Question

O valor da integral definida abaixo é;

Answer 1

Olá,
Segue o cálculo:
$\int\limits^0_{-3} {x^2-4x+7} \, dx = \int\limits^0_{-3} {x^2} \, dx - \int\limits^0_{-3} {4x} \, dx + \int\limits^0_{-3} {7} \, dx = \\ \frac{x^3}{3}| \frac{0}{-3} - 2x^2| \frac{0}{-3}+7x| \frac{0}{-3} = \\ \frac{(0)^3}{3}- \frac{(-3)^3}{3}-2*((0)^2-(-3)^2)+7*((0)-(-3)) = \\ 0+9-2*(0-9)+7*(0+3) = \\ +9+18+21= \\ 48$

Espero ter ajudado.

Answer 2

Bom dia João!


Solução!


$\displaystyle \int_{a}^{b}f(a x^{2} +bx+c)=f(b)-f(a)$



$\displaystyle \int_{-3}^{0} (x^{2} -4x+7)dx\\\\\\ I= \frac{ x^{3} }{3}- \dfrac{4 x^{2} }{2}+7x \Bigg|_{-3}^{0}\\\\\\\\ I=\left ( \dfrac{ x^{3} }{3} - \dfrac{4 x^{2} }{2}+7x\right )- \left ( \dfrac{ x^{3} }{3} - \dfrac{4 x^{2} }{2}+7x \right ) \\\\\\\\ I=\left ( \dfrac{( 0)^{3} }{3} - \dfrac{4 (0)^{2} }{2}+7(0)\right )- \left ( \dfrac{ (-3)^{3} }{3} - \dfrac{4 (-3)^{2} }{2}+7(-3) \right )$



$I=\left ( \dfrac{( 0)^{3} }{3} - \dfrac{4 (0)^{2} }{2}+7(0)\right )- \left ( \dfrac{ -27}{3} - \dfrac{4 .9}{2}-21 \right ) \\\\\\\\ I=\left ( 0\right )- \left (-9 - 18-21 \right ) \\\\\\\\ I=\left ( 0\right )- \left (-48 \right ) \\\\\\\\ I=0+48\\\\\\ I=48$


$\boxed{Resposta:\displaystyle \int_{-3}^{0} (x^{2} -4x+7)dx=48}}$

Bom dia!

Bons estudos!