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The integral ∫_(βˆ’βˆž) ^(0) 2^(π‘Ÿ) dπ‘Ÿ is convergent. What does it converge to?

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Video Transcript

The integral from negative infinity to zero of two to the power of π‘Ÿ with respect to π‘Ÿ is convergent. What does it converge to?

Now, we know a formula to evaluate integrals of this form. It tells us that the integral from negative infinity to π‘Ž of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑑 tends to negative infinity of the integral from 𝑑 to π‘Ž of 𝑓 of π‘₯ with respect to π‘₯. In our case, we can see that π‘Ž is equal to zero. We are in fact integrating with respect to π‘Ÿ instead of respect to π‘₯. Therefore, our function 𝑓 of π‘₯, or in our case, 𝑓 of π‘Ÿ, is two to the power of π‘Ÿ. We can substitute these values into our formula. We now have that the integral from negative infinity to zero of two to the power of π‘Ÿ with respect to π‘Ÿ is equal to the limit as 𝑑 tends to negative infinity of the integral from 𝑑 to zero of two to the power of π‘Ÿ with respect to π‘Ÿ. Let’s start by finding this integral.

We can consider the antiderivative. If we try differentiating two to the power of π‘Ÿ with respect to π‘Ÿ, then we obtain the natural logarithm of two multiplied by two to the power of π‘Ÿ. Since the natural logarithm of two is a constant, we can divide by it on both sides and we can put it inside the differential. And we obtain that the derivative of two to the power of π‘Ÿ of the natural logarithm of two is equal to two to the power of π‘Ÿ. And so, here, we found the antiderivative of two to the power of π‘Ÿ. It’s two to the power of π‘Ÿ over the natural logarithm of two.

And we can use this to say that the integral from 𝑑 to zero of two to the power of π‘Ÿ with respect to π‘Ÿ is equal to two to the power of π‘Ÿ over the natural logarithm of two from 𝑑 to zero. Since zero is the upper bound, when we substitute it in, the sign remains the same, giving us two to the power of zero over the natural logarithm of two. And since 𝑑 is the lower bound, when we substitute it in, we must change the sign, giving us negative two to the power of 𝑑 over the natural logarithm of two. Since anything to the power of zero is one, we can rewrite to use the power of zero as one.

Therefore, we evaluated that this integral is equal to one over the natural logarithm of two minus two to the power of 𝑑 over the natural logarithm of two. And we can substitute this value for our integral back into our limit. We’re able to use the properties of limits in order to split this limit up. We have for the limit of a difference of functions is equal to the difference of the limits of the functions, giving us that our limit is equal to the limit as 𝑑 tends to negative infinity of one over the natural logarithm of two minus the limit as 𝑑 tends to negative infinity of two to the power of 𝑑 over the natural logarithm of two.

Now, in our first limit here, we’re taking the limit of one over the natural logarithm of two, which has no 𝑑 dependence. Therefore, this limit is simply equal to one over the natural logarithm of two. We’re able to apply another limit property to the second limit. And that is that the limit of a quotient of functions is equal to the quotient of the limit of the functions. Now in the denominator here, we’re taking the limit of a constant function. Therefore, the denominator is simply equal to the natural logarithm of two. In the numerator, we have the limit as 𝑑 tends to negative infinity of two to the power of 𝑑. Here, 𝑑 is getting more and more negative. Therefore, two to the power of 𝑑 will be getting closer and closer to zero. And so, we can say that the limit as 𝑑 tends to negative infinity of two to the power of 𝑑 must be zero. Here, we reach our solution which is that the integral from negative infinity to zero of two to the power of π‘Ÿ with respect to π‘Ÿ converges to β€” and is therefore equal to β€” one over the natural logarithm of two.

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