Find the area between the curves y^2 - y + x = 0 and y^2 = x + 3.
A=∫ab∣f(x)−g(x)∣dxA=\int _a^b|f\left(x\right)-g\left(x\right)|dxA=∫ab∣f(x)−g(x)∣dx
=∫−3−2∣f3(x)−f4(x)∣dx+∫−2−34∣f2(x)−f3(x)∣dx+∫−3414∣f1(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=∫−3−2∣f3(x)−f4(x)∣dx+∫−2−43∣f2(x)−f3(x)∣dx+∫−4341∣f1(x)−f2(x)∣dx
=∫−3−2∣x+3−(−x+3)∣dx+∫−2−34∣1−1−4x2−x+3∣dx+∫−3414∣1+1−4x2−1−1−4x2∣=\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|=∫−3−2∣∣x+3−(−x+3)∣∣dx+∫−2−43∣∣21−1−4x−x+3∣∣dx+∫−4341∣∣21+1−4x−21−1−4x∣∣
=43+6124+43=\frac{4}{3}+\frac{61}{24}+\frac{4}{3}=34+2461+34
=12524=\frac{125}{24}=24125
A= 5.2083
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