# What is #f(x) = int x^3-2x+e^x dx# if #f(2) = 5 #?

therefore from the first equation

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To find ( f(x) = \int x^3 - 2x + e^x , dx ) given ( f(2) = 5 ):

First, find the antiderivative of ( x^3 - 2x + e^x ):

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

Combine these antiderivatives:

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

Given ( f(2) = 5 ):

[ 5 = \frac{1}{4}(2^4) - 2^2 + e^2 + C ] [ 5 = \frac{1}{4}(16) - 4 + e^2 + C ] [ 5 = 4 - 4 + e^2 + C ] [ 5 = e^2 + C ]

Subtract ( e^2 ) from both sides:

[ C = 5 - e^2 ]

Thus, ( f(x) = \frac{1}{4}x^4 - x^2 + e^x + (5 - e^2) ).

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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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