### Video Transcript

The integral from negative β to zero of π§ times π to the power of two π§ with respect to π§ is convergent. What does it converge to?

The question gives us a definite integral which we are told is convergent. We need to find what this integral converges to. Since weβre talking about a definite integral, we should check the continuity of our integrand over the limits of our integral. We can see that our integrand is continuous for all real numbers. This is because π§ is continuous for all real numbers, since itβs a linear function. And π to the power of two π§ is continuous for all real numbers, since itβs an exponential function. So our integrand is the product of two functions which are continuous for all real numbers.

Now that weβve checked the continuity of our integral, letβs recall what it means for the lower limit of our integral to be negative β. We say the integral from negative β to π of some function π of π₯ with respect to π₯ is equal to the limit as π‘ approaches negative β of the integral from π‘ to π of π of π₯ with respect to π₯, where π‘ is less than π. And itβs worth noting if this limit does not exist, we say our integral is divergent. If this limit does exist, we say our integral is convergent and is equal to the value of this limit. Itβs worth reiterating at this point, in our case, our integrand was continuous for all real numbers, so we can just calculate this definite integral directly.

However, sometimes our integrand will not be continuous at some values. When this happens, weβll just need to use even more of our laws of improper integrals to evaluate the integral. So letβs start evaluating our integral. In our case, the upper limit of our integral is zero. So weβll set π equal to zero. Next, weβll set our integrand to be π§π to the power of two π§. Weβll do everything with respect to π§ in this case. So, by using our laws of improper integrals, we have the integral from negative β to zero of π§ times π to the power of two π§ with respect to π§ is equal to. The limit as π‘ approaches negative β of the integral from π‘ to zero of π§ times π to the power of two π§ with respect to π§, if this limit exists.

And in our case, the question tells us that our integral is convergent, so our limit should converge. We now see we need to evaluate the integral of π§ times π to the power of two π§ with respect to π§. This is the product of two functions. So weβll do this by using integration by parts. We recall integration by parts tells us the integral from π to π of π’ times π£ prime with respect to π§ is equal to π’ times π£ evaluated at the limits of our integral π and π minus the integral from π to π of π’ prime times π£ with respect to π§. Our integrand is π§ times π to the power of two. Weβll set π’ to be our function π§ since differentiating π§ makes our function more simple.

Differentiating this with respect to π§, we get π’ prime is equal to one. And to find our function π£, we need to integrate π to the power of two π§. To do this, we need to divide by the coefficient of π§ in our exponent. This gives us π£ is equal to one-half times π to the power of two π§. Weβre now ready to use integration by parts to evaluate our definite integral. Doing this, we get the integral from π‘ to zero of π§ times π to the power of two π§ with respect to π§ is equal to. π§ times a half multiplied by π to the power of two π§ evaluated at the limits of our integral π‘ and zero. Minus the integral from π‘ to zero of one times a half multiplied by π to the power of two π§ with respect to π§.

Evaluating this expression at the limits of our integral and simplifying our integrand. We get zero times one-half multiplied by π to the power of two times zero minus π‘ over two times π to the power of two π‘ minus the integral from π‘ to zero of one-half π to the power of two π§ with respect to π§. The first term in this expression contains a factor of zero, so itβs equal to zero. Next, weβll evaluate the integral in this expression. Again, to evaluate this integral, we need to divide by the coefficient of π§ in the exponent. This gives us one-quarter π to the power of two π§ evaluated at the limits of our integral, π‘ and zero.

Evaluating this at the limits of our integral, we get negative π‘ over two times π to the power of two π‘ minus one-quarter π to the power of two times zero minus a quarter π to the power of two π‘. We see π to the power of two times zero is π to the zeroth power, which is equal to one. The last thing weβll do is distribute the negative one over our parentheses. This gives us negative π‘ over two times π to the power of two π‘ minus a quarter plus one-quarter π to the power of two π‘. And remember, this is an expression for the definite integral inside of our limit. So we can just write this into our limit. So we just need to evaluate the limit of this expression as π‘ approaches negative β.

Weβll do the limit as π‘ approaches negative β of each of these terms separately. Letβs start with negative π‘ over two times π to the power of two π‘. If we tried substituting π‘ is equal to negative β directly into this expression, we get β times zero. This is an indeterminate form. So we canβt determine the limit using this method. Instead, to evaluate this limit, weβll need to be a little bit creative. We will write this as negative π‘ divided by two times π to the power of negative two π‘. We have the quotient of two functions. And as π‘ is approaching negative β, our numerator approaches β and our denominator approaches β. These are both continuous functions with derivatives not equal to zero. So we can attempt to evaluate this limit by using LβHΓ΄pitalβs rule.

So by using LβHΓ΄pitalβs rule, the limit of this term is equal to the limit as π‘ approaches negative β of negative π‘ prime divided by two times π to the power of negative two π‘ prime. And thatβs only if this limit exists or itβs equal to positive or negative β. So weβll differentiate our numerator and our denominator with respect to π‘. Differentiating the numerator, we get negative one. And differentiating the denominator, we get negative four π to the power of negative two π‘.

We can evaluate this limit directly. However, it might be easier to take π to the power of negative two π‘ into our numerator. Using our laws of exponents to take this into our numerator, we get the limit as π‘ approaches negative β of negative π to the power of two π‘ divided by negative four. And now we can just evaluate this limit. Our numerator is approaching zero, but our denominator remains constant. So this limit is equal to zero. So the limit of the first term in our expression is equal to zero. The limit of our second term is much simpler. Since negative one-quarter is a constant, it doesnβt change as the value of π‘ changes. So the limit of our second term is just negative one-quarter.

Finally, we need to evaluate the limit as π‘ approaches negative β of one-quarter times π to the power of two π‘. In actual fact, we already evaluated this limit. When we were using LβHΓ΄pitalβs rule, we showed the limit as π‘ approaches β of negative π to the power of two π‘ divided by negative four is equal to zero. But we can just cancel the negative one in our numerator and our denominator. So the limit of our third term as π‘ approaches negative β is equal to zero. So weβve shown our limit converges and is equal to negative one-quarter. Remember, this is the same as saying that our integral is convergent and is equal to negative one-quarter.

Therefore, weβve shown the integral from negative β to zero of π§ times π to the power of two π§ with respect to π§ is convergent. And itβs equal to negative one-quarter.