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