 Let's chat a little bit more about properties of definite integrals in the context of our lecture 45 here in our series In this situation suppose we know the integral of f from 0 to 10 is 17 And we know the integral from 0 to 8 of f is equal to 12 Can we calculate the integral from 8 to 10? Well the thing to notice here if we want to do the integral from 8 to 10 of f of x dx here This is the same thing as the integral of 0 to 10 f of x dx plus the integral From 0 to 8 f of x dx So we saw we saw in a previous lecture here that if we want to integrate from a to b of any function f of x dx This is the same thing as integrate from a to c f of x dx plus the integral from c to the c to 10 to b right here. Sorry if f of x dx So actually I kind of fudged a little bit what I wrote before This wouldn't be from if we want to do 8 to 10. We should be a little bit more careful with our bounds So let's try that again This time around we actually would want to do 8 to 0 and Then 0 to 10 so here we're using these values where 8 was the original value a 10 was the original value b and then we're introducing this this alternate this this extras term see right here and So we know how to integrate from 0 to 10. That's just this thing right here, but how do we integrate from? 0 to 8 we have Sorry, we actually have 0 a we don't have 8 to 0, but we can actually flip this second one around That is this one right here Comes down and if we flip around the order so we get negative 0 to 8 put the smaller one on the bottom f of x dx plus the integral from 0 to 10 of f of x dx Like so Then using the information we know before the integral from 0 to 8 of f of x was 12 So we get negative 12 and then we add to that the integral from 0 to 10 of f of x which is 17 and so we can see that 17 take away 12 is going to equal 5 Which is the the integral right here so using those properties of integration be very helpful and calculating these integrals in a similar calculation We see right here Let me add to that list a little bit some properties of integration Something we might call the comparison tests for definite integrals and the comparison test basically if I were to illustrate it We'll say something like the following Let's look at property 6 right here if f of x is greater than equal to 0 for all x inside of an interval So it's it's a non-negative function then the air then the integral from a to b of f of x dx will also be greater than equal to 0 So the idea is that we have a function f and it sits entirely above The x-axis or maybe it touches the x-axis in certain places Well, if we take our values a to b and we look at the area under the curve This is going to be a positive area. That's all that property 6 is saying if you're above the x-axis The area of the curve will be positive For the next picture for for number 7 right here If f of x is greater than or equal to g of x for all x inside that interval Then the integral from a to b of f of x dx will be greater than equal to integral from a to b of g of x dx there and so imagine something like the following we have some function f right here We have some smaller function. Let's say g Right g sits below f of x and then Let's take our x-axis right here. So if we pick some values a and b what can we say about the area? Well, the area that sits below The area that sits below f is going to be bigger than the area that sits below g because f is bigger than g So we get that type of comparison going on right there and in the last condition Give some explanation to what that's going on there is if if little m is less than or equal to f of x and capital M is greater than or equal to And capital M is greater equal to f of x for all x is there Then we're gonna get that m little m times b minus a is less than equal to the integral Which is less than equal to m minus b minus a this this right here 8 is just a special case of 7 right here The idea is if you have a function You have some function f and we have a domain on the x-axis here go from a to b Let's pick the maximum and minimum values on this function, right? So our maximum value would be about right here. So this is our capital M And if we take the horizontal line associated to y equals M We get this flat line like this if we look for the smallest value there So on this section be about right here little m and if we take the horizontal line associated to that See right there and then we're gonna slice it according to x equals a and x equals b What seems to be going on right here? Well, if you look at the area below the red curve This is a rectangle and that's a rectangle whose height is m little m and whose length is b minus a And so the area of that rectangle will be m times b minus a what you see right here Well, if you treat y equals M as a constant function that constant function sits below the function f of x So by property 7 the area under the constant function will be the area It'll be less than the area under f and that area just so happens to be a rectangle And then if we reverse directions with capital M capital M is for maximum little m is for minimum here If you take capital M. Well the area under the y equals capital M. That is also again a rectangle its height Its height here is capital M and whose width is still b minus a so the area of that rectangle will be bigger than the area of this This integral right here. Let me kind of show you how one could use that to help you out with some estimation here So let's say we look at the integral from 0 to 1 of e to the negative x squared dx If you remember from a previous video Integrating exponential functions is extremely difficult even with the forthcoming fundamental theorem of calculus This would still be a very challenging problem, but what we can do is make some estimations If we take f of x to equal e to the negative x squared Well, we can say is the following on the interval 0 to 1 this function Is always decreasing Just to give you an idea of the graph if we were to graph this thing This this graph would look something like the following right? This isn't perfectly drawn to scale, but this will suffice for us And so we're looking at an interval like something like this from 0 to 1 Just apologize for my crudeness there, but we're gonna get this decreasing function So since the function is decreasing what this tells us is that f of x will be It'll be greater than or equal to the point on the right f of 1 Whoops that looks like f of 11 f of 1 right there Because as it's going as it's going down the curve f of 1 down here is the smallest point on that section and Likewise It's gonna be smaller than f of 0 because again as this function is decreasing The biggest point is gonna be at f of 0 right there And so we can do these calculations f of 0 if we plug 0 into the function We get e to the 0 which equals 1 on the other side We're gonna end up with e to the negative 1 squared and so this ends up with just 1 over e as our bounds So applying this the property from before the integral from 0 to 1 of e to the negative x squared dx This will sit below 1 times 1 minus 0 the length of the interval But it'll sit above 1 over e times 1 minus 0 as the length of the interval is 1 in both situations We get our integral from 0 to 1 of e to the negative x squared dx It sits below 1, but it sits above 1 over e For which 1 over e Is approximately 0.367 so that is a pretty wide range of what we have right there But it still gives us some estimates on how big this integral is so even if we don't calculate it exactly We can estimate the area under this curve Using using these these type of comparison to like test here And so that's actually gonna include lecture 45 in our series math 12 10 calculus one Thanks for watching today. If you like these videos, please like them below Leave comments if you have any questions subscribe If you want to see any if you want to see more videos like this in the future And I will hopefully see you all next time. Keep on calculating everyone. Bye