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How do you evaluate #int_1^2 e^(1-x)dx#?

Answer 1

Aurora Alvarado

#1-e^(-1) ~= 0.63212#

#int_1^2 e^(1-x)dx#

Let #u = 1-x -> (du)/dx=-1#

#I = int-e^u du#

#= -e^u#

Undo substitution:

#=-e^(1-x)]_1^2#

#= -e^(1-2)+e^(1-1) #

#=1-e^(-1)#

#~= 0.63212#

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Answer 2

Emma Aldridge

To evaluate ∫₁² e^(1-x)dx, we can first integrate the function e^(1-x) with respect to x. The integral of e^(1-x)dx is -e^(1-x), according to the rules of integration. Then, we evaluate this antiderivative from the lower limit of integration, which is 1, to the upper limit of integration, which is 2.

Therefore, ∫₁² e^(1-x)dx = [-e^(1-x)] evaluated from 1 to 2 = [-e^(1-2)] - [-e^(1-1)] = [-e^0] - [-e^0] = [-1] - [-1] = -1 + 1 = 0

So, the value of the given definite integral is 0.

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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.