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> | ||
<math> \int_{0}^{9} \left(1+\sqrt{x} - \frac{3+x}{3}\right)dx = x + | <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
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 }