qual o valor da area destacada na figura abaixo;
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Solução!
![A=\displaystyle \int_{1}^{4} (\sqrt{x})dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}} dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\
A=\frac{( x)^{ \frac{3}{2}} }{ \frac{3}{2}}~~\bigg|_{1}^{4} \\\\\\\\\\
A= \frac{ \sqrt{x^{3} } }{ \frac{3}{2} } ~~\bigg|_{1}^{4} \\\\\\\](https://tex.z-dn.net/?f=A%3D%5Cdisplaystyle+%5Cint_%7B1%7D%5E%7B4%7D+%28%5Csqrt%7Bx%7D%29dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D%5Cdisplaystyle+%5Cint_%7B1%7D%5E%7B4%7D+%5Cfrac%7B%28+x%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D%2B1+%7D+%7D%7B+%5Cfrac%7B1%7D%7B2%7D%7D+dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D%5Cdisplaystyle+%5Cint_%7B1%7D%5E%7B4%7D+%5Cfrac%7B%28+x%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D%2B1+%7D+%7D%7B+%5Cfrac%7B1%7D%7B2%7D%2B1%7D+dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D%5Cdisplaystyle+%5Cint_%7B1%7D%5E%7B4%7D+%5Cfrac%7B%28+x%29%5E%7B+%5Cfrac%7B1%7D%7B2%7D%2B1+%7D+%7D%7B+%5Cfrac%7B1%7D%7B2%7D%2B1%7D+dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D%5Cfrac%7B%28+x%29%5E%7B+%5Cfrac%7B3%7D%7B2%7D%7D+%7D%7B+%5Cfrac%7B3%7D%7B2%7D%7D%7E%7E%5Cbigg%7C_%7B1%7D%5E%7B4%7D++%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%5Cfrac%7B+%5Csqrt%7Bx%5E%7B3%7D+%7D+%7D%7B+%5Cfrac%7B3%7D%7B2%7D+%7D+%7E%7E%5Cbigg%7C_%7B1%7D%5E%7B4%7D++%5C%5C%5C%5C%5C%5C%5C+%0A%0A%0A%0A%0A%0A+%0A "A=\displaystyle \int_{1}^{4} (\sqrt{x})dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}} dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\\
A=\displaystyle \int_{1}^{4} \frac{( x)^{ \frac{1}{2}+1 } }{ \frac{1}{2}+1} dx\\\\\\\\\
A=\frac{( x)^{ \frac{3}{2}} }{ \frac{3}{2}}~~\bigg|_{1}^{4} \\\\\\\\\\
A= \frac{ \sqrt{x^{3} } }{ \frac{3}{2} } ~~\bigg|_{1}^{4} \\\\\\\")
![A= \frac{2}{3} \sqrt{ x^{3} } ~~\bigg|_{1}^{4} \\\\\\\
A= (\frac{2}{3} \sqrt{ x^{3} })-(\frac{2}{3} \sqrt{ x^{3} }) \\\\\\\
A= (\frac{2}{3} \sqrt{(4)^{3} })-(\frac{2}{3} \sqrt{(1)^{3} })\\\\\\\
A= (\frac{2}{3} \sqrt{64 })-(\frac{2}{3} \sqrt{1})\\\\\\\
A= (\frac{2}{3}.8)-(\frac{2}{3}.1)\\\\\\\
A= (\frac{16}{3})-(\frac{2}{3})\\\\\\\
A= (\frac{16-2}{3})\\\\\\\
A= \frac{14}{3}u.a](https://tex.z-dn.net/?f=A%3D++%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B+x%5E%7B3%7D+%7D+%7E%7E%5Cbigg%7C_%7B1%7D%5E%7B4%7D+%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B+x%5E%7B3%7D+%7D%29-%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B+x%5E%7B3%7D+%7D%29+%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B%284%29%5E%7B3%7D+%7D%29-%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B%281%29%5E%7B3%7D+%7D%29%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B64+%7D%29-%28%5Cfrac%7B2%7D%7B3%7D+%5Csqrt%7B1%7D%29%5C%5C%5C%5C%5C%5C%5C+%0AA%3D+%28%5Cfrac%7B2%7D%7B3%7D.8%29-%28%5Cfrac%7B2%7D%7B3%7D.1%29%5C%5C%5C%5C%5C%5C%5C+%0AA%3D+%28%5Cfrac%7B16%7D%7B3%7D%29-%28%5Cfrac%7B2%7D%7B3%7D%29%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%28%5Cfrac%7B16-2%7D%7B3%7D%29%5C%5C%5C%5C%5C%5C%5C+%0AA%3D+%5Cfrac%7B14%7D%7B3%7Du.a "A= \frac{2}{3} \sqrt{ x^{3} } ~~\bigg|_{1}^{4} \\\\\\\
A= (\frac{2}{3} \sqrt{ x^{3} })-(\frac{2}{3} \sqrt{ x^{3} }) \\\\\\\
A= (\frac{2}{3} \sqrt{(4)^{3} })-(\frac{2}{3} \sqrt{(1)^{3} })\\\\\\\
A= (\frac{2}{3} \sqrt{64 })-(\frac{2}{3} \sqrt{1})\\\\\\\
A= (\frac{2}{3}.8)-(\frac{2}{3}.1)\\\\\\\
A= (\frac{16}{3})-(\frac{2}{3})\\\\\\\
A= (\frac{16-2}{3})\\\\\\\
A= \frac{14}{3}u.a")
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Solução!
Boa noite!
Bons estudos!
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