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What is #f(x) = int 3x^2-4 dx# if #f(2) = 2 #?

Answer 1

Evelyn Agard

#f(x)=x^3-4x+2#

First, find the antiderivative of the function using the product rule in reverse:

#intax^ndx=(ax^(n+1))/(n+1)+C#

Thus,

#f(x)=(3x^(2+1))/(2+1)-(4x^(0+1))/(0+1)+C=(3x^3)/3-(4x)/1+C=x^3-4x+C#

Since #f(x)=x^3-4x+C#, and we know that #f(2)=2#, we can determine #C# (the constant of integration).

#f(2)=2^3-4(2)+C=2#

#8-8+C=2#

#C=2#

Thus,

#f(x)=x^3-4x+2#

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Answer 2

Axel Alvarez

To find ( f(x) = \int (3x^2 - 4) , dx ) if ( f(2) = 2 ), you first need to find the indefinite integral of ( 3x^2 - 4 ) with respect to ( x ), which is ( x^3 - 4x + C ), where ( C ) is the constant of integration. Then, evaluate ( f(x) ) at ( x = 2 ) and equate it to 2:

( f(2) = (2)^3 - 4(2) + C = 2 )

Solve for ( C ):

( 8 - 8 + C = 2 )

( C = 2 )

So, the integral ( f(x) ) is ( x^3 - 4x + 2 ).

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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.