# #int_0^oo 17 e^(-10s)ds = # ?

Considering

then

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To evaluate the integral ∫₀^∞ 17e^(-10s)ds, we use the formula for the integral of an exponential function. The integral converges, so we proceed with the evaluation:

∫₀^∞ 17e^(-10s)ds = lim┬(b→∞)〖∫_0^b 17e^(-10s)ds〗

Using the integral of e^(-10s), which is -1/10e^(-10s), we get:

= lim┬(b→∞)[17(-1/10)e^(-10s)]₀^b

= lim┬(b→∞)[(-17/10)e^(-10b) - (-17/10)e^(0)]

As b approaches infinity, e^(-10b) approaches 0, so the term (-17/10)e^(-10b) approaches 0. Thus, the integral simplifies to:

= (-17/10)e^(0) - 0

= -17/10 × 1

= -17/10

Therefore, ∫₀^∞ 17e^(-10s)ds = -17/10.

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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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