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

Question

Evaluate the improper integrals in Exercises 53–62.

∫ from 3 to ∞ of (2 / (u² − 2u)) du

Accepted Answer

Welcome back everyone. In this problem, we want to evaluate the improper integral of 3 divided by x2 minus 3x with respect to x between the limits of 4 and infinity. A says it's LN4, B LN3, C LN 3 halves, and D LN 2/3. Now before we evaluate this integral, let's first try to break down our integrand using partial fraction decomposition. OK. Now we can do that because the integrand is a rational function. So setting up the partial fractions, OK, by partial fractions. Then we can say that 3 divided by x2 minus 3 x will be equal to a divided by x. Plus p divided by x minus 3. Now to find the constants A and B, we can multiply by the common denominator x multiplied by x minus 3 and notice here we rewrote it as x multiplied by x minus 3 because that's what we would get if we factor x from our expression, right. So when we do that. OK, then we end up with 3 being equal to a multiplied by x minus 3. Plus b multiplied by x Now, to find the constants, well, to find B, let's let X equal 3, OK. Because if we let X be equal to 3 here. Then that means 3 will be equal to a multiplied by 0. Plus B multiplied by 3, which means that 3 equals 3B. And as such, B will be equal to 1. Now to find A, let's let x equal 0. And if x equals 0, then 3 will be equal to a multiplied by -3 plus b multiplied by 0. Which means 3 equals -3a and as such a will be equal to -1, so that means we can rewrite 3 divided by x2 minus 3 x as 1 divided by x minus 3 minus 1 divided by x. Now we can integrate each term. So if we're doing that, that is, we're finding the general antiderivative, OK, it's going to be the integral of 1 divided by x minus 3 minus 1 divided by x with respect to x, and if we integrate both of those with respect to x, it's going to be the ln of the absolute value of x minus 3 minus the Ln of the absolute value of x plus c. And know using the logarithmic property that if we subtract two logs it's the same as dividing their arguments we could rewrite this as the ln of the absolute value of x minus 3 divided by x or The LN of x minus 3 divided by x plus c between our limits of integration. So what does this mean for us? Well, no, we're basically saying. That we need to evaluate this natural log between our limits of 4 and infinity and now since the upper bound is infinity, we can set up the improper integral limit by replacing infinity with a variable b and taking the limit as b approaches infinity. In other words, we can rewrite our integral, OK. We can rewrite our integral. As the limit as b approaches infinity between the limits of b and 4 of 1 divided by x minus 3 minus 1 divided by x with respect to x, which we know the integrals of both of those, so really then evaluating the antiderivative at the limits of 4 and b, we're finding the limit as b approaches infinity of. LN of X minus 3 divided by X between 4 and B. No, here applying the fundamental theorem of calculus, that's really saying that we find the limit as b approaches infinity of the natural log of b minus 3 divided by b. Minus the natural log of 4 minus 3 divided by 4. Now what will both of these limits be? Well, for a first term, as b approaches infinity, the fraction b minus 3 divided by b, which we could rewrite as 1 minus 3 divided by b, would approach 1. So really that first term here would be the natural log of 1. For a second term for my here as b approaches infinity, it doesn't affect our argument since it's a constant so this would be minus the natural log of 1/4. Now if we simplify that, we know that the natural log of 1 is 0, OK, and for a second term we could rewrite negative Ln of 1/4 as LN4, OK, because if we remember. Let me put that here. If we remember, recall that negative ln of 1 divided by x equals ln of x based on the properties of logs. I know 0 plus LN4 would simply leave us with LN4. OK? In other words, our improper integral would be the natural log of 4. 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 3 to ∞ of (2 / (u² − 2u)) du

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Evaluate the improper integrals in Exercises 53–62.
∫ from 3 to ∞ of (2 / (u² − 2u)) du