Question

Ajudem nessas integrais duplas aí galera

Answer 1

Resposta:

Explicação passo-a-passo:

a)

$\int_0^1\int_0^2(2x+y)\;dxdy=\int_0^1\left[x^2+xy\right]_0^2dy$

$\int_0^1\int_0^2(2x+y)\;dxdy=\int_0^12^2+2y-(0^2+0\cdot y)\;dy$

$\int_0^1\int_0^2(2x+y)\;dxdy=\int_0^12y+4\;dy$

$\int_0^1\int_0^2(2x+y)\;dxdy=[y^2+4y]_0^1$

$\int_0^1\int_0^2(2x+y)\;dxdy=1^2+4-(0^2+4\cdot 0)$

$\int_0^1\int_0^2(2x+y)\;dxdy=5$

b)

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\int_0^2\left[2y^2+x^3y\right]_{x^2}^{2x}\;dx$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\int_0^22\cdot(2x)^2+x^3\cdot2x-[2\cdot(x^2)^2+x^3\cdot x^2]\;dx$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\int_0^28x^2+2x^4-2x^4-x^5\;dx$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\int_0^2-x^5+8x^2\;dx$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\left[-\frac{x^6}{6}+\frac{8x^3}{3}\right]_0^2$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=-\frac{2^6}{6}+\frac{8\cdot2^3}{3}$

$\int_0^2\int_{x^2}^{2x}(x^3+4y)\;dydx=\frac{32}{3}$

c)

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=\int_0^4[y\sqrt{16-x^2}]_0^{\sqrt{16-x^2}}\;dx$

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=\int_0^4\sqrt{16-x^2}\cdot\sqrt{16-x^2}\;dx$

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=\int_0^416-x^2\;dx$

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=\left[-\frac{x^3}{3}+16x\right]_0^4$

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=-\frac{4^3}{3}+16\cdot4$

$\int_0^4\int_0^{\sqrt{16-x^2}}\sqrt{16-x^2}\;dydx=\frac{128}{3}$

d)

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=\int_{\pi/2}^\pi[2\sin x]_{\pi/6}^y\;dy$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=\int_{\pi/2}^\pi2\sin y-2\sin\pi/6\;dy$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=\int_{\pi/2}^\pi2\sin y-2\cdot\frac{1}{2}\;dy$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=\int_{\pi/2}^\pi2\sin y-1\;dy$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=[-y-2\cos y]_{\pi/2}^\pi$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=-\pi-2\cos\pi-\left(-\frac{\pi}{2}-2\cos\frac{\pi}{2}\right)$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=-\pi+2-\left(-\frac{\pi}{2}-2\cdot 0\right)$

$\int_{\pi/2}^\pi\int_{\pi/6}^y2\cos x\;dxdy=2-\frac{\pi}{2}$

e)

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\int_{-1}^1[3xy+y^2]_{x^3}^{x+1}\;dx$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\int_{-1}^13x\cdot(x+1)+(x+1)^2-[3x\cdot x^3+(x^3)^2]$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\int_{-1}^13x(x+1)+(x+1)^2-3x^4-x^6\;dx$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\int_{-1}^13x^2+3x+x^2+2x+1-3x^4-x^6\;dx$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\int_{-1}^1-x^6-3x^4+4x^2+5x+1\;dx$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\left[-\frac{x^7}{7}-3\frac{x^5}{5}+4\frac{x^3}{3}+5\frac{x^2}{2}+x\right]_{-1}^1$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=-\frac{1}{7}-\frac{3}{5}+\frac{4}{3}+\frac{5}{2}+1-(\frac{1}{7}+\frac{3}{5}-\frac{4}{3}+\frac{5}{2}-1)$

$\int_{-1}^1\int_{x^3}^{x+1}3x+2y\;dydx=\frac{334}{105}$