o valor da integral
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É dada a seguinte integral definida:
Vamos fazer uma substituição:
Substituindo na integral dada:
![\displaystyle I=\int_{-1}^0x^2\sqrt{x^3+1}\,dx\\\\
I=\int_{u_1}^{u_2}x^2\sqrt{u}\cdot\dfrac{1}{3x^2}du\\\\
I=\dfrac{1}{3}\int_{u_1}^{u_2}\sqrt{u}\,du\\\\
I=\dfrac{1}{3}\int_{u_1}^{u_2}u^{\frac{1}{2}}\,du\\\\
I=\dfrac{1}{3}\left[\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{u_1}^{u_2}\\\\
I=\dfrac{1}{3}\left[\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\
I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}](https://tex.z-dn.net/?f=%5Cdisplaystyle+I%3D%5Cint_%7B-1%7D%5E0x%5E2%5Csqrt%7Bx%5E3%2B1%7D%5C%2Cdx%5C%5C%5C%5C%0A+I%3D%5Cint_%7Bu_1%7D%5E%7Bu_2%7Dx%5E2%5Csqrt%7Bu%7D%5Ccdot%5Cdfrac%7B1%7D%7B3x%5E2%7Ddu%5C%5C%5C%5C%0A+I%3D%5Cdfrac%7B1%7D%7B3%7D%5Cint_%7Bu_1%7D%5E%7Bu_2%7D%5Csqrt%7Bu%7D%5C%2Cdu%5C%5C%5C%5C%0A+I%3D%5Cdfrac%7B1%7D%7B3%7D%5Cint_%7Bu_1%7D%5E%7Bu_2%7Du%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%2Cdu%5C%5C%5C%5C%0A+I%3D%5Cdfrac%7B1%7D%7B3%7D%5Cleft%5B%5Cdfrac%7Bu%5E%7B%5Cfrac%7B1%7D%7B2%7D%2B1%7D%7D%7B%5Cfrac%7B1%7D%7B2%7D%2B1%7D%5Cright%5D_%7Bu_1%7D%5E%7Bu_2%7D%5C%5C%5C%5C%0A+I%3D%5Cdfrac%7B1%7D%7B3%7D%5Cleft%5B%5Cdfrac%7Bu%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D_%7Bu_1%7D%5E%7Bu_2%7D%5C%5C%5C%5C%0A+I%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5Bu%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D_%7Bu_1%7D%5E%7Bu_2%7D "\displaystyle I=\int_{-1}^0x^2\sqrt{x^3+1}\,dx\\\\
I=\int_{u_1}^{u_2}x^2\sqrt{u}\cdot\dfrac{1}{3x^2}du\\\\
I=\dfrac{1}{3}\int_{u_1}^{u_2}\sqrt{u}\,du\\\\
I=\dfrac{1}{3}\int_{u_1}^{u_2}u^{\frac{1}{2}}\,du\\\\
I=\dfrac{1}{3}\left[\dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]_{u_1}^{u_2}\\\\
I=\dfrac{1}{3}\left[\dfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\
I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}")
Voltando à variável x:
![I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\
I=\dfrac{2}{9}\left[(x^3+1)^{\frac{3}{2}}\right]_{-1}^{0}\\\\
I=\dfrac{2}{9}\left[(0^3+1)^{\frac{3}{2}}-((-1)^3+1)^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}\left[(0+1)^{\frac{3}{2}}-(-1+1)^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}[1-0]\\\\
I=\dfrac{2}{9}\cdot1\\\\
I=\dfrac{2}{9}\\\\
\displaystyle
\boxed{\int_{-1}^0x^2\sqrt{x^3+1}\,dx=\dfrac{2}{9}}\Longrightarrow \text{Letra }\bold{D}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5Bu%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D_%7Bu_1%7D%5E%7Bu_2%7D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5B%28x%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D_%7B-1%7D%5E%7B0%7D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5B%280%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D-%28%28-1%29%5E3%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5B%280%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D-%28-1%2B1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5Cleft%5B1%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D-0%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5B1-0%5D%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5Ccdot1%5C%5C%5C%5C%0AI%3D%5Cdfrac%7B2%7D%7B9%7D%5C%5C%5C%5C%0A%5Cdisplaystyle%0A%5Cboxed%7B%5Cint_%7B-1%7D%5E0x%5E2%5Csqrt%7Bx%5E3%2B1%7D%5C%2Cdx%3D%5Cdfrac%7B2%7D%7B9%7D%7D%5CLongrightarrow+%5Ctext%7BLetra+%7D%5Cbold%7BD%7D "I=\dfrac{2}{9}\left[u^{\frac{3}{2}}\right]_{u_1}^{u_2}\\\\
I=\dfrac{2}{9}\left[(x^3+1)^{\frac{3}{2}}\right]_{-1}^{0}\\\\
I=\dfrac{2}{9}\left[(0^3+1)^{\frac{3}{2}}-((-1)^3+1)^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}\left[(0+1)^{\frac{3}{2}}-(-1+1)^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}\left[1^{\frac{3}{2}}-0^{\frac{3}{2}}\right]\\\\
I=\dfrac{2}{9}[1-0]\\\\
I=\dfrac{2}{9}\cdot1\\\\
I=\dfrac{2}{9}\\\\
\displaystyle
\boxed{\int_{-1}^0x^2\sqrt{x^3+1}\,dx=\dfrac{2}{9}}\Longrightarrow \text{Letra }\bold{D}")
Vamos fazer uma substituição:
Substituindo na integral dada:
Voltando à variável x:
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