# What is #f(x) = int e^(2x)-2e^x+3x dx# if #f(4 ) = 2 #?

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To find ( f(x) = \int e^{2x} - 2e^x + 3x , dx ) given that ( f(4) = 2 ):

First, calculate the indefinite integral of ( e^{2x} - 2e^x + 3x ) with respect to ( x ).

[ \int (e^{2x} - 2e^x + 3x) , dx ]

[ = \frac{1}{2} e^{2x} - 2e^x + \frac{3}{2}x^2 + C ]

Then, use the given condition to solve for the constant ( C ):

[ f(4) = 2 ]

[ \frac{1}{2} e^{2 \cdot 4} - 2e^4 + \frac{3}{2}(4)^2 + C = 2 ]

[ \frac{1}{2} e^{8} - 2e^4 + 24 + C = 2 ]

[ \frac{1}{2} e^{8} - 2e^4 + 24 + C - 2 = 0 ]

[ \frac{1}{2} e^{8} - 2e^4 + 22 + C = 0 ]

[ C = -\frac{1}{2} e^{8} + 2e^4 - 22 ]

Finally, substitute ( C ) back into the indefinite integral to obtain ( f(x) ):

[ f(x) = \frac{1}{2} e^{2x} - 2e^x + \frac{3}{2}x^2 -\frac{1}{2} e^{8} + 2e^4 - 22 ]

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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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