# How do you evaluate the integral #int e^-xdx# from 0 to #oo#?

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To evaluate the integral (\int_{0}^{\infty} e^{-x} , dx), we apply the fundamental theorem of calculus.

[ \int_{0}^{\infty} e^{-x} , dx = \lim_{t \to \infty} \int_{0}^{t} e^{-x} , dx ]

[ = \lim_{t \to \infty} \left[ -e^{-x} \right]_{0}^{t} ]

[ = \lim_{t \to \infty} \left( -e^{-t} - (-e^{0}) \right) ]

[ = \lim_{t \to \infty} \left( -e^{-t} - (-1) \right) ]

[ = -\lim_{t \to \infty} e^{-t} - (-1) ]

[ = 0 - (-1) ]

[ = 1 ]

Therefore, the value of the integral (\int_{0}^{\infty} e^{-x} , dx) is 1.

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To evaluate the integral ∫e^(-x)dx from 0 to ∞, you can use the definite integral property.

∫e^(-x)dx from 0 to ∞ = lim(b→∞) ∫e^(-x)dx from 0 to b

= lim(b→∞) [-e^(-x)] evaluated from 0 to b

= lim(b→∞) [-e^(-b) - (-e^(0))]

= lim(b→∞) [-e^(-b) - (-1)]

As b approaches infinity, e^(-b) approaches 0, so the integral becomes:

= -(-1) = 1

Therefore, the value of the integral of e^(-x) from 0 to ∞ is 1.

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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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