6.1 Areas Between Curves/7: Difference between revisions

Latest revision as of 19:51, 20 September 2022

$f(x)=x$

$(0,0)$ $(1,1)$

$g(x)=x^{2}$

$\int _{0}^{1}\left(R-r\right)dx$

$R=x$

$r=x^{2}$

$\int _{0}^{1}\left(x-x^{2}\right)dx$

${\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}{\bigg |}_{0}^{1}$

$={\frac {1^{2}}{2}}-{\frac {1^{3}}{3}}-{\frac {0^{2}}{2}}-{\frac {0^{3}}{3}}$

$={\frac {1}{6}}$

@@ Line 1: / Line 1: @@
 <math>f(x)=x</math>
+<math>(0,0)</math>
+<math>(1,1)</math>
 <math>g(x)=x^2</math>
-<math>\int_{a}^{b}\(R-r)dx</math>
+<math>\int_{0}^{1}\left(R-r\right)dx</math>
 <math>R=x</math>
 <math>r=x^2</math>
-<math>\int_{0}^{1}\(x-x^2)dx</math>
+<math>\int_{0}^{1}\left(x-x^2\right)dx</math>
 <math>\frac{x^2}{2}-\frac{x^3}{3}\bigg|_{0}^{1}</math>
-<math>\frac{1^2}{2}-\frac{1^3}{3}-\frac{0^2}{2}-\frac{0^3}{3}
-<math\>\frac{1}{6}/math>
+<math>=\frac{1^2}{2}-\frac{1^3}{3}-\frac{0^2}{2}-\frac{0^3}{3}</math>
+<math>=\frac{1}{6}</math>
+[[File:Desmos-graph200.png|left|200 px|]]