The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₀² dx / √|x − 1|

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of dx divided by the square root of the absolute value of x minus 2 between the bones of 0 and 4. Now to evaluate this improper integral, we must first identify the point of discontinuity, and if we think about it here, the integral becomes undefined at x equals 2 because that would make our denominator in our expression zero, and that lies directly in the middle of our interval from 0 to 4. So we need to split our integral. OK, so that means then. OK, because of the absolute value and the vertical asymptote at x equals 2, we need to rewrite our integral. Or sorry, or yes, or integral as the integral between 0 and 2 of d x divided by the square root of the absolute value of x minus 2. Plus the integral between 2 to 4 of d x divided by the square root of the absolute value of x minus 2. Now let's try to remove the absolute value for the interval from 0 to 2 x minus 2 would be negative. So the absolute value of x minus 2 would be the negative value of x minus 2, OK, which if we simplify that we can rewrite it as 2 minus x. On the other hand, between 2 and 4. X minus 2 would be positive, so. Our absolute value of term or function would be x minus 2. So this would be our improper integral split up at x equals 2. So now we need to evaluate each improper integral, and we can use the limits to approach the discontinuity at x equals 2. So let's focus on the first part of our integral, OK, now solving for The integral between 0 and 2 of d x divided by the square root of 2 minus x we can help or or integrating by substitution can help. So let's let u be equal to 2 minus x. That means the derivative of u with respect to x equals -1. So d negative duu OK would be equal to dx. So this means then that the integral or indefinite integral of 1 divided by the square root of 2 minus x with respect to x could be rewritten as the integral of 1 or the negative value of the integral. Of 1 divided by a root u with respect to u, which we can rewrite as the negative value of the integral of u to the negative 12 d u, which if we integrate u it's going to be u to the power of 1/2 divided by 1/2, which would simplify it to -2 multiplied by a root u plus c and we know that u is 2 minus x. So since this is our indefinite integral, OK, then applying it to our definite integral between 0 and 2. OK. Then here this is going to be equal to. The limit, as our variable b approaches 2 from the left of the integral from 0 to b of -2 multiplied by root 2 minus x, OK. Oh sorry, let me rewrite our expression here. My apologies. It's 1 divided by root 2 minus x with respect to x, which would be. The limit as B approaches 2 from the left of -2 multiplied by root 2 minus X, OK, between the bones of 0 and B. I know when we apply that. Then it's going to become the limit as B approaches 2 from the left of -2 multiplied by root 2 minus B, OK, minus. -2 multiplied by root 2 minus 0, OK. And no, this would simplify to 0 + 2 root 2, or simply 2 root 2. So this is our improper integral for our first expression. Next, let's do it for the 2nd. So in other words, no, we're solving for. The integral between 2 and 4 of 1 divided by the square root of x minus 2 with respect to x. Now, if we let you can solve by substitution again, so if we let you equal x minus 2, then that means the derivative of you with respect to x is 1, so DU would be equal to dx. So our indefinite integral. OK, of 1 divided by the square root of x minus 2 with respect to x could be rewritten as the integral of 1 divided by root u with respect to u. Which is the integral of u to the negative half with respect to you, and when we integrate that, it will be u to the 12 multiplied or divided by a half, so we can rewrite that as 2 root u plus c, which similar to what we had in our previous problem, we could replace our u, so that's 2 multiplied by root 2 minus x, OK, or sorry, root x minus 2. Plus C. Know that we have that, OK, then if we were to find our definite integral, we could rewrite it then, so this is our integral between 2 and 4 of d x divided by or sorry. One divided by. Root x minus 2 D X And that is going to be our limit as A approaches 2 from the right of 2 multiplied by root x minus 2. OK, between the bones of A and 4. That is going to become the limit as A approaches 2 from the right, OK, of 2 root 4 minus 2 minus 2 multiplied by a root a minus 2, and this, let me make sure I put, I show it clearly here is a minus 2, and this would simplify to 2 root 2 minus 0 or simply 2 root 2. So now that we've evaluated both of our limits, when we combine them, OK, then this means our integral. OK. Between. 0 and 4. Of d x divided by the square root of the absolute value of x minus 2 would be equal to 2 root 2 plus 2 root 2. Which equals 4 root to. So this is our improper integra. Thanks a lot for watching everyone. I hope this video helped.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₀² dx / √|x − 1|

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The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀² dx / √|x − 1|