intervalo.

Sabendo que o TFC - Teorema Fundamental do Cálculo, diz:

   $\Large\displaystyle\begin{gathered}\tt S = \int\limits_{a}^{b}f(x)\,dx = F(x)|_{a}^{b} = F(b) - F(a)\end{gathered}$

Então, temos:

        $\Large\displaystyle\begin{gathered}\tt S = \int\limits_{0}^{1}(-x^{2} + 1)\,dx\end{gathered}$

            $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(\frac{-x^{2 + 1}}{2 + 1} + x + c\bigg)\bigg|_{0}^{1}\end{gathered} }$

            $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{x^{3}}{3} + x + c\bigg)\bigg|_{0}^{1}\end{gathered} }$

            $\Large\displaystyle\text{ \begin{gathered}\tt = \bigg(-\frac{1^{3}}{3} + 1 + c\bigg) - \bigg(-\frac{0^{3}}{3} + 0 + c\bigg)\end{gathered} }$

            $\Large\displaystyle\begin{gathered}\tt = \bigg(-\frac{1}{3} + 1 + c\bigg) - (-0 + 0 + c)\end{gathered}$

             $\Large\displaystyle\begin{gathered}\tt = -\frac{1}{3} + 1 + c - c\end{gathered}$

             $\Large\displaystyle\begin{gathered}\tt = -\frac{1}{3} + 1\end{gathered}$

             $\Large\displaystyle\begin{gathered}\tt = \frac{-1 + 3}{3}\end{gathered}$

             $\Large\displaystyle\begin{gathered}\tt = \frac{2}{3}\end{gathered}$

Portanto, a área é:

         $\Large\displaystyle\begin{gathered}\tt S = \frac{2}{3}\,u.\,a.\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

Saiba mais:

1. https://brainly.com.br/tarefa/43044491
2. https://brainly.com.br/tarefa/2409823
3. https://brainly.com.br/tarefa/43129156
4. https://brainly.com.br/tarefa/4620403
5. https://brainly.com.br/tarefa/12599396
6. https://brainly.com.br/tarefa/32696440
7. https://brainly.com.br/tarefa/51525738
8. https://brainly.com.br/tarefa/52611877
9. https://brainly.com.br/tarefa/53293778

 $\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$