Integration (1 Viewer)

[rainnwind]

At any point on the curve y=f(x) the second derivative is given by f"(x)=2x-1. The tangent at the point (0,1) on the curve has gradient 1. Find the equation of the curve.

Ans: y= (x^3)/3 - (x^2)/2 + x + 1


Thanks in advance.

 

[Trebla]

At any point on the curve y=f(x) the second derivative is given by f"(x)=2x-1. The tangent at the point (0,1) on the curve has gradient 1. Find the equation of the curve.

Ans: y= (x^3)/3 - (x^2)/2 + x + 1


Thanks in advance.

f''(x) = 2x - 1
f'(x) = x² - x + c₁
But f'(0) = 1 (as gradient is 1 at x = 0)
=> c₁ = 1
.: f'(x) = x² - x + 1
f(x) = x³/3 - x²/2 + x + c₂
f(0) = 1 (as (0,1) satisfies the equation) => c₂ = 1
.: f(x) = x³/3 - x²/2 + x + 1

 

[sl123]

Integrate second derivative.

This will give you: y'=x^2 - x + C1

Since gradient on curve is 1, therefore first derivative = 1.

Hence, C1= 1 (as x = 0)

therefore y'=x^2 -x +1


eqn of curve must therefore be the integral of the first derivative:

x^3/3 -x^2 +x +c2
since y=1
therefore c2=1

Ans: y= (x^3)/3 - (x^2)/2 + x + 1


Hope you can understand my work, quite hard to explain through typing


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LOL^^ too slow