vault backup: 2023-03-24 12:29:32

AMiAL/Ćwiczenia/Zadania/Całki_Zast/Zadanie 2.md

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+# Obliczyć pole obszaru zawartego między wykresami funkcji $f$ oraz $g$.
+## 1.
+$$\begin{gathered}
+f(x)=x^{2}\\
+g(x)=x\\
+\\
+x-x^{2}=0
+\\
+\int^{1}_{0}x-x^{2}dx=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{2}-{\frac{1}{3}}=\frac{1}{6}
+\end{gathered}$$
+```desmos-graph
+left=-4; right=4;
+top=4; bottom=-4;
+---
+f(x)=x^2
+g(x)=x
+```
+## 2.
+$$\begin{gather}
+f(x)=x^{2}\\
+g(x)=3-x^{2}\\
+3-x^{2}-x^{2}=3-2x^{2}=0\\
+\Delta=24\\
+\sqrt{\Delta}=2\sqrt{6}\\
+x=\pm\frac{\sqrt{6}}{2}\\
+\int^{\frac{\sqrt{6}}{2}}_{-\frac{\sqrt{6}}{2}}3-2x^{2}dx=\left[3x- \frac{2x^{3}}{3}\right]^{\frac{\sqrt{6}}{2}}_{-\frac{\sqrt{6}}{2}}=\frac{3\sqrt{6}}{2}- \frac{12\sqrt{6}}{3}-\left(-\frac{3\sqrt{6}}{2}+\frac{12\sqrt{6}}{3}\right)=\\=\frac{6\sqrt{6}}{2}- \frac{24\sqrt{6}}{3}=3\sqrt{6}-8\sqrt{6}=-5\sqrt{6}
+\end{gather}$$
+```desmos-graph
+left=-4; right=4;
+top=4; bottom=-4;
+---
+f(x)=x^2
+g(x)=3-x^2
+```