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

1 Answer
paho
Dec 20, 2017

#f(x) = x^3/3 - (3 x^2) / 2 - 2x + 11/6#

Explanation:

Let's first solve the integral:

# int x^2 - 3x - 2 dx = int x^2 dx - 3 int x dx - 2 int 1 dx = x^3/3 - (3 x^2) / 2 - 2x + C #

We know that #f(-1) = 2#. Therefore, we set #x = -1# and solve the following equation:

#- 1/3 -3/2 + 2 + C = 2#.

Solving this, we find that #C = 11/6#, and thus

#f(x) = x^3/3 - (3 x^2) / 2 - 2x + 11/6#.

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