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Current time:0:00Total duration:11:27

Improper integral with two infinite bounds

Video transcript

right here we have the graph of y is equal to 250 over 25 plus x squared and what I'm curious about in this video is the total area under this curve and above the x-axis so I'm talking about everything everything that I am shading in white here including what we can't see to those we keep moving to the right and we keep moving to the left so I'm talking about from X goes from from X at negative infinity all the way to X at infinity so first how would we actually denote this well it would be an improper integral we would denote this area as the indefinite integral from X is equal to negative infinity to X is equal to infinity of our function 250 over 25 plus 25 plus x squared DX now we've already seen improper integrals where one of our boundaries with infinity but how do you do it when you have one boundary at positive infinity and one boundary at negative infinity you can't take a limit to two different things and so the way that we're going to tackle this is to actually break up this area into two different improper integrals one improper integral that describes this area right over here in blue from negative infinity to zero so we'll say this is equal to the improper integral that goes from negative infinity to zero of 250 over 25 plus x squared DX plus the improper integral that goes from 0 to positive infinity so plus the improper or the definite integral from 0 to positive infinity of 250 over over 25 plus x squared 25 plus x squared DX and now we can start to make sense of this so what we have in blue what we have in blue can be rewritten this is equal to the limit as n approaches negative infinity of the definite integral from n to 0 of 250 over 25 plus x squared DX plus plus and I'm running out of real estate here plus the limit as it's already used let me use em now the limit as M approaches positive infinity of the definite integral from 0 to M of 250 over 25 plus x squared DX so now all we have to do is evaluate these definite integrals and to do that we just have to figure out an antiderivative of 250 over 25 plus x squared so let's try to figure out what that is so an anti-ager I'll do it over here on the left so the antiderivative where you need to figure out the antiderivative of 250 over 25 plus x squared and it might already jump out at you that trig substitution might be a good thing to do you see this pattern of a squared plus x squared where in this case a would be 5 so we can make the substitution that X is equal to a tangent theta 5 tangent theta and since we're going to have to reverse substitute later on we can also put in the constraint well we say x over 5 is equal to tangent of theta which is completely consistent with this first statement and so if we wanted to have data expressed as a function of X and we can put the constraint that theta is equal to arctangent arctangent of X over 5 so once again this is completely consistent with this over here X can be 5 tangent of theta and theta can be 5 it can be equal to arctangent of x over 5 so now let's do the substitution actually before that we also have to figure out what DX is equal to so DX DX is equal to I'll do it right over here DX well the derivative of this with respect to theta is 5 secant squared theta D theta so now we're ready to substitute back in so all of this business is going to be equal to is going to be equal to 250 times DX well 250 times DX is 250 times 5 secant squared theta D theta so that's 250 DX is this business up here all of that over 25 plus x squared well x squared is going to be 25 five tan square root of theta 25 tangent squared of theta and now we can try to simplify all of this business this is equal to let's see this is equal to 250 times five secant squared theta over twenty five times one plus tangent squared theta D theta so 25 goes into 250 ten times and then one plus tangent squared theta is secant squared theta you can prove it for yourself if you rights tan squared theta as sine squared theta over cosine squared theta add that to 1 which is the same thing as cosine squared theta over cosine squared theta and then you can use some basic basic trig identities to realize that this this is equal to secant squared theta which simplifies our expression a good bit you get secant squared theta over secant squared theta which is just 1 and so you're just left with the 10 times the 550d theta so this is equal to 50 I'll take the 50 outside the integral 50 D theta which is equal to 50 theta 50 theta we can write the plus C just to show that this these are all of the antiderivatives but we only need kind of the most basic antiderivative to evaluate these definite integrals but we right now only have it in terms of theta let's write this in terms of X we set the constraint that theta is equal to arctangent of x over 5 so this is equal to 50 arctangent arc tangent of x over 5 plus C these are all the anti derivatives we can pick C is equal to 0 to find n antiderivative of these things to evaluate the definite integrals so let's do that so what we have in blue we can rewrite we can rewrite as the limit as n approaches negative infinity of of the antiderivative of this or an antiderivative of this which we could say is 50 arctangent arc tangent of x over 5 and we're going to evaluate it at 0 and N and to this we're going to add the limit as M approaches positive infinity of 50 just antiderivative this 50 arctangent not act tangent arc tangent arc tangent of x over 5 evaluated from 0 to M from 0 from 0 to M and let me put a little parentheses right around this x over 5 so what's this going to be this is going to be 50 so let me write it it's going to be the limit as n approaches negative infinity of 50 arc tangent of 0 over 5 minus 50 arc arc tangent of n n over 5 and then to that we're going to add to that we're going to add them give myself even a little bit more space the limit limit as M approaches infinity of 50 arc tan arc tan of M over 5 minus 50 arc 10 I see where I think you see where all this is going minus 50 arc tan of 0 over 5 so now we can evaluate these things and to help us evaluate them let's think about a unit circle so let's think about a unit circle so it can really visualize the arctangent function so the tangent one way to think about the tangent is the slope of the line that is on the angle or that helps define the terminal side of an angle so for example if we have an angle that looks like this this is an angle between the positive x-axis and this line right over here if we have an angle just like that the tangent of that angle is going to be the slope of this line and so one way to think about it is okay if I'm thinking about if I if I want the arc tangent of 0 it's essentially saying okay let's get a line where the term or an angle where the terminal side has slope 0 well an angle where the terminal side has slope 0 is an angle of 0 so arc tangent of 0 is 0 radians so this 50 times 0 that's just going to be 0 now we could also write that over here this is just going to be zero so what we're left with what we're left with is the limit the limit as n approaches negative infinity of negative 50 arctangent arc tangent of n over 5 plus plus the limit as M approaches positive infinity of 50 arctangent arctangent of M over 5 so let's think about what these are going to be so the limit as n approaches negative infinity one way to view that is the limit as the slope of this terminal side approaches negative infinity so that's getting more and more negative more and more negative it approaches it's that negative in finger it's approaching negative infinity negative infinity when our angle when our angle right over here is negative PI over 2 radians negative PI over 2 radians so arc tan of or you could say the limit of the limit of arc tan of n over 5 as n approaches negative infinity this part right over here is going to approach as n approaches negative infinity it's going to be negative PI over 2 and then we're just going to multiply that times the negative 50 so this is going to be negative 50 times negative PI over 2 negative PI over 2 which is going to be positive 25 pi so this is going to be 25 pi and similarly our 10 of M over 5 is M approaches infinity well as m approaches infinity the slope of the terminal side of the angle approaches infinity so the slope is getting higher higher it approaches infinity as it gets more close to closer and closer to vertical so the arc tan of M over 5 is M approaches infinity is going to be positive PI over 2 this is an angle of positive PI over 2 so we see an orange right over here it's going to be positive 50 times pi over 2 which is going to be 25 pi so the area that we have in blue going back to our original problem the area that we have in blue is 25 pi the area that we have in orange is 25 pi and so if we wanted to answer our original question what's the total area under this curve which is kind of a cool and a cool question to try to answer we get it's 25 pi put plus 25 pi which is 50 50 pi and we're done