Evaluate the improper integrals in Exercises 53–62.
∫ fr...

Question

Evaluate the improper integrals in Exercises 53–62.

∫ from 0 to 2 of (1 / (y − 1)^(2/3)) dy

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of 1 divided by x minus 2 2/3 with respect to x between the bones of 1 and 3. A says it's 6, B says it's 3, C, 4, and D says it's 2. Now this integral is improper because the integral has a vertical asymptote at x equals 2, which lies directly in the middle of the interval from 1 to 3. So to solve this correctly, we have to split the integral into two parts at the point of discontinuity and evaluate each as a limit. Now, if we were to write that note, we're basically saying that since our function here where we're letting F of X be equal to 1 divided by x minus 2 to 2/3, so since this is undefined. At x equals 2. Then basically we're rewriting our integral between 1 and 3 of our function. OK. As the integral between 1 and 2 of 1 divided by x minus 2 to the power of 2/3 with respect to x. Which is f x that I refer to from herein plus the integral from 2 to 3 of f of x with respect to x. So now we need to evaluate each of these as a limit. Let's rewrite them using limits. Now, in this case, OK, that means we can rewrite it as the limit, OK, as a approaches 2 from the left of the integral between 1 and a of x minus 2 to the power of -2/3 with respect to x. Plus the limit. As B approaches 2. From the right, of the integral between b and 3 of x minus 2 to the power of -2/3 with respect to x, where here a is a variable approaching 2 from the left and b is a variable approaching 2 from the right. Now, how can we figure this out? Well, we can find the antiderivative using the power root. Now if we do that, then the integral of our expression x minus 2 to 2/3 with respect to x. is going to be equal to x minus 2 to the power of 1/3 divided by 1/3 plus c, which we can rewrite as 3 multiplied by x minus 2 to the power of 1/3 plus c. OK, I know if we apply the fundamental theorem of calculus to both parts of our of of our integral here for our limits, OK, then. This would imply That our limit here as A approaches 2 from the left of 3 multiplied by x minus 2 to the power of 1/3, between 1 and a would be equal to our limit, OK. Of 3 multiplied by a minus 2 1/3 minus 3 multiplied by 1 minus 2 1/3. So let's see if we can evaluate that. Now if we think about, if we think about it here, OK, then as A approaches 2, then a minus 2 will approach 0. So this is going to be 3 multiplied by 0 to 1/3 minus 3 multiplied by -1 1/3. And now when we solve here that will be 3 minus 0, OK, which would be equal to 3. So here The, sorry, my apologies, that would be 0 minus 3 multiplied by -1, which would be equal to 3. So it's still 3, but uh let me just, it's just written in a different way, really. Now if we focus on our second part, then we'd be finding the limit as b approaches 2 from the right of 3 multiplied by x minus 2 to the power of a third between B and 3, OK. No, that's gonna be our limit as B approaches 2. Of 3 multiplied by 3 minus 2. To 1/3, OK, -3 multiplied by B minus 2 1/3. And know when we go ahead and solve here. We get this to be 3 multiplied by 1 1/3 minus 3 multiplied by 0, where 1 1/3 is 1 and 3 multiplied by that equals 3 and 3 multiplied by 0 equals 0, so it's really 3 minus 0, which is 3. No, if we add the results from both integral pieces, then this means that our integral between 1 and 3 of 1 divided by x minus 2 to the power of 2/3 with respect to x would be equal to 3 + 3 our first part plus our second part, which equals 6. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the improper integrals in Exercises 53–62.∫ from 0 to 2 of (1 / (y − 1)^(2/3)) dy

Watch next

Evaluate the improper integrals in Exercises 53–62.
∫ from 0 to 2 of (1 / (y − 1)^(2/3)) dy