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

Question

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

∫₀^∞ 2e^(−θ) sinθ dθ

Accepted Answer

Welcome back everyone. In this problem we want to evaluate the improper integral of e to the negative x plus x dx between the bones of zero and infinity. A says it's a 12, B2, C 1, and D says the integral diverges. Now how can we evaluate this improper integral? Well, first we can start by rewriting the integral with a limit. So if we take our integral here between 0 and infinity. Then we could rewrite it as the limit as b approaches infinity of the integral between 0 and b of e to the negative x plus x with respect to x. So if we can figure out what this integral is, then we should be able to find its limit as b approaches infinity. So let's first try by finding the indefinite integral. Uh, what do we know here? Does it look like anything you've seen before? Well, recall, that generally the integral of e to the ax multiplied by the cosine of bx with respect to x is equal to eax divided by a 2 + b2 multiplied by a cost of bx plus b sin of b x. Plus C So in this case, given A equals -1 because E is raised to the power of x, so the coefficient of X is -1, right? And B equals 1 because the coefficient of x in cost x would be 1. Then that means Or indefinite integral of e to the negative x cos x dx. Will be equal to e to the negative x divided by -12 plus 12 multiplied by -1 x + 1 sin x all plus c. And if we simplify that, it will be a half of e to the negative x multiplied by sin x minus X plus c. Now, since we know what the indefinite integral is, we can evaluate the in sorry, we know what the indefinite integral is, so we can evaluate the definite integral. So that means this integral between 0 and negativeb then OK, would be equal to e to the negative x divided by 2 multiplied by sin x minus x, OK. Between the bounds of 0 and the B. And if we Apply or substitute our upper and lower limits, then it's going to be a half of e to the negative B, OK, multiplied by sin b minus cost b. OK Minus A half of e to the 0 or E 0 multiplied by the sin of 0, OK, minus the cosine of 0. Now, where do we go from here? Well, in this case, if we go ahead and simplify. Then we're going to end up with e to the negativeb divided by 2 multiplied by the sin of b minus the cosine of b for our first term, but in our second term e to the 0 will be 1, OK, so that's minus 5, OK. Multiplied by the sin of 0, which is 0 minus the cosine of 0, which is 1, so this would be -52 multiplied by -1, or we could rewrite all of this as plus 1/2 because -1 multiplied by -122 equals 5. And now that we have our definite integral, we can take the limit as B approaches infinity. So essentially as B approaches infinity we're looking for the limit. Yeah, of our expression. E to the negative B. E to the negativeb multiplied by the sin of b minus the cosine of b all divided by 2 plus 1/2. Now what would this be? Well, notice that as b approaches infinity, the factor e to the negative b would tend to zero because the exponential function decays rapidly for large values of b. On the other hand, the expression sin b minus plus b oscillates between 2 and 2 for all b. As both sine and cosine functions are bounded, but no, since e to the negative b tends to zero and the oscillatory term remains bounded, that means the product of these two terms would tend to zero as b approaches infinity. In other words, that limit would be equal to zero for a second term, a half, the limit as b approaches infinity of a half would be a half, since a half is a constant. So this is really 0 + 1/22, which would be equal to 1/2. Therefore, our improper integral of e to the negative x x dx between 0 and infinity is a 12. If we look back at our answer twices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₀^∞ 2e^(−θ) sinθ dθ

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The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀^∞ 2e^(−θ) sinθ dθ