 This lesson is on improper integrals. What is an improper integral? Well, we have several different types. The first type is when we go from some constant a to infinity, and we try to find an answer for that a to infinity. The second type is when we go from negative infinity to some constant we can call b. The third type is when we go from negative infinity to infinity. Or we have other types where we go from a to b. One being where it's discontinuous at the end point a or the end point b. Or the other type is where it's discontinuous somewhere between a and b. So let's look at several different types here. Type a or the infinity type. When we go from a to infinity or negative infinity to b or negative infinity to infinity. Our example. Determine the integral from negative one to infinity of dx over x to the fourth. Well the way we do these is to write the limit as b approaches infinity and replace our infinity with a b. Because that is the proper way to write improper integrals. And we have our dx over x to the fourth. Doing the antiderivative of dx over x to the fourth we get negative one third x cubed in the denominator. And we're going from negative one to b and we have our limit as b approaches infinity. Well that becomes the limit as b approaches infinity of negative one third times one over b cubed minus one over negative one. And that equals well when b approaches infinity one over b squared means nothing. The one over negative one becomes positive one. So we have negative one third times positive one which is negative one third. Very simple procedure. Just make sure when you're doing these on a quiz or a test or some examination you have to show you work. You put that b approaching infinity limit in there. Let's go to another example. This one says determine the integral from negative infinity to infinity of dx over one plus x squared. We have to break this one up so a convenient place would be at zero. So let's do the limit as b approaches negative infinity of the integral of b to zero of dx over one plus x squared. Add to that the limit. I'll use another variable but b will still work for this one. b approaches positive infinity from zero to b of dx over one plus x squared. Take the antiderivative of dx over one plus x squared and that's the arctangent of x. And remember we have the limit as b approaches negative infinity. So this one goes from b to zero plus limit as b approaches infinity of the arctangent of x from zero to b. So when we put the zero in on the first one we get the arctangent of zero which is zero minus the arctangent of b as b approaches negative infinity. Well if you think of your graph of your arctangent it looks like this. So that would be negative pi over two. We would add to that the limit as b approaches infinity of arctangent of b which is in this case going to positive infinity which means it's positive pi over two plus the arctangent of zero which is zero. Add these two together we'll have plus pi over two plus pi over two is equal to pi. So the answer on this one is pi. A little different from what you might think as you just look at it and think of integrals and one adding out negatives and positives and becoming a zero. This actually adds up to pi. Let's go on. This type of integral is called type b or interval ab. This is where you have a discontinuity at a or b or somewhere in between. Let's look at an example of this. Integral of negative two to zero of dx over the square root of x plus two. This is where we have a discontinuity at the endpoint x equals negative two. So we change this to read limit as b approaches negative two of the integral b to zero dx over square root of x plus two. The anti derivative of this is square root of x plus two times two. And we do limit as b approaches negative two from b to zero. Substituting in the numbers we put in the zero first and we get two square roots of two minus and when we put b in and it is negative two we actually get a zero for the square root. It's minus zero so that answer becomes two square roots of two. We knew we were going to get an answer for this because we have a square root in the denominator and when we take the anti derivative of a square root in the denominator it becomes a square root in the numerator. So think through those possibilities as you do this type of integration. Let's try another type b. This says the integral from negative two to two of one over x to the two-thirds power. This has a discontinuity when x is equal to zero so we have to break it up. So we say the limit as b approaches zero of negative two to b of one over x to the two-thirds dx plus the limit as b approaches zero of b to two of one over x. The two-thirds dx. Evaluating the first one we get x add one to the two-thirds so we get x to the one-third times three from negative two to b where b is approaching zero. Add to that limit as b approaches zero of three x to the one-third from b to two. Again if we substitute in the b at zero we get zero minus three times negative two to the one-third power. Add to that limit as b approaches zero so if we substitute first the two in we get three times two to the one-third power minus zero. The negative three times negative two gives us a positive. Remember we can factor out this negative on this particular problem because of the fact that negative two to the one-third power can be found. So we will have three times two to the one-third power plus three times two to the one-third power so three and three is six times two to the one-third power. This concludes your lesson on improper integrals.