How do you evaluate the integral #inte^(-x) dx#?

Answer 1
The answer is #I=-e^(-x)+C#.

This integral can be solved by a substitution:

#u=-x# #du=-dx# #-du=dx#

So, now we can substitute:

#int e^(-x)dx = int e^u (-du)# #=-int e^u du# #=-e^u + C#
and substitute back: #=-e^(-x) + C#

For simple looking integrands, you should try a quick check to see if substitution works before trying harder integration methods.

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Answer 2

To evaluate the integral ( \int e^{-x} , dx ), use the following steps:

  1. Recognize that ( \int e^{-x} , dx ) is of the form of the exponential function.
  2. Apply the integration formula for the exponential function: ( \int e^{ax} , dx = \frac{1}{a} e^{ax} + C ), where ( a ) is a constant and ( C ) is the constant of integration.
  3. In this case, ( a = -1 ), so ( \int e^{-x} , dx = \frac{1}{-1} e^{-x} + C ).
  4. Simplify to get ( \int e^{-x} , dx = -e^{-x} + C ).

Thus, the integral of ( e^{-x} ) with respect to ( x ) is ( -e^{-x} + C ), where ( C ) is the constant of integration.

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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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