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

Question

Evaluate the improper integrals in Exercises 53–62.

∫ from 0 to 3 of (1 / √(9 − x²)) dx

Accepted Answer

Welcome back everyone. In this problem we want to find the value of the integral of 1 divided by the square root of 25 minus x 2 with respect to x between the limits of 0 and 5. Now first, let's think about our integrand here. What do we know? Well, it's an improper integral, so we must account for the vertical asymptote at the upper bound. Now the integrand f of x, so let's refer to it as f of x, which is really 1 divided by the square root of 25 minus x 2. OK, we know that it's continuous. On the interval from open bracket 0 to 5 closed parentheses, so including 0 but not 5, OK, but. It has an. Infinite discontinuity. An infinite discontinuity. OK, at XE equals 5. And we know that because, OK, the limit as X approaches 5 from the left. OK, of FFX. Is going to be equal to infinity, huh? So that means then that this is a type 2 improper integral and we have to rewrite the integral using a limit. So let's let T, OK, let T be a variable. Approaching 5 from the left. OK, cooking crab. From the left. Well, in that case, that means that or or integral then or definite integral of F of X. OK, with respect to X. Will be equal to the limit ast approaches 5 from the left of the integral between 0 and the t of 1 of sorry F of X with respect to x, OK. So I'm just referring to it as FFX just to make it easier to refer to, OK. Now how can we integrate here? Well, recall, OK, the general antiderivative, OK, which we know that the integral here of 1 divided by the square root of a 2 minus x2 with respect to x is equal to the inverse sign of x divided by a plus c. So in this case if we rewrite f of x as 1 divided by the square root of 52 minus x2 because 25 is 52, then its integral is going to be the inverse sign of x divided by 5. Plus C. And now, in this case. We can apply the fundamental theorem of calculus to evaluate the integral from 0 to T because no, this means we can write this as our limit, OK, as T approaches 5 from the left of the inverse sign of x divided by 5, OK, between 0 and t. When we expand that here, we'll have our limit ast approaches 5 from the left, OK, of the inverse sign then oft divided by 5 minus the inverse sign of 0 divided by 5. So now let's find the limit ast approaches 5 from the left for both. Now for our first term, as T approaches 5 from the left, T divided by 5 will approach 1 from the left. So really that's going to be the inverse sign of one. Likewise, on the other side, we don't have any variables in our argument for the inverse sign, and 0 divided by 5 is 0, so that would really be the inverse sign of 0. And now we know the inverse sin of 1 is a half of pi and the inverse sign of 0 is 0. Therefore, this means then that our integral between 0 and 5 of 1 divided by the square root of 25 minus x 2 with respect to x is equal to a half of pi. Thanks a lot for watching everyone. I hope this video helped.

Evaluate the improper integrals in Exercises 53–62.∫ from 0 to 3 of (1 / √(9 − x²)) dx

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Evaluate the improper integrals in Exercises 53–62.
∫ from 0 to 3 of (1 / √(9 − x²)) dx