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

1 Answer
Pitufox27
Oct 10, 2016

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.

Explanation:

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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