A curve has dy/dx=3x^2-2x. The curve passes through the point (2;5). What is the equation of the curve?

Expert Answers

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To find the curve we integrate the given dy/dx = 3x^2 - 2x.

Int [ 3x^2 - 2x dx ]

=> 3*x^3 / 3 - 2x^2 / 2 + C

=> x^3 -  x^2 + C

As the curve passes through (2 , 5)

5 = 2^3 - 2^2 + C

=> 5 = 8 - 4 + C

=> C = 1

The required curve is y = x^3 - x^2 + 1

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