Answer to Question #317119 in Civil and Environmental Engineering for Rodz

Question #317119

Find the area between the curves y^2 - y + x = 0 and y^2 = x + 3.


1
Expert's answer
2022-03-24T01:41:12-0400

A=abf(x)g(x)dxA=\int _a^b|f\left(x\right)-g\left(x\right)|dx


=32f3(x)f4(x)dx+234f2(x)f3(x)dx+3414f1(x)f2(x)dx=\int _{-3}^{-2}\left|f_3\left(x\right)-f_4\left(x\right)\right|dx+\int _{-2}^{-\frac{3}{4}}\left|f_2\left(x\right)-f_3\left(x\right)\right|dx+\int _{-\frac{3}{4}}^{\frac{1}{4}}\left|f_1\left(x\right)-f_2\left(x\right)\right|dx


=32x+3(x+3)dx+234114x2x+3dx+34141+14x2114x2=\int _{-3}^{-2}\left|\sqrt{x+3}-\left(-\sqrt{x+3}\right)\right|dx+\int _{-2}^{-\frac{3}{4}}\left|\frac{1-\sqrt{1-4x}}{2}-\sqrt{x+3}\right|dx+\int _{-\frac{3}{4}}^{\frac{1}{4}}\left|\frac{1+\sqrt{1-4x}}{2}-\frac{1-\sqrt{1-4x}}{2}\right|


=43+6124+43=\frac{4}{3}+\frac{61}{24}+\frac{4}{3}


=12524=\frac{125}{24}


A= 5.2083


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