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

Let's first evaluate the indefinite integral:

Therefore, the original integral is equal to:

Therefore,

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

First, integrate the function ( 3x^3 - 2x + xe^{x-2} ) with respect to ( x ) to find ( f(x) ):

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

[ = \frac{3}{4}x^4 - x^2 + \int xe^{x-2} , dx ]

To integrate ( xe^{x-2} ), we use integration by parts:

Let ( u = x ) and ( dv = e^{x-2} , dx ).

Then, ( du = dx ) and ( v = e^{x-2} ).

Applying integration by parts formula:

[ \int u , dv = uv - \int v , du ]

[ = x \cdot e^{x-2} - \int e^{x-2} , dx ]

[ = x \cdot e^{x-2} - e^{x-2} + C ]

So, combining everything:

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

Given that ( f(1) = 3 ), we substitute ( x = 1 ) and solve for ( C ):

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

[ 3 = \frac{3}{4} - 1 + e^{-1} - e^{-1} + C ]

[ 3 = \frac{3}{4} - 1 ]

[ 3 = -\frac{1}{4} + C ]

[ C = 3 + \frac{1}{4} ]

[ C = \frac{13}{4} ]

Therefore, ( f(x) = \frac{3}{4}x^4 - x^2 + x \cdot e^{x-2} - e^{x-2} + \frac{13}{4} ).

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