Пошаговое объяснение:
[tex]y^2=4x\ \ \ \ x+y=3\ \ \ \ y\geq 0\ \ \ \ y=0\ \ \ \ S=?\\y^2=4x\\y=\sqrt{4x}=2\sqrt{x} \\y=3-x.\ \ \ \ \ \\ 2\sqrt{x} =3-x\\(2\sqrt{x})^2=(3-x)^2\\ 4x=9-6x+x^2\\x^2-10x+9=0\\D=64\ \ \ \ \ \sqrt{D}=8\\ x_1=1\ \ \ \ y_1=3-1=2\\x_2=9\ \ \ \ y_2=3-9=-6\notin\\0=3-x\\x=3\\[/tex]
[tex]S=\int\limits^1_0 {(2\sqrt{x} -0)} \, dx+\int\limits^3_1 {(3-x-0)} \, dx=2*\int\limits^1_0 {\sqrt{x} } \, dx +\int\limits^3_1 {(3-x)} \, dx =\\=2*\frac{2}{3} *\sqrt{x^3} \ |_0^1+\int\limits^3_1 {3} \, dx -\int\limits^3_1 {x} \, dx= \frac{4}{3} *\sqrt{1}-\frac{4}{3}*\sqrt{0} +3x\ |_1^3-\frac{x^2}{2}\ |_1^3=\\ =\frac{4}{3}+3*3-3*1-(\frac{3^2}{2} -\frac{1^2}{2})=1\frac{1}{3} +9-3-(\frac{9}{2}-\frac{1}{2} )=7\frac{1}{3}-4=3\frac{1}{3} .[/tex]
Ответ: S=3,333333 кв. ед.
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Пошаговое объяснение:
[tex]y^2=4x\ \ \ \ x+y=3\ \ \ \ y\geq 0\ \ \ \ y=0\ \ \ \ S=?\\y^2=4x\\y=\sqrt{4x}=2\sqrt{x} \\y=3-x.\ \ \ \ \ \\ 2\sqrt{x} =3-x\\(2\sqrt{x})^2=(3-x)^2\\ 4x=9-6x+x^2\\x^2-10x+9=0\\D=64\ \ \ \ \ \sqrt{D}=8\\ x_1=1\ \ \ \ y_1=3-1=2\\x_2=9\ \ \ \ y_2=3-9=-6\notin\\0=3-x\\x=3\\[/tex]
[tex]S=\int\limits^1_0 {(2\sqrt{x} -0)} \, dx+\int\limits^3_1 {(3-x-0)} \, dx=2*\int\limits^1_0 {\sqrt{x} } \, dx +\int\limits^3_1 {(3-x)} \, dx =\\=2*\frac{2}{3} *\sqrt{x^3} \ |_0^1+\int\limits^3_1 {3} \, dx -\int\limits^3_1 {x} \, dx= \frac{4}{3} *\sqrt{1}-\frac{4}{3}*\sqrt{0} +3x\ |_1^3-\frac{x^2}{2}\ |_1^3=\\ =\frac{4}{3}+3*3-3*1-(\frac{3^2}{2} -\frac{1^2}{2})=1\frac{1}{3} +9-3-(\frac{9}{2}-\frac{1}{2} )=7\frac{1}{3}-4=3\frac{1}{3} .[/tex]
Ответ: S=3,333333 кв. ед.