# What is #f(x) = int e^(5x-1)+x dx# if #f(2) = 3 #?

using the chain rule. (This is left as an exercise for the reader to show)

by inspection therefore we have

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To find ( f(x) = \int e^{5x - 1} + x , dx ) given that ( f(2) = 3 ), we first need to find the antiderivative of the function ( e^{5x - 1} + x ), and then use this antiderivative to evaluate ( f(x) ) at ( x = 2 ).

The antiderivative of ( e^{5x - 1} ) is ( \frac{1}{5} e^{5x - 1} ) and the antiderivative of ( x ) is ( \frac{1}{2}x^2 ). Therefore, the antiderivative of ( e^{5x - 1} + x ) is ( \frac{1}{5} e^{5x - 1} + \frac{1}{2}x^2 ).

Now, to find ( f(x) ), we integrate this antiderivative:

[ f(x) = \frac{1}{5} \int e^{5x - 1} dx + \frac{1}{2} \int x , dx ] [ f(x) = \frac{1}{5} \left( \frac{1}{5} e^{5x - 1} \right) + \frac{1}{2} \left( \frac{1}{2} x^2 \right) + C ] [ f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + C ]

Given that ( f(2) = 3 ), we can find the value of ( C ):

[ 3 = \frac{1}{25} e^{5(2) - 1} + \frac{1}{4} (2)^2 + C ] [ 3 = \frac{1}{25} e^9 + \frac{1}{4} \times 4 + C ] [ 3 = \frac{1}{25} e^9 + 1 + C ]

Therefore, ( C = 3 - \frac{1}{25} e^9 - 1 ).

Now we substitute this value of ( C ) back into the equation for ( f(x) ):

[ f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + 3 - \frac{1}{25} e^9 ]

So, ( f(x) = \frac{1}{25} e^{5x - 1} + \frac{1}{4} x^2 + 3 - \frac{1}{25} e^9 ).

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To find ( f(x) = \int e^{5x-1} + x , dx ) given that ( f(2) = 3 ), we first need to find the indefinite integral of ( e^{5x-1} + x ) with respect to ( x ). Then, we can evaluate the function at ( x = 2 ) and solve for the constant of integration.

Given: [ f(x) = \int e^{5x-1} + x , dx ]

First, let's find the indefinite integral: [ \int e^{5x-1} + x , dx ]

[ = \int e^{5x-1} , dx + \int x , dx ]

[ = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + C ]

Now, we evaluate the function at ( x = 2 ) using the given information that ( f(2) = 3 ): [ f(2) = \frac{1}{5} e^{5(2)-1} + \frac{2^2}{2} + C ]

[ 3 = \frac{1}{5} e^{9} + 2 + C ]

[ 3 - 2 = \frac{1}{5} e^{9} + C ]

[ 1 = \frac{1}{5} e^{9} + C ]

[ C = 1 - \frac{1}{5} e^{9} ]

Now, we have found the constant of integration (( C )), so we can write the function ( f(x) ) as: [ f(x) = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + \left(1 - \frac{1}{5} e^{9}\right) ]

Therefore, ( f(x) = \frac{1}{5} e^{5x-1} + \frac{x^2}{2} + 1 - \frac{1}{5} e^{9} ).

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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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