Question

preciso com urgência
calcule as integrais abaixo

Answer 1

Pretendemos calcular o integral:

$\displaystyle\int\limits_0^{\pi/2}\int\limits_0^1 y\sin x \textrm{ d}y\textrm{ d}x.$

Começamos então por separar os integrais:

$\displaystyle\int\limits_0^{\pi/2}\sin x \textrm{ d}x \times \int\limits_0^1 y\textrm{ d}y.$

Calculamos agora cada um deles:

$\displaystyle\int\limits_0^{\pi/2}\sin x \textrm{ d}x = -\cos x \Big\vert_0^{\pi/2} = -\underbrace{\cos\left(\dfrac{\pi}{2}\right)}_{=0} + \underbrace{\cos 0}_{=1} = 1.$

$\displaystyle\int\limits_0^1 y\textrm{ d}y = \dfrac{y^2}{2}\Big\vert_0^1 = \dfrac{1^2}{2}-\dfrac{0^2}{2} = \dfrac{1}{2}.$

Assim, obtemos por fim:

$\displaystyle\int\limits_0^{\pi/2}\int\limits_0^1 y\sin x \textrm{ d}y\textrm{ d}x = \dfrac{1}{2}.$

Para o segundo integral, pretendemos calcular:

$\displaystyle\iint\limits_R (x+y)\textrm{ d}A,$

onde $R$ é a região limitada por $y=x^2$ e $y=2x$.

Primeiro determinamos os pontos de interseção entre as duas curvas:

$x^2 = 2x \iff x^2 - 2x = 0 \iff x(x-2) = 0 \iff x = 0 \textrm{ ou } x=2.$

Para cada valor de $x\in[0,2]$, os valores de $y$ varia entre $x^2$ e $2x$, ou seja, $y\in[x^2,2x]$. Assim, podemos escrever o integral na forma:

$\displaystyle\iint\limits_R (x+y)\textrm{ d}A = \displaystyle\int\limits_0^2\left[\int\limits_{x^2}^{2x} (x+y)\textrm{ d}y\right]\textrm{d}x.$

Calculamos primeiro o integral interior:

$\displaystyle\int\limits_{x^2}^{2x} (x+y)\textrm{ d}y = \left[xy + \dfrac{y^2}{2}\right]_{y=x^2}^{y=2x} = x(2x) + \dfrac{(2x)^2}{2} - x\times x^2 - \dfrac{(x^2)^2}{2} =\\\\= 2x^2 + 2x^2 - x^3 - \dfrac{x^4}{2} = -\dfrac{x^4}{2}-x^3+4x^2.$

Podemos agora calcular o integral exterior:

$\displaystyle\int\limits_0^2\left(-\dfrac{x^4}{2}-x^3+4x^2\right)\textrm{d}x = \left[-\dfrac{1}{2}\dfrac{x^5}{5}-\dfrac{x^4}{4}+4\dfrac{x^3}{3}\right]_0^2 = -\dfrac{1}{2}\dfrac{2^5}{5}-\dfrac{2^4}{4} + 4\dfrac{2^3}{3}=\\\\=-\dfrac{2^4}{5}- \dfrac{2^4}{4}+ \dfrac{2}{3}2^4 = 2^4\left(\dfrac{2}{3}-\dfrac{1}{5}-\dfrac{1}{4}\right) = 16\times\dfrac{13}{60} = \dfrac{52}{15}.$

Obtemos então o valor final:

$\displaystyle\iint\limits_R (x+y)\textrm{ d}A = \dfrac{52}{15}.$