# What is #F(x) = int x-xe^(-x) dx# if #F(0) = 2 #?

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To find ( F(x) ), we need to find the indefinite integral of ( x - xe^{-x} ) with respect to ( x ). The integral can be found by integrating each term separately.

[ \int x , dx = \frac{x^2}{2} + C_1 ]

[ \int xe^{-x} , dx ]

To find the integral of ( xe^{-x} ), we can use integration by parts. Let ( u = x ) and ( dv = e^{-x} , dx ). Then, ( du = dx ) and ( v = -e^{-x} ).

[ \int xe^{-x} , dx = -xe^{-x} - \int -e^{-x} , dx ]

[ = -xe^{-x} + e^{-x} + C_2 ]

Therefore, the integral of ( x - xe^{-x} ) is ( \frac{x^2}{2} - xe^{-x} + e^{-x} + C ), where ( C = C_1 + C_2 ).

Given that ( F(0) = 2 ), we can find ( C ):

[ F(0) = \frac{0^2}{2} - 0e^{-0} + e^{-0} + C = 2 ]

[ 0 + 1 + C = 2 ]

[ C = 1 ]

Thus, ( F(x) = \frac{x^2}{2} - xe^{-x} + e^{-x} + 1 ).

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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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