Find the equation of a curve that passes through the point (1;4) and dy/dx=4x^3+4x

Expert Answers

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We have dy/dx = 4x^3 + 4x.

dy/dx = 4x^3 + 4x

=> dy = (4x^3 + 4x) dx

Integrate both the sides

Int [ dy ] = Int [ (4x^3 + 4x) dx ]

=> y = 4x^4 / 4 + 4*x^2 / 2

=> y = x^4 + 2*x^2 + C

As the curve passes through (1 , 4)

4 = 1^4 + 2*1^2 + C

=> 4 = 1 + 2 + C

=> C = 1

This gives y = x^4 + 2*x^2 + 1

The required equation of the curve is y = x^4 + 2*x^2 + 1

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