Question

Considere uma função f: [0,6] ⟶ ℝ, tal que f (x) = −|2x − 6| + 6, com o gráfico representado na figura abaixo. Qual a integral de f(x) dx?​

Answer 1

• A integral de f(x) dx , definida pelo intervalo [0,6] será igual 18.

Dada a integral modular definida: $\large\displaystyle\begin{aligned} \int_{0}^{6} -|2x-6| + 6\ dx\end{aligned}$. Aplicando a definição de modulo, temos que:

$\large\displaystyle\begin{aligned} f(x) = |x| \Rightarrow \begin{cases} x, se \ x\geq 0\\ -x, se \ x \leq 0\end{cases}\end{aligned}$

Logo, temos pela definição de módulo que:

$\large\displaystyle\begin{aligned} f(x) = |2x-6|\Rightarrow \begin{cases} 2x-6, se \ x\geq 0\\ -(2x-6), se \ x \leq 0\end{cases}\end{aligned}$

• Sabendo disso, teremos que calcular a seguinte integral:

$\large\displaystyle\begin{aligned}\int_{0}^{3} \; [- (-2x + 6) + 6] dx + \int_{3}^{6} \; [- (2x - 6) + 6] dx\end{aligned}$

Ficando assim:

$\large\displaystyle\begin{aligned}\int_{0}^{3} \; [- (-2x + 6) + 6] dx + \int_{3}^{6} \; [- (2x - 6) + 6] dx\end{aligned}$

$\large\displaystyle\begin{aligned}\int_{0}^{3} \; [2x ] dx + \int_{3}^{6} \; [ -2x +12] dx\end{aligned}$

$\large\displaystyle\begin{aligned} \left.\left(\frac{2x^2}{2} \right)\right| _0^3 + \int_{3}^{6} -2x dx + \int^6_3 12dx \end{aligned}$

$\large\displaystyle\begin{aligned} \left.\left(\frac{2x^2}{2} \right)\right| _0^3 + \left. \left( \frac{-2x^2}{2} \right)\right|^6_3+ \left.\left( 12x \right)\right| ^6_3\end{aligned}$

$\large\displaystyle\begin{aligned} \left(\frac{2(3)^2}{2} - \frac{2(0)^2}{2} \right) + \left( \frac{-2(6)^2}{2} - \frac{-2(3)^2}{2} \right)+ \left( 12(6) - 12(3) \right)\end{aligned}$

$\large\displaystyle\begin{aligned} \left(\frac{2(3)^2}{2} \right) + \left( -36+9\right)+ \left( 36 \right)\end{aligned}$

$\large\displaystyle\begin{aligned} \left(9\right) + \left( -27\right)+ \left( 36 \right)\end{aligned}$

$\large\displaystyle\text{ \begin{aligned}\therefore \boxed{\boxed{\green{ 18}}}\end{aligned} }$

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