6.1 Areas Between Curves/12: Difference between revisions

From Burton Tech. Points Wiki
Jump to navigation Jump to search
(Created page with "<math> \begin{align} & \color{red}\mathbf{y=x^2} & \color{royalblue}\mathbf{y=4x-x^2} \\ & x=0 & x=2 \\ \end{align} </math> <math>\int_{0}^{2} \left|(8-x^2) - (x^2)\right|dx </math> <math> \begin{align} 8-x^2 &= x^2 \\ -2x^2 &= -8 \\ x^2 &= 4 \\ x &= \pm2 \end{align} </math> <math>\int_{-3}^{3} \left|(8-x^2) - (x^2)\right|dx = \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx + \int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx + \int_{2}^{3}\left((x^2)-(8-x^2)\right)dx =...")
 
No edit summary
 
(26 intermediate revisions by the same user not shown)
Line 1: Line 1:
[[File:Desmos graph 6.1.12.png|right|350px|]]
<math>
<math>
\begin{align}
\begin{align}
Line 11: Line 12:




<math>\int_{0}^{2} \left|(8-x^2) - (x^2)\right|dx </math>
<math>\int_{0}^{2} \left[(4x-x^2) - (x^2)\right] dx</math>


<math>  
<math>  
\begin{align}
\begin{align}
8-x^2 &= x^2 \\
4-x^2 &= x^2 \\
-2x^2 &= -8 \\
4x-2x^2 &= 0 \\
  x^2 &= 4 \\
  2x(2-x) &= 0 \\
x &= \pm2
x &= 0& x = 2
\end{align}
\end{align}
</math>
</math>




<math>\int_{-3}^{3} \left|(8-x^2) - (x^2)\right|dx = \int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx + \int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx + \int_{2}^{3}\left((x^2)-(8-x^2)\right)dx = \frac{14}{3} + \frac{64}{3} + \frac{14}{3} = \frac{92}{3}</math>
<math>\int_{0}^{2} \left[(4x-x^2) - (x^2)\right]dx = \int_{0}^{2} (4x)dx</math>
 


<math>
<math>
\begin{align}
\begin{align}
\int_{-3}^{-2}\left((x^2)-(8-x^2)\right)dx &= \int_{-3}^{-2}\left(2x^2-8)\right)dx \\[2ex]
&= \left[\frac{4x^2}{2}\right]\Bigg|_{0}^{2} \\[2ex]
 
&= \left[2x^2\right]\Bigg|_{0}^{2} \\[2ex]
&= \left[\frac{2x^3}{3}-8x\right]\Bigg|_{-3}^{-2} \\[2ex]
&= \left[2(2)^2\right]-\left[2(0)^2\right] = 8-0 \\[2ex]
 
&= 8
&= \left[\frac{2(-2)^3}{3}-8(-2)\right]-\left[\frac{2(-3)^3}{3}-8(-3)\right] \\[2ex]
 
&= \left[\frac{-16}{3}+16\right]-\left[\frac{-54}{3}+24\right] = \frac{38}{3}-8 \\[2ex]
 
&= \frac{14}{3}
 
\end{align}
</math>
 
 
<math>
\begin{align}
 
\int_{-2}^{2} \left((8-x^2) - (x^2)\right)dx &= \int_{-2}^{2}\left(8-2x^2\right)dx \\[2ex]
&= \left[8x-\frac{2x^3}{3}\right]\Bigg|_{-2}^{2} \\[2ex]
&= \left[8(2)-\frac{2(2)^3}{3}\right] - \left[8(-2)-\frac{2(-2)^3}{3}\right] \\[2ex]
&= \left[16-\frac{16}{3}\right]-\left[-16+\frac{16}{3}\right] = 32-\frac{32}{3} \\[2ex]
&= \frac{64}{3}


\end{align}
\end{align}
</math>
</math>

Latest revision as of 23:30, 22 September 2022

Desmos graph 6.1.12.png



Retrieved from "https://wiki.btechp.org/index.php?title=6.1_Areas_Between_Curves/12&oldid=4867"