• Matéria: Matemática
  • Autor: oliveiraengelet
  • Perguntado 9 anos atrás

alguem pode me ajudar na integral definida ?

Anexos:

Respostas

respondido por: Lukyo
1
\displaystyle\int_a^b (Px^2+Qx+R)\,dx


Regra para encontrar primitivas de potências:

\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}\,,~~~~\text{com }n\ne -1.


Primitivando a função da integral, e avaliando nos extremos:

\displaystyle\int_a^b (Px^2+Qx+R)\,dx\\\\\\ =\left.\left(\dfrac{Px^{2+1}}{{2+1}}+\dfrac{Qx^{1+1}}{1+1}+Rx\right)\right|_a^b\\\\\\ =\left.\left(\dfrac{Px^3}{3}+\dfrac{Qx^2}{2}+Rx\right)\right|_a^b\\\\\\ =\left(\dfrac{P\cdot b^3}{3}+\dfrac{Q\cdot b^2}{2}+R\cdot b\right)-\left(\dfrac{P\cdot a^3}{3}+\dfrac{Q\cdot a^2}{2}+R\cdot a\right)\\\\\\ =\dfrac{P\cdot (b^3-a^3)}{3}+\dfrac{Q\cdot (b^2-a^2)}{2}+R\cdot (b-a)\\\\\\\\ \therefore~~\boxed{\begin{array}{c}\displaystyle\int_a^b (Px^2+Qx+R)\,dx=\dfrac{P\cdot (b^3-a^3)}{3}+\dfrac{Q\cdot (b^2-a^2)}{2}+R\cdot (b-a) \end{array}}

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