# What is #f(x) = int (x-2)^2-5x+4 dx# if #f(2) = 1 #?

#f(x) = 1/3 (x-2)^3 -5/2 x^2 + 4x + 3 #

is the standard integral , where c is the constant of integration

These can be applied 'term by term' to the above integral

using f(2) = 1 , allows c , to be found

0 - 10 + 8 + c = 1 → c = 3

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To find ( f(x) ), integrate ( (x - 2)^2 - 5x + 4 ) with respect to ( x ).

Given that ( f(2) = 1 ), you can find the constant of integration.

( f(x) = \int (x - 2)^2 - 5x + 4 , dx )

( f(x) = \int (x^2 - 4x + 4) - 5x + 4 , dx )

( f(x) = \int x^2 - 9x + 8 , dx )

( f(x) = \frac{1}{3}x^3 - \frac{9}{2}x^2 + 8x + C )

Given ( f(2) = 1 ),

( 1 = \frac{1}{3}(2)^3 - \frac{9}{2}(2)^2 + 8(2) + C )

( 1 = \frac{8}{3} - 9(2) + 16 + C )

( 1 = \frac{8}{3} - 18 + 16 + C )

( 1 = \frac{8}{3} - 2 + C )

( 1 = \frac{8}{3} - \frac{6}{3} + C )

( 1 = \frac{2}{3} + C )

( C = 1 - \frac{2}{3} )

( C = \frac{3}{3} - \frac{2}{3} )

( C = \frac{1}{3} )

Therefore, ( f(x) = \frac{1}{3}x^3 - \frac{9}{2}x^2 + 8x + \frac{1}{3} ).

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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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