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

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

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To find the function ( f(x) ), we need to integrate the given function ( x^2 - 3x - 2 ) with respect to ( x ). Then, we can use the given condition ( f(-1) = 2 ) to determine the constant of integration.

The integral of ( x^2 - 3x - 2 ) with respect to ( x ) is ( \frac{1}{3}x^3 - \frac{3}{2}x^2 - 2x + C ), where ( C ) is the constant of integration.

Given that ( f(-1) = 2 ), we substitute ( x = -1 ) into the function and equate it to 2: [ f(-1) = \frac{1}{3}(-1)^3 - \frac{3}{2}(-1)^2 - 2(-1) + C = 2 ]

Solving this equation for ( C ), we get: [ \frac{-1}{3} - \frac{3}{2} + 2 + C = 2 ] [ C = 2 - \left( \frac{-1}{3} - \frac{3}{2} + 2 \right) ] [ C = 2 - \frac{-1}{3} + \frac{3}{2} - 2 ] [ C = \frac{2}{3} + \frac{3}{2} - 2 ] [ C = \frac{4}{6} + \frac{9}{6} - \frac{12}{6} ] [ C = \frac{1}{6} ]

Therefore, the function ( f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 - 2x + \frac{1}{6} ).

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