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

Answer 1

Ezra Adams

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6#

Let #I=int(e^(2x)-e^x+x)dx=e^(2x)/2-e^x+x^2/2+C#, #C# is a constant of integration.

To determine #C#, we are given the cond. that #f(4)=2#.

Now, #f(x)=I=e^(2x)/2-e^x+x^2/2+C.# Hence, #f(4)=2 rArr e^8/2-e^4+8+C=2 rArr C=e^4-e^8/2-6. #

Sub.ing this value of #C# in #I#, we get,

#f(x)=e^(2x)/2-e^x+x^2/2+e^4-e^8/2-6#

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Answer 2

Scarlett Allison

To find ( f(x) = \int e^{2x} - e^x + x , dx ) given that ( f(4) = 2 ), we need to evaluate the definite integral and then solve for the constant of integration using the given information.

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

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

Given ( f(4) = 2 ), we substitute ( x = 4 ) into the equation:

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

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

Now, solve for ( C ):

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

Therefore, the function ( f(x) ) is:

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

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Answer from HIX Tutor

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.