Question

valor da integral
3 2
∫ ∫ ₍ x . y ₎ dy dx
0 0

Answer 1

Resposta:

Obs.: lembre-se do teorema fundamental do cálculo, no qual:

$\boxed{\int\limits^{\sf b}_{\sf a}\sf f(x)\,dx=F(x)\bigg|^{\sf b}_{\sf a}=F(b)-F(a)}$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\displaystyle\int\limits^{\sf2}_{\sf0}\sf(x\cdot y)\,dydx$

Integrando em relação a y primeiro (considere x uma constante):

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\Bigg[\sf x\cdot\displaystyle\int\limits^{\sf2}_{\sf0}\sf(y)\,dy\Bigg]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\Bigg[\sf x\cdot\bigg|\displaystyle\int\sf(y)\,dy\:\bigg|^{\sf2}_{\sf0}\Bigg]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\Bigg[\sf x\cdot\bigg|\dfrac{y^2}{2}\bigg|^{\sf2}_{\sf0}\Bigg]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\bigg[\sf x\cdot\bigg(\dfrac{2^2}{2}-\dfrac{0^2}{2}\bigg)\bigg]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\bigg[\sf x\cdot\bigg(\dfrac{4}{2}-\dfrac{0}{2}\bigg)\bigg]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}\big[\sf x\cdot(2-0)\big]dx$

$\sf V=\displaystyle\int\limits^{\sf3}_{\sf0}(\sf 2x)\,dx$

Por fim, integre em relação a x:

$\sf V=\displaystyle\int\sf(2x)\,dx\:\bigg|^{\sf3}_{\sf0}$

$\sf V=x^2\:\big|^{\sf3}_{\sf0}$

$\sf V=3^2-0^2$

$\sf V=9-0$

$\red{\boxed{\sf V=9}}\rightarrow\sf valor~da~integral.$