Determine a área da região compreendida entre a reta y=6 e a curva y=x2 para -6 ≤ x ≤ 6
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Boa tarde Marli!
Solução!
Primeiro vamos determinar o intervalo de integração!
![y= x^{2} \\\\\\
y=6\\\\\\
x^{2} =6\\\\\
x= \sqrt{6}\\\\\\
Logo!
\\\\\\
intervalo=\ [0,\sqrt {6} ]](https://tex.z-dn.net/?f=y%3D+x%5E%7B2%7D+%5C%5C%5C%5C%5C%5C%0Ay%3D6%5C%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D+%3D6%5C%5C%5C%5C%5C%0Ax%3D+%5Csqrt%7B6%7D%5C%5C%5C%5C%5C%5C%0ALogo%21%0A%5C%5C%5C%5C%5C%5C%0Aintervalo%3D%5C+%5B0%2C%5Csqrt+%7B6%7D+%5D "y= x^{2} \\\\\\
y=6\\\\\\
x^{2} =6\\\\\
x= \sqrt{6}\\\\\\
Logo!
\\\\\\
intervalo=\ [0,\sqrt {6} ]")
![A=\displaystyle \int_{0}^{ \sqrt{6}} (x^{2}) dx\\\\\\\\\\\
A=\displaystyle \int_{0}^{ \sqrt{6}} ( \frac{ x^{2+1} }{2+1} ) dx= \frac{ x^{3} }{3}+c \\\\\\\\\\
A= \frac{ x^{3}}{3}~~\bigg|^{ \sqrt{6} }_{0}\\\\\\\
A=\dfrac{ x^{3}}{3}-\frac{ x^{3}}{3}\\\\\\
A=\dfrac{ (\sqrt{6}) ^{3}}{3})-(\frac{(0)^{3}}{3}\\\\\\
A= \frac{(2,4)^{3} }{3}- \frac{(0)^{3} }{3}\\\\\\\\
A= \dfrac{13,824}{3} -0\\\\\\
\boxed{A=4,6~u~a}](https://tex.z-dn.net/?f=A%3D%5Cdisplaystyle+%5Cint_%7B0%7D%5E%7B+%5Csqrt%7B6%7D%7D++%28x%5E%7B2%7D%29+dx%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A+A%3D%5Cdisplaystyle+%5Cint_%7B0%7D%5E%7B+%5Csqrt%7B6%7D%7D++%28+%5Cfrac%7B+x%5E%7B2%2B1%7D+%7D%7B2%2B1%7D+%29+dx%3D+%5Cfrac%7B+x%5E%7B3%7D+%7D%7B3%7D%2Bc+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0AA%3D+%5Cfrac%7B+x%5E%7B3%7D%7D%7B3%7D%7E%7E%5Cbigg%7C%5E%7B+%5Csqrt%7B6%7D+%7D_%7B0%7D%5C%5C%5C%5C%5C%5C%5C%0A%0AA%3D%5Cdfrac%7B+x%5E%7B3%7D%7D%7B3%7D-%5Cfrac%7B+x%5E%7B3%7D%7D%7B3%7D%5C%5C%5C%5C%5C%5C%0A%0AA%3D%5Cdfrac%7B+%28%5Csqrt%7B6%7D%29+%5E%7B3%7D%7D%7B3%7D%29-%28%5Cfrac%7B%280%29%5E%7B3%7D%7D%7B3%7D%5C%5C%5C%5C%5C%5C%0AA%3D+%5Cfrac%7B%282%2C4%29%5E%7B3%7D+%7D%7B3%7D-+%5Cfrac%7B%280%29%5E%7B3%7D+%7D%7B3%7D%5C%5C%5C%5C%5C%5C%5C%5C%0A+%0AA%3D+%5Cdfrac%7B13%2C824%7D%7B3%7D+-0%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BA%3D4%2C6%7Eu%7Ea%7D+%0A%0A++ "A=\displaystyle \int_{0}^{ \sqrt{6}} (x^{2}) dx\\\\\\\\\\\
A=\displaystyle \int_{0}^{ \sqrt{6}} ( \frac{ x^{2+1} }{2+1} ) dx= \frac{ x^{3} }{3}+c \\\\\\\\\\
A= \frac{ x^{3}}{3}~~\bigg|^{ \sqrt{6} }_{0}\\\\\\\
A=\dfrac{ x^{3}}{3}-\frac{ x^{3}}{3}\\\\\\
A=\dfrac{ (\sqrt{6}) ^{3}}{3})-(\frac{(0)^{3}}{3}\\\\\\
A= \frac{(2,4)^{3} }{3}- \frac{(0)^{3} }{3}\\\\\\\\
A= \dfrac{13,824}{3} -0\\\\\\
\boxed{A=4,6~u~a}")
Boa tarde!
Bons estudos!
Solução!
Primeiro vamos determinar o intervalo de integração!
Boa tarde!
Bons estudos!
nannymaria:
Determine a área da região compreendida entre a reta y=6 e a curva y=x²/6 para -6 ≤ x ≤ 6.
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