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How do you evaluate the definite integral #int e^(-x) dx# from #[0,2]#?

Answer 1

Elijah Abram

#int_0^2 e^-x dx = [-e^-x]_0^2 = -e^-2 +e^0 = 1-1/e^2#

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Answer 2

Aurora Alvarado

To evaluate the definite integral ( \int_0^2 e^{-x} , dx ), you can use the fundamental theorem of calculus. First, find the antiderivative of ( e^{-x} ), which is ( -e^{-x} ). Then, evaluate the antiderivative at the upper limit of integration (2) and subtract the value of the antiderivative at the lower limit of integration (0).

[ \int_0^2 e^{-x} , dx = \left[-e^{-x}\right]_0^2 = -e^{-2} - (-e^0) = -e^{-2} + 1 ]

So, ( \int_0^2 e^{-x} , dx = 1 - 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.