Ensino Superior ⇒ Integral Tópico resolvido

[Rick777]

Sabendo que
[tex3]y'=\frac{y}{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]y'=\frac{y}{2}[/tex3]

[tex3]y=\frac{3}{4}(x-2)^{2}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]y=\frac{3}{4}(x-2)^{2}[/tex3]

Calcule a integral abaixo

[tex3]\int\limits_{0}^{2}y'ydx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{0}^{2}y'ydx[/tex3]

[ragefloyd]

[tex3]\int\limits_{0}^{2}y'y\ dx=\int\limits_{0}^{2}\frac{y}{2}y\ dx=\frac12\int\limits_{0}^{2}y^2\ dx=\frac12\int\limits_{0}^{2}\left[\frac{3}{4}(x-2)^2\right]^2\ dx=\frac12\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}x^4-8x^3+24x^2-32x+16\ dx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{0}^{2}y'y\ dx=\int\limits_{0}^{2}\frac{y}{2}y\ dx=\frac12\int\limits_{0}^{2}y^2\ dx=\frac12\int\limits_{0}^{2}\left[\frac{3}{4}(x-2)^2\right]^2\ dx=\frac12\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}x^4-8x^3+24x^2-32x+16\ dx[/tex3]

[tex3]=\frac{9}{32}\int\limits_{0}^{2}x^4\ dx-8\int\limits_{0}^{2}x^3\ dx+24\int\limits_{0}^{2}x^2\ dx-32\int\limits_{0}^{2}x\ dx+\int\limits_{0}^{2}16\ dx=\frac{9}{32}\left[ \frac{1}{5}x^5 - \frac{8}{4}x^4 + \frac{24}{3}x^3 - \frac{32}{2}x^2 + 16x \right]_0^2[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]=\frac{9}{32}\int\limits_{0}^{2}x^4\ dx-8\int\limits_{0}^{2}x^3\ dx+24\int\limits_{0}^{2}x^2\ dx-32\int\limits_{0}^{2}x\ dx+\int\limits_{0}^{2}16\ dx=\frac{9}{32}\left[ \frac{1}{5}x^5 - \frac{8}{4}x^4 + \frac{24}{3}x^3 - \frac{32}{2}x^2 + 16x \right]_0^2[/tex3]

[tex3]=\frac{9}{32}\left[ \left(\frac{2^5}{5}-2\cdot2^4+8\cdot2^3 - 16\cdot 2^2+16\cdot2 \right) - 0 \right]=\frac{9}{32}\left( \frac{32}{5}-32+64-64+32 \right)=\boxed{\frac{9}{5}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]=\frac{9}{32}\left[ \left(\frac{2^5}{5}-2\cdot2^4+8\cdot2^3 - 16\cdot 2^2+16\cdot2 \right) - 0 \right]=\frac{9}{32}\left( \frac{32}{5}-32+64-64+32 \right)=\boxed{\frac{9}{5}}[/tex3]

[PedroLucas]

[tex3]\int\limits_{0}^{2}y'y\ dx=\int\limits_{0}^{2}\frac{y}{2}y\ dx=\frac12\int\limits_{0}^{2}y^2\ dx=\frac12\int\limits_{0}^{2}\left[\frac{3}{4}(x-2)^2\right]^2\ dx=\frac12\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}x^4-8x^3+24x^2-32x+16\ dx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\int\limits_{0}^{2}y'y\ dx=\int\limits_{0}^{2}\frac{y}{2}y\ dx=\frac12\int\limits_{0}^{2}y^2\ dx=\frac12\int\limits_{0}^{2}\left[\frac{3}{4}(x-2)^2\right]^2\ dx=\frac12\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}x^4-8x^3+24x^2-32x+16\ dx[/tex3]

[tex3]=\frac{9}{32}\int\limits_{0}^{2}x^4\ dx-8\int\limits_{0}^{2}x^3\ dx+24\int\limits_{0}^{2}x^2\ dx-32\int\limits_{0}^{2}x\ dx+\int\limits_{0}^{2}16\ dx=\frac{9}{32}\left[ \frac{1}{5}x^5 - \frac{8}{4}x^4 + \frac{24}{3}x^3 - \frac{32}{2}x^2 + 16x \right]_0^2[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]=\frac{9}{32}\int\limits_{0}^{2}x^4\ dx-8\int\limits_{0}^{2}x^3\ dx+24\int\limits_{0}^{2}x^2\ dx-32\int\limits_{0}^{2}x\ dx+\int\limits_{0}^{2}16\ dx=\frac{9}{32}\left[ \frac{1}{5}x^5 - \frac{8}{4}x^4 + \frac{24}{3}x^3 - \frac{32}{2}x^2 + 16x \right]_0^2[/tex3]

[tex3]=\frac{9}{32}\left[ \left(\frac{2^5}{5}-2\cdot2^4+8\cdot2^3 - 16\cdot 2^2+16\cdot2 \right) - 0 \right]=\frac{9}{32}\left( \frac{32}{5}-32+64-64+32 \right)=\boxed{\frac{9}{5}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]=\frac{9}{32}\left[ \left(\frac{2^5}{5}-2\cdot2^4+8\cdot2^3 - 16\cdot 2^2+16\cdot2 \right) - 0 \right]=\frac{9}{32}\left( \frac{32}{5}-32+64-64+32 \right)=\boxed{\frac{9}{5}}[/tex3]

Acho que aqui ficaria mais rápido pelo método da substituição:
[tex3]\frac{1}{2}\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}(x-2)^4\ dx \\
u = x-2\\
du = 1\cdot dx[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{1}{2}\int\limits_{0}^{2}\frac{9}{16}(x-2)^4\ dx=\frac{9}{32}\int\limits_{0}^{2}(x-2)^4\ dx \\
u = x-2\\
du = 1\cdot dx[/tex3]

Substituindo na integral sem mexer nos limites superiores e inferiores (quero voltar no final):
[tex3]\frac{9}{32}\int\limits_{0}^{2}u^4\ du = \frac{9}{32}\cdot\frac{u^5}{5} = \frac{9}{160}\cdot (x-2)^5 ]_{0}^{2}
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{9}{32}\int\limits_{0}^{2}u^4\ du = \frac{9}{32}\cdot\frac{u^5}{5} = \frac{9}{160}\cdot (x-2)^5 ]_{0}^{2}
[/tex3]

Substituindo os limites superior e inferior:
[tex3]\frac{9}{160}\cdot (x-2)^5 ]_{0}^{2} = \frac{9}{160}[\cdot (2-2)^5 - (0-2)^5] = \frac{9}{\cancel{160}^5}\cdot \cancel{32} = \boxed{\boxed{\frac{9}{5}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{9}{160}\cdot (x-2)^5 ]_{0}^{2} = \frac{9}{160}[\cdot (2-2)^5 - (0-2)^5] = \frac{9}{\cancel{160}^5}\cdot \cancel{32} = \boxed{\boxed{\frac{9}{5}}}[/tex3]

[ragefloyd]

PedroLucas De fato bem melhor! Obrigado por apontar.

Sendo bem sincero, eu não tenho tanta experiência com integrais, mas eu vi um que sabia resolver então postei minha solução hsuahsuah