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