O valor da integral é:
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Soluções para a tarefa
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Vamos resolver primeiro a integral de dentro, a qual está entre colchetes:
![I_1=\displaystyle\int^{x}_{x^2}dy=[y]^{x}_{x^2}\\\\
I_1=x-x^2](https://tex.z-dn.net/?f=I_1%3D%5Cdisplaystyle%5Cint%5E%7Bx%7D_%7Bx%5E2%7Ddy%3D%5By%5D%5E%7Bx%7D_%7Bx%5E2%7D%5C%5C%5C%5C%0AI_1%3Dx-x%5E2 "I_1=\displaystyle\int^{x}_{x^2}dy=[y]^{x}_{x^2}\\\\
I_1=x-x^2")
Agora podemos resolver a integral "de fora":
![I_2=\displaystyle\int^{1}_{0}\left[\displaystyle\int_{x^2}^{x}dy\right]dx
=\displaystyle\int^{1}_{0}(x-x^2)dx\\\\
I_2=\displaystyle\int^{1}_{0}x\,dx-\displaystyle\int^{1}_{0}x^2\,dx\\\\
I_2=\left[\dfrac{x^2}{2}\right]^{1}_{0}-\left[\dfrac{x^3}{3}\right]^{1}_{0}\\\\
I_2=\left[\left(\dfrac{1^2}{2}\right)-\left(\dfrac{0^2}{2}\right)\right]-\left[\left(\dfrac{1^3}{3}\right)-\left(\dfrac{0^3}{3}\right)\right]](https://tex.z-dn.net/?f=I_2%3D%5Cdisplaystyle%5Cint%5E%7B1%7D_%7B0%7D%5Cleft%5B%5Cdisplaystyle%5Cint_%7Bx%5E2%7D%5E%7Bx%7Ddy%5Cright%5Ddx%0A%3D%5Cdisplaystyle%5Cint%5E%7B1%7D_%7B0%7D%28x-x%5E2%29dx%5C%5C%5C%5C%0AI_2%3D%5Cdisplaystyle%5Cint%5E%7B1%7D_%7B0%7Dx%5C%2Cdx-%5Cdisplaystyle%5Cint%5E%7B1%7D_%7B0%7Dx%5E2%5C%2Cdx%5C%5C%5C%5C%0AI_2%3D%5Cleft%5B%5Cdfrac%7Bx%5E2%7D%7B2%7D%5Cright%5D%5E%7B1%7D_%7B0%7D-%5Cleft%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%5Cright%5D%5E%7B1%7D_%7B0%7D%5C%5C%5C%5C%0AI_2%3D%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%5E2%7D%7B2%7D%5Cright%29-%5Cleft%28%5Cdfrac%7B0%5E2%7D%7B2%7D%5Cright%29%5Cright%5D-%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%5E3%7D%7B3%7D%5Cright%29-%5Cleft%28%5Cdfrac%7B0%5E3%7D%7B3%7D%5Cright%29%5Cright%5D "I_2=\displaystyle\int^{1}_{0}\left[\displaystyle\int_{x^2}^{x}dy\right]dx
=\displaystyle\int^{1}_{0}(x-x^2)dx\\\\
I_2=\displaystyle\int^{1}_{0}x\,dx-\displaystyle\int^{1}_{0}x^2\,dx\\\\
I_2=\left[\dfrac{x^2}{2}\right]^{1}_{0}-\left[\dfrac{x^3}{3}\right]^{1}_{0}\\\\
I_2=\left[\left(\dfrac{1^2}{2}\right)-\left(\dfrac{0^2}{2}\right)\right]-\left[\left(\dfrac{1^3}{3}\right)-\left(\dfrac{0^3}{3}\right)\right]")
![I_2=\left[\left(\dfrac{1}{2}\right)-\left(\dfrac{0}{2}\right)\right]-\left[\left(\dfrac{1}{3}\right)-\left(\dfrac{0}{3}\right)\right]\\\\
I_2=\left[\left(\dfrac{1}{2}\right)-0\right]-\left[\left(\dfrac{1}{3}\right)-0\right]\\\\
I_2=\left(\dfrac{1}{2}\right)-\left(\dfrac{1}{3}\right)=\dfrac{3-2}{6}\\\\
\boxed{I_2=\dfrac{1}{6}}\Longrightarrow\text{Letra }\bold{D}](https://tex.z-dn.net/?f=I_2%3D%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29-%5Cleft%28%5Cdfrac%7B0%7D%7B2%7D%5Cright%29%5Cright%5D-%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%5Cright%29-%5Cleft%28%5Cdfrac%7B0%7D%7B3%7D%5Cright%29%5Cright%5D%5C%5C%5C%5C%0AI_2%3D%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29-0%5Cright%5D-%5Cleft%5B%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%5Cright%29-0%5Cright%5D%5C%5C%5C%5C%0AI_2%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29-%5Cleft%28%5Cdfrac%7B1%7D%7B3%7D%5Cright%29%3D%5Cdfrac%7B3-2%7D%7B6%7D%5C%5C%5C%5C%0A%5Cboxed%7BI_2%3D%5Cdfrac%7B1%7D%7B6%7D%7D%5CLongrightarrow%5Ctext%7BLetra+%7D%5Cbold%7BD%7D "I_2=\left[\left(\dfrac{1}{2}\right)-\left(\dfrac{0}{2}\right)\right]-\left[\left(\dfrac{1}{3}\right)-\left(\dfrac{0}{3}\right)\right]\\\\
I_2=\left[\left(\dfrac{1}{2}\right)-0\right]-\left[\left(\dfrac{1}{3}\right)-0\right]\\\\
I_2=\left(\dfrac{1}{2}\right)-\left(\dfrac{1}{3}\right)=\dfrac{3-2}{6}\\\\
\boxed{I_2=\dfrac{1}{6}}\Longrightarrow\text{Letra }\bold{D}")
Agora podemos resolver a integral "de fora":
Respondido por
0
Resposta:
reposta: 1/6 corrigido pelo AVA.
Explicação passo a passo:
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