Math, 04.05.2021Jelanny

Find the area of the plane region R as described ​


Find the area of the plane region R as described ​

Answers

The area of quadrilateral R is 8 square units.

Step-by-step explanation:

See the attached photo for solution.


Find the area of the plane region r as described :  r is the quadrilateral with vertices at the orig

The area is \frac{19}{6} squared unit.

Step-by-step explanation:

x-y=2\\-y=2-x\\y=x-2\\

x=y^{2} \\\sqrt{x} =\sqrt{y^{2} } \\y=\sqrt{x}

x-2=\sqrt{x} \\(x-2)^{2} =(\sqrt{x})^{2}\\x^{2} -4x+4=x\\x^{2} -5x+4=0\\(x-4)(x-1)=0\\x=4,1

So a=4 and b=1

A=|\int\limits^a_b {\sqrt{x} } \, dx -\int\limits^a_b {x-2} \, dx |\\A=|\int\limits^4_1 {\sqrt{x} } \, dx -\int\limits^4_1 {x-2} \, dx|\\A=||\frac{2}{3} x^{\frac{3}{2} } |_{1} ^{4} -|\frac{1}{2} x^{2} -2x|_{1} ^{4}|\\\\

A=|[(\frac{2}{3} (4)^{\frac{3}{2} })-(\frac{2}{3} (1)^{\frac{3}{2} }) ] -[(\frac{1}{2} (4)^{2}-2(4))-(\frac{1}{2} (1)^{2}-2(1)]|\\\\A=|[(\frac{2}{3} (8))-(\frac{2}{3} (1)) ] -[(\frac{1}{2} (16)-8)-(\frac{1}{2}-2)]|\\\\A=|[(\frac{16}{3})-(\frac{2}{3}) ] -[(8-8)-(-\frac{3}{2} )]|\\\\A=|[(\frac{14}{3}) ] -[(0)+\frac{3}{2} ]|\\\\A=|\frac{14}{3} -\frac{3}{2}|\\\\A=|\frac{19}{6} |\\\\A=\frac{19}{6}

So the area is \frac{19}{6} squared unit.

The area is 6\frac{2}{3}.

Step-by-step explanation:

A=\int\limits^1__{} -1 }xdx-\int\limits^1__{} -1 }x^{2} +3dx\\\\A=|\frac{1}{2} x^{2} |^{1} _{-1} -|\frac{1}{3}  x^{3} +3x|^{1} _{-1}\\\\A=[(\frac{1}{2} (1)^{2} )-(\frac{1}{2} (-1)^{2})]-[(\frac{1}{3}  (1)^{3} +3(1))-(\frac{1}{3}  (-1)^{3} +3(-1))]\\\\A=[(\frac{1}{2}  )-(\frac{1}{2})]-[(\frac{1}{3} +3)-(-\frac{1}{3} -3)]\\\\A=-(-\frac{2}{3} -6)\\\\A=6\frac{2}{3}

The area is 12 square unit.

Step-by-step explanation:

A=|\int\limits^2_0 {3x^{3}-x^{2} -10x } \, dx -\int\limits^2_0 {2x-x^{2} } \, dx |\\A=||\frac{3}{4} x^{4} -\frac{1}{3} x^{3} -5x^{2} |_{0} ^{2} -|x^{2} -\frac{1}{3} x^{3} |_{0} ^{2} |\\\\A=|[\frac{3}{4} (2)^{4} -\frac{1}{3} (2)^{3} -5(2)^{2} ] -[(2)^{2} -\frac{1}{3} (2)^{3} ]|\\\\A=|[\frac{3}{4} (16) -\frac{1}{3} (8) -5(4) ] -[(4) -\frac{1}{3} (8) ]|\\\\A=|[12 -\frac{8}{3}  -20 ] -[4 -\frac{8}{3} ]|\\\\A=|12 -\frac{8}{3}  -20  -4 +\frac{8}{3} |\\A=|-12|\\A=12

So the area is 12 square unit.

The area is 1/12 square unit.

Step-by-step explanation:

A=\int\limits^1_0 {x^{2} } \, dx -\int\limits^1_0 {x^{3} } \, dx \\A=[\frac{1}{3}x^{3}  ]_{0} ^{1} -[\frac{1}{4}x^{4}  ]_{0} ^{1} \\A=(\frac{1}{3} )-(\frac{1}{4} )\\A=\frac{1}{12}

By doing that in sketch paper, the answer is 12 squared unit.

11/3 quare units

Step-by-step explanation:

A=|\int\limits^2_1 {2x-x^{2} } \, dx -\int\limits^2_1 {x^{2}-4x+4 } \, dx |\\A=||x^{2} -\frac{1}{3} x^{3}|^{2} _{1} -|\frac{1}{3} x^{3} -2x^{2} +4x|^{2} _{1} |\\\\A=|[((2)^{2} -\frac{1}{3} (2)^{3})-((1)^{2} -\frac{1}{3} (1)^{3})] -||\frac{1}{3} x^{3} -2x^{2} +4x|^{2} _{1} |\\\\A=|[(4 -\frac{8}{3})-(1 -\frac{1}{3})]-||\frac{1}{3} x^{3} -2x^{2} +4x|^{2} _{1} |\\\\

A=|(3 -\frac{7}{3})-[(\frac{1}{3} (2)^{3} -2(2)^{2} +4(2))-(\frac{1}{3} (1)^{3} -2(1)^{2} +4(1))] |\\A=|\frac{2}{3}-[(\frac{1}{3} (8) -2(4) +4(2))-(\frac{1}{3} (1) -2(1) +4)] |\\\\A=|\frac{2}{3}-[(\frac{8}{3} -8 +8)-(\frac{1}{3} -2+4)] |\\\\A=|\frac{2}{3}-[(\frac{8}{3})-(\frac{1}{3}+2)] |\\A=|\frac{2}{3}-(\frac{8}{3}-\frac{1}{3}-2) |\\\\\\A=|\frac{2}{3}-(\frac{7}{3}+2) |\\A=|\frac{2}{3}-\frac{7}{3}-2 |\\\\A=|-\frac{5}{3}-2 |\\\\A=|-\frac{11}{3}|\\A=\frac{11}{3}

So the area is 11/3.



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