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

Answer 1

#f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+3.768#

First integrate to obtain #f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+c# Then substitute #f(2)=3# to obtain #3=1/2e^3-1/3e^4+e^2+c# Rearrange for c then evaluate: #c=3-1/2e^3+1/3e^4-e^2=3.768# (4s.f.) Thus the original function is #f(x)=1/2e^(2x-1)-1/3e^(3x-2)+e^x+3.768#
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Answer 2

To find ( f(x) = \int e^{2x-1} - e^{3x-2} + e^x , dx ) given ( f(2) = 3 ), you need to find the indefinite integral of the function and then use the given condition to solve for the constant of integration. However, the given function is a sum of exponentials, so you can integrate each term separately and then add them up. After integrating each term, you can plug in the upper limit ( x = 2 ) and solve for the constant of integration to find the value of ( f(2) ).

Let's integrate each term separately:

  1. Integral of ( e^{2x-1} ): [ \int e^{2x-1} , dx = \frac{1}{2}e^{2x-1} + C_1 ]

  2. Integral of ( e^{3x-2} ): [ \int e^{3x-2} , dx = \frac{1}{3}e^{3x-2} + C_2 ]

  3. Integral of ( e^x ): [ \int e^x , dx = e^x + C_3 ]

Now, summing up the integrals with their respective constants of integration: [ f(x) = \frac{1}{2}e^{2x-1} - \frac{1}{3}e^{3x-2} + e^x + C ]

Now, use ( f(2) = 3 ) to solve for ( C ): [ f(2) = \frac{1}{2}e^{2(2)-1} - \frac{1}{3}e^{3(2)-2} + e^2 + C = 3 ] [ \frac{1}{2}e^3 - \frac{1}{3}e^4 + e^2 + C = 3 ] [ \frac{1}{2}e^3 - \frac{1}{3}e^4 + e^2 + C - 3 = 0 ]

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

So, the value of the constant of integration is: [ C = 3 - \frac{1}{2}e^3 + \frac{1}{3}e^4 - e^2 ]

Thus, the function ( f(x) ) is: [ f(x) = \frac{1}{2}e^{2x-1} - \frac{1}{3}e^{3x-2} + e^x + 3 - \frac{1}{2}e^3 + \frac{1}{3}e^4 - e^2 ]

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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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