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How do you integrate #int e^(-2x)dx# from #[0,1]#?

Answer 1

Scarlett Allison

In general #int(e^(alphax))dx = 1/alphae^(alphax)#

#int_0^1e^(-2x)dx = -1/2e^(-2x)|_0^1=-1/2 (e^-2-1)#

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Answer 2

Luke Alvarez

To integrate ( \int e^{-2x} , dx ) from ( x = 0 ) to ( x = 1 ), you can use the formula for integrating exponential functions. The integral of ( e^{-2x} ) with respect to ( x ) is ( -\frac{1}{2}e^{-2x} ), so the definite integral from ( x = 0 ) to ( x = 1 ) is ( -\frac{1}{2}e^{-2x} ) evaluated from ( x = 0 ) to ( x = 1 ).

Evaluating this, we get:

[ \left[-\frac{1}{2}e^{-2x}\right]_{0}^{1} = -\frac{1}{2}e^{-2(1)} - \left(-\frac{1}{2}e^{-2(0)}\right) ]

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

[ = \frac{1}{2} - \frac{1}{2}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.