Video: Determining Whether the Integrals of Polynomial Function Multiplied by Exponential Functions with Negative Exponents Are Convergent or Divergent and Finding Their Values

The integral ∫_(βˆ’βˆž) ^(∞) π‘₯𝑒^(βˆ’π‘₯Β²) dπ‘₯ is convergent. What does it converge to?

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Video Transcript

The integral between negative infinity and positive infinity of π‘₯ multiplied by 𝑒 to the power of negative π‘₯ squared with respect to π‘₯ is convergent. What does it converge to?

We have a formula which can help us to evaluate integrals of this form. The formula tells us that the integral from negative infinity to positive infinity of 𝑓 of π‘₯ with respect to π‘₯ is equal to the limit as 𝑠 tends to negative infinity of the integral from 𝑠 to 𝑐 of 𝑓 of π‘₯ with respect to π‘₯ plus the limit as 𝑑 tends to positive infinity of the integral from 𝑐 to 𝑑 of 𝑓 of π‘₯ with respect to π‘₯, where 𝑐 can be any real number. Now, in our case, 𝑓 of π‘₯ is equal to π‘₯ multiplied by 𝑒 to the power of negative π‘₯ squared. So we can substitute this into our formula. Now, we can pick 𝑐 to be any real number. Let’s choose 𝑐 to be zero. However, you can pick any value of 𝑐 that you like.

Let’s now try to solve our integrals. And we can do this by trying to find the antiderivative of π‘₯ multiplied by 𝑒 to the negative π‘₯ squared. We notice that the derivative of 𝑒 to the power of negative π‘₯ squared is equal to negative two π‘₯𝑒 to the power of negative π‘₯ squared. Now, we can divide both sides of this equation by the constant term negative two. And we obtain this. On the left-hand side, we have a constant term multiplying a derivative. Using the derivative rule for scalar multiplication, we can move the negative one-half inside the derivative. And so, we found our antiderivative of π‘₯ multiplied by 𝑒 to the power of negative π‘₯ squared. It’s negative one-half 𝑒 to the power of negative π‘₯ squared.

Using this antiderivative, we’re able to evaluate the integrals inside our limits. Next, we can substitute in our upper and lower bounds. We obtained the limit as 𝑠 tends to negative infinity of negative one-half multiplied by 𝑒 to the power of negative zero squared minus negative one-half multiplied by 𝑒 to the power of negative 𝑠 squared. Plus the limit as 𝑑 tends to positive infinity of negative one-half multiplied by 𝑒 to the power of negative 𝑑 squared minus negative one-half multiplied by 𝑒 to the power of negative zero squared. Since 𝑒 to the power of negative zero squared is simply 𝑒 to the power of zero, which is just one, the negative one-half multiplied by 𝑒 to the power of negative zero squared terms is simply negative one-half. And now, we can sort out the double negative signs which will give us a positive sign.

Next, we can use one of the properties of limits, which tells us that the limit of a sum of functions is equal to the sum of the limits of their functions. And so, we’re able to split our limits up like this. Here, we have two limits of constant terms. So these limits are simply equal to that constant term. The first one being negative one-half and the second just being one-half. Therefore, when we add these two terms together, they will cancel out with one another. Now, we just need to worry about these two remaining limits. Let’s consider the one on the right first.

As 𝑑 gets larger and larger and larger, negative 𝑑 squared will be getting larger and larger but in the negative direction. Therefore, 𝑒 to the power of negative 𝑑 squared will be getting closer and closer to zero. And so, we can say that the limit as 𝑑 tends to infinity of negative one-half 𝑒 to the power of negative 𝑑 squared is equal to zero. Similarly, as 𝑠 gets larger and larger in the negative direction, 𝑠 squared will be getting larger and larger in the positive direction. And so, negative 𝑠 squared will be getting larger in the negative direction. And when this happens, 𝑒 to the power of negative 𝑠 squared gets closer and closer to zero. Therefore, this limit is also equal to zero.

Since the negative one-half and the positive one-half canceled out with one another and these two limits are both equal to zero, therefore, we can say that the integral from negative infinity to positive infinity of π‘₯ multiplied by 𝑒 to the power of negative π‘₯ squared with respect to π‘₯ converges to β€” and is therefore equal to β€” zero.

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