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Question Number 66731 by aditya730222@gmail.com last updated on 19/Aug/19

$$y=x^2-3xy=2xfindarea$$

Commented by Tony Lin last updated on 19/Aug/19

$$x^2-3x=2x$$$$x(x-5)=0$$$${\int }_0^5[2x-(x^2-3x)]dx$$$$={\int }_0^5(-x^2+5x)dx$$$$={[-\frac{1}{3}{x}^3+\frac{5}{2}{x}^2]}_0^5$$$$=\frac{125}{6}$$

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