# 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
```