What is #f(x) = int -x^2+x-4 dx# if #f(2) = -1 #?

Answer 1

We must evaluate the primitive and then substitute #x# value to obtain #- 1#. This becomes a first degree equation over C and we can obtain the value of the constant of integration.

First we solve the primitive:

#f (x) = int (- x^2+x-4) dx=- 1/3 x^3 + 1/2 x^2 - 4x + C#
Then, if #f(2)=- 1#, substituing the x value in the expression of the primitive and equals to the #f (x)# value, we obtain:
#f (2) = - 1/3 (2)^3 + 1/2 (2)^2 - 4 (2) + C = - 1#

Thus:

#- 8/3 + 2 - 8 + C = -1 rArr C = 23/3#
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Answer 2

To find the value of ( f(x) = \int -x^2 + x - 4 , dx ) given that ( f(2) = -1 ), you need to evaluate the definite integral of the given function from an arbitrary constant ( a ) to ( x ) and then use the given condition to solve for ( a ).

Starting with the indefinite integral:

[ \int -x^2 + x - 4 , dx = -\frac{1}{3}x^3 + \frac{1}{2}x^2 - 4x + C ]

Now, substitute ( x = 2 ) and ( f(2) = -1 ) into the equation and solve for ( C ):

[ -\frac{1}{3}(2)^3 + \frac{1}{2}(2)^2 - 4(2) + C = -1 ] [ -\frac{8}{3} + 2 - 8 + C = -1 ] [ C = -1 + \frac{8}{3} - 2 + 8 ] [ C = \frac{1}{3} ]

So, the value of ( C ) is ( \frac{1}{3} ).

Thus, the function ( f(x) = \int -x^2 + x - 4 , dx ) with ( f(2) = -1 ) is:

[ f(x) = -\frac{1}{3}x^3 + \frac{1}{2}x^2 - 4x + \frac{1}{3} ]

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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.

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.

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.

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.

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