Question Video: Determining Whether the Integrals of Exponential Functions with Negative Exponents Are Convergent or Divergent and Finding Their Values Mathematics • Higher Education

The integral ∫_(2)^(∞) 𝑒^(βˆ’5𝑝) d𝑝 is convergent. What does it converge to?

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Video Transcript

The integral between two and ∞ of 𝑒 to the negative five 𝑝 with respect to 𝑝 is convergent. What does it converge to?

Because this integral has an infinite limit, we call this an improper integral. We have a general result that tells us that the integral between π‘Ž and ∞ of 𝑓 of π‘₯ with respect to π‘₯ is the limit as 𝑑 approaches ∞ of the integral between π‘Ž and 𝑑 of 𝑓 of π‘₯ with respect to π‘₯. And we know that this integral is convergent, which means the integral must approach a specific value. And that’s what we’re going to find.

What this result tells us is that we can replace the infinite limit in our integral with a variable 𝑑 and then take the limit as 𝑑 approaches ∞ of our integral. So we’re going to find the limit as 𝑑 approaches ∞ of the integral between two and 𝑑 of 𝑒 to the power of negative five 𝑝 with respect to 𝑝. To do this, we recall the general rule that the integral of 𝑒 to the π‘˜π‘₯ power with respect to π‘₯ is equal to one over π‘˜ multiplied by 𝑒 to the π‘˜π‘₯ power plus a constant of integration 𝑐. So the integral of 𝑒 to the power of negative five 𝑝 with respect to 𝑝 is one over negative five 𝑒 to the power of negative five 𝑝.

Note that because we have limits of integration for this question, we don’t need to include a constant of integration. Let’s write this as a single faction, negative 𝑒 to the power of negative five 𝑝 over five. And now, let’s apply these limits. And when we do this, we have to remember that our integral is negative. So when we apply the limits, we’re going to be subtracting a negative. So we can replace this with an add. And we can replace negative five multiplied by two with negative 10. But we still need to apply this limit.

If we rewrite the term negative 𝑒 to the power of negative five 𝑑 over five, using the fact that 𝑒 to the negative five 𝑑 is the same as one over 𝑒 to the five 𝑑, and then we recognize that we can simplify this to negative one over five 𝑒 to the five 𝑑. And we know the fact that the limit as 𝑑 approaches ∞ of 𝑒 to the 𝑑 power is ∞. So the denominator approaches ∞. So the term approaches zero. So in fact, when we apply the limit to this bracket, what we have left is 𝑒 to the negative 10 power over five, which shows that this limit approaches a specific value just as we expected.

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