Question

O valor da integral definida

Answer 1

Vamos resolver : 

$\int\limits^1_0 {(x^{3}-4x^{2}}+1)} \, dx \\\\\\( \frac{x^{4}}{4}- \frac{4x^{3}}{3} +x)\\\\\\d.e.f.i.n.i.n.d.o...\\\\\\| \frac{1^{4}}{4} - \frac{4.1^{3}}{3} +1|-| \frac{0^{4}}{4} - \frac{4.0^{3}}{3} +0|\\\\\\| \frac{1}{4} - \frac{4}{3} +1|-| \frac{0}{4} - \frac{0}{3} |\\\\\\| \frac{3}{12} - \frac{16}{12} + \frac{12}{12} |-|0|\\\\\\- \frac{13}{12} + \frac{12}{12} =\boxed{\boxed{\boxed{- \frac{1}{12}\ }}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ok$

Answer 2

✅ Após resolver os cálculos, concluímos que a integral definida no referido intervalo de integração é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf \int_{0}^{1} (x^{3} - 4x^{2} + 1)\,dx = -\frac{1}{12}\:\:\:}}\end{gathered} }$

Portanto, a opção correta é:

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

Seja a integral definida:

       $\Large\displaystyle\begin{gathered}\tt \int_{0}^{1} (x^{2} - 4x^{2} + 1)\,dx\end{gathered}$

Sabendo que a integral definida pode ser resumida na seguinte fórmula:

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

Então, temos:

$\Large\displaystyle\begin{gathered}\tt \int_{0}^{1} (x^{3} - 4x^{2} + 1)\,dx = \int_{0}^{1} x^{3} + \int_{0}^{1} - 4x^{2}\,dx + \int_{0}^{1} 1\,dx\end{gathered}$

                                              $\Large\displaystyle\begin{gathered}\tt = \int_{0}^{1} x^{3}\,dx - 4\int_{0}^{1} x^{2}\,dx + 1\int_{0}^{1} dx\end{gathered}$

                                              $\Large\displaystyle\text{ \begin{gathered}\tt = \left[\frac{x^{3 + 1}}{3 + 1} - 4\cdot\frac{x^{2 + 1}}{2 + 1} + x + c\right]\bigg|_{0}^{1}\end{gathered} }$

                                              $\Large\displaystyle\begin{gathered}\tt = \left[\frac{1}{4}x^{4} - \frac{4}{3}x^{3} + x + c\right]\bigg|_{0}^{1}\end{gathered}$

        $\Large\displaystyle\begin{gathered}\tt = \left[\frac{1}{4}\cdot1^{4} - \frac{4}{3}\cdot1^{3} + 1 + c\right] - \left[\frac{1}{4}\cdot0^{4} - \frac{4}{3}\cdot0^{3} + 0 + c\right]\end{gathered}$

                                             $\Large\displaystyle\begin{gathered}\tt = \frac{1}{4} - \frac{4}{3} + 1 + c - 0 + 0 - 0 - c\end{gathered}$

                                             $\Large\displaystyle\begin{gathered}\tt = \frac{1}{4} - \frac{4}{3} + 1\end{gathered}$

                                             $\Large\displaystyle\begin{gathered}\tt = \frac{3 - 16 + 12}{12}\end{gathered}$

                                             $\Large\displaystyle\begin{gathered}\tt = - \frac{1}{12}\end{gathered}$

✅ Portanto, a integral procurada é:

     $\Large\displaystyle\begin{gathered}\tt \int_{0}^{1} (x^{3} - 4x^{2} + 1)\,dx = -\frac{1}{12}\end{gathered}$

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Veja a solução gráfica da questão, representada na figura.