Question

Obtenha a area entre os graficos das funçoes f(x)= x³ - x²+3 e g (x)= -x/2+1/2 no intervalo [-1,1]

Answer 1

$\boxed{\begin{array}{l}\displaystyle\sf A=\int_{-1}^1\bigg[(x^3-x^2+3)-\bigg(-\dfrac{x}{2}+\dfrac{1}{2}\bigg)\bigg]\,dx\\\\\displaystyle\sf A=\int_{-1}^1\bigg[x^3-x^2+3+\dfrac{x}{2}-\dfrac{1}{2}\bigg]\,dx\\\\\displaystyle\sf A=\int_{-1}^1\bigg[x^3-x^2+\dfrac{x}{2}+\dfrac{5}{2}\bigg]dx\end{array}}$

$\boxed{\begin{array}{l}\sf A=\bigg[\dfrac{1}{4}x^4-\dfrac{1}{3}x^3+\dfrac{1}{4}x^2+\dfrac{5}{2}x\bigg]_{-1}^1\\\\\sf A=\dfrac{1}{4}\cdot 1^4-\dfrac{1}{3}\cdot 1^3+\dfrac{1}{4}\cdot 1^2+\dfrac{5}{2}\cdot1-\bigg[\dfrac{1}{4}\cdot(-1)^4-\dfrac{1}{3}\cdot(-1)^3+\dfrac{1}{4}\cdot(-1)^2+\dfrac{5}{2}\cdot(-1)\bigg]\end{array}}$

$\large\boxed{\begin{array}{l}\sf A=\backslash\!\!\!\!\dfrac{1}{4}-\dfrac{1}{3}+\backslash\!\!\!\dfrac{1}{4}+\dfrac{5}{2}-\backslash\!\!\!\!\dfrac{1}{4}-\dfrac{1}{3}-\backslash\!\!\!\dfrac{1}{4}+\dfrac{5}{2}\\\\\sf A=5-\dfrac{2}{3}\\\\\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{13}{3}}}}}\blue{\checkmark} \end{array}}$