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What is #f(x) = int (3x-2)^2-5x+1 dx# if #f(2) = 1 #?

Answer 1

#f(x)=3x^3-17/2 x^2+5x+1#

The indefinite integral is

#int ((3x-2)^2-5x+1)\ dx=int (9x^2-17x+5)\ dx#

#=3x^3-17/2 x^2+5x+C#

Therefore, #f(x)=3x^3-17/2 x^2+5x+C# for some #C#. Since #f(2)=1#, we must choose the value of #C# so that #1=3*2^3-17/2 * 2^2+5*2+C#. This implies that #1=24-34+10+C# so that #C=1#.

Hence, the answer is #f(x)=3x^3-17/2 x^2+5x+1#.

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Answer 2

Scarlett Allison

To find ( f(x) ), integrate the given function ( \int (3x-2)^2 - 5x + 1 , dx ) with respect to ( x ), then use the given condition ( f(2) = 1 ) to solve for the constant of integration.

Given that ( f(2) = 1 ), substitute ( x = 2 ) into the function and solve for the constant of integration.

[ f(2) = \int_{c}^{2} (3x-2)^2 - 5x + 1 , dx = 1 ]

After solving the integral, you'll have ( f(x) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.