Improper Integrals Example 1 Part 2 Video

This video from IntegralCALC shows you how to solve part 2 of the Integrals Example 1 Math problem.
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Improper Integrals Example 1 Part 2 So the first thing we do, we always plug in the top term first. No matter what it is we’re putting the top first and then the bottom into everything on this side. We’re plugging in for x always. We’re going to have negative and I’m just going to put this in here, infinity times e-3 times infinity over 3 minus e-3 times infinity over 9. So we plugged in infinity to everything on this side of the line here and then we have to plug in 0. So we plugged in the top term then we put a big minus sign, always a minus sign when you plug in this first term here and you do minus and then you plug in the second term. So minus and then this is going to be 0 this time, minus 0 e-3 times 0 over 3 minus e-3 times 0 over 9. You plug in infinity to everything, that’s the first term minus plug in 0 to everything that’s the second term and that’s our whole solution. All we have to do now is simplify. So simplification, the way that I deal with infinity in a problem like this is I plug in a really, really big number to my calculator until like I would take a hundred thousand or a million, usually something like that, maybe 10 million or a hundred million if we have to and plug it in and see what I get out because it’s harder for me to visualize. So we’re going to take the first term. This actually ends up being 0, this first term. If you take a million or 10 million and you plug in here and you raise e to that super huge negative number because this is a negative 3 up here that ends up being 0. And if that’s 0 then this whole term here is 0. This term, same thing, you plug in a really big number there, e tot eh 300 million ends up being 0. So this is 0. Let’s go ahead and write this out. What we’re ending up with here is 0 minus 0 and then minus, let’s do our bracket for here. Of course we’ve got a 0 out in front here e 0 is just 1 but then 0×1 is 0, so this whole term ends up being 0 and then we have minus, in this term here of course -3×0 is 0. Anything to the 0 power is 1, so e to the 0 is 1 and we’ve got one over 9. So we end up with 1/9 and then if we simplify that of course these go away. We’ve got 0 minus 0 minus 0 and then since we have minus a negative, this is a plus 1/9 and this all goes away and 1/9 is our final answer. I hope you followed that. We will do couple more samples. Thanks guys!

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