6.1 Areas Between Curves/10: Difference between revisions

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\begin{align}
\begin{align}
&1+\sqrt{x} = \frac{3+x}{3} \\
&1+\sqrt{x} = \frac{3+x}{3} \\
& 1+\sqrt{x}-\frac{3+x}{3} = 0 \\
& 1+\sqrt{x}-\frac{3+x}{3} = 0 \\
& \frac{3+3\sqrt{x}}{3}-\frac{3+x}{3} = 0 \\
& \frac{3+3\sqrt{x}}{3}-\frac{3+x}{3} = 0 \\
& 3+3\sqrt{x}-3+x = 0 \\
& 3+3\sqrt{x}-3+x = 0 \\
& 3\sqrt{x}+x = 0 \\
& 3\sqrt{x}+x = 0 \\
& 3\sqrt{x} = -x \\
& 3\sqrt{x} = -x \\
& 9x = x^2 \\
& 9x = x^2 \\
& 9x-x^2 = 0 \\
& 9x-x^2 = 0 \\
& x(9-x) = 0 \\
& x(9-x) = 0 \\
& x = 0,9  
& x = 0,9  
\end{align}
\end{align}
</math>
</math>


\left[8x-\frac{2x^3}{3}\right]\Bigg|_{-2}^{2} \\[2ex]
 
<math> \int_{0}^{9} \left(1+\sqrt{x} - \frac{3+x}{3}\right)dx = x + [\frac{2x^\frac{3}{2}{3}]\Bigg|_{0}^{9} - [\frac{1}{3}\]\Bigg|_{0}^{9}3+x </math>
<math> \int_{0}^{9} \left(1+\sqrt{x} - \frac{3+x}{3}\right)dx = x + \frac{2x^\frac{3}{2}}3\Bigg|_{0}^{9} - \frac{1}3\int_{0}^{9}3+xdx = x + \frac{2x^\frac{3}{2}}3\Bigg|_{0}^{9} - \frac{{1}}3(\frac{{x^2}}3+3x)\Bigg|_{0}^{9} = 9 + \frac{{2(27)}}3 - \frac{{9^2}}2(3) - 9 = \frac{54}}3 -  \frac{81}}6 =  \frac{108}}6 - \frac{81}}6 = \frac{27}}6 =  \frac{9}}2 </math>

Revision as of 19:23, 20 September 2022

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Failed to parse (syntax error): {\displaystyle \int_{0}^{9} \left(1+\sqrt{x} - \frac{3+x}{3}\right)dx = x + \frac{2x^\frac{3}{2}}3\Bigg|_{0}^{9} - \frac{1}3\int_{0}^{9}3+xdx = x + \frac{2x^\frac{3}{2}}3\Bigg|_{0}^{9} - \frac{{1}}3(\frac{{x^2}}3+3x)\Bigg|_{0}^{9} = 9 + \frac{{2(27)}}3 - \frac{{9^2}}2(3) - 9 = \frac{54}}3 - \frac{81}}6 = \frac{108}}6 - \frac{81}}6 = \frac{27}}6 = \frac{9}}2 }