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

Answer 1

The function #f# is:

#f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})#.

By integrating the value of #f#:
#f (x) = int (e^{2 x - 1} - e^{1 - 2x} + e^x) dx =#
#= int e^{2x-1} dx - int e^{1-2x} dx + int e^x dx =#
#= 1/2 e^{2x-1} - (- 1/2) e^{1 - 2x} + e^x + C =#
#= 1/2 (e^{2x-1} + e^{1-2x}) + e^x + C#
Then, substituting the known point of #f#:
#f (2) = 1/2 (e^3 + e^{- 3}) + e^2 + C = 3#,
we can obtain the value of the constant #C#:
#C = 3 - e^2 - 1/2 (e^3 + e^{- 3})#.
The function #f# will have the following formula:
#f (x) = 1/2 (e^{2x-1} + e^{1-2x}) + e^x + 3 - e^2 - 1/2 (e^3 + e^{- 3})#.
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Answer 2

To find (f(x) = \int (e^{2x-1} - e^{1-2x} + e^x) , dx) given that (f(2) = 3), we integrate the function term by term and then apply the initial condition to solve for the constant of integration.

  1. Integrate (e^{2x-1} - e^{1-2x} + e^x):

[ \int e^{2x-1} , dx - \int e^{1-2x} , dx + \int e^x , dx ]

For (\int e^{2x-1} , dx), use a substitution (u = 2x - 1 \Rightarrow du = 2dx):

[ \frac{1}{2}\int e^u , du = \frac{1}{2}e^u + C_1 = \frac{1}{2}e^{2x-1} ]

For (\int e^{1-2x} , dx), use a substitution (v = 1-2x \Rightarrow dv = -2dx):

[ -\frac{1}{2}\int e^v , dv = -\frac{1}{2}e^v + C_2 = -\frac{1}{2}e^{1-2x} ]

For (\int e^x , dx):

[ \int e^x , dx = e^x + C_3 ]

Combining these:

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

  1. Apply the initial condition (f(2) = 3):

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

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

Calculate the values:

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

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

[ f(x) = \frac{1}{2}e^{2x-1} - \frac{1}{2}e^{1-2x} + e^x + 3 - \left(\frac{1}{2}e^{3} - \frac{1}{2}e^{-3} + e^{2}\right) ]

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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.

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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