If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:5:27

- [Voiceover] So what we're
gonna do in this video is several examples where
we evaluate expressions with definite integrals and so right over here we
have the definite integral from negative two to three of 2 f(x)dx plus the definite integral
from 3 to seven of 3 f(x)dx and all we know about f(x) is the graph of y=f(x) from negative, from x equals negative six to x equals seven. They also give us the areas between f(x), y equals f(x) and the x axis. The negative areas show that our function is below the x axis and so given that, can we evaluate that and like always, pause the video and see if
you can do it on your own. Well the first thing that
my brain wants to do is I want to, I want to take these
constants out of the integral because then once they're out and I'm just taking the
straight up definite integrals of f(x), I can relate that to the areas over here and I know I can do that and this is a very common
integration property and applies to definite
and indefinite integrals but if I'm taking the integral of k f(x)dx, this is the same thing as k
times the integral of f(x)dx. So let's just apply that property there which is really you're
taking the scaler outside of the integral. To say this is going to be the same thing as two times the definite
integral from negative two to three of f(x)dx plus three times the integral from
three to seven of f(x)dx. All right. Now can we evaluate these things? So what is this going to be? The definite integral
from negative two to three of f(x)dx. Well we can view that as the area between the curves y
equals f(x) and the x axis, between x equals negative
two and x equals three. So between x equals two and x equals three, they give us the area
between y equals f(x) and the x axis. It is seven. So this thing over here is seven and then we have the
integral from three to seven of f(x) so we're gonna go from three to seven and once again, well this is going to evaluate to a negative value because f(x) is below the x axis there and it's going to evaluate
to negative three. So this is all going to be this is going to be two
times seven so 14 plus, 14 plus three times negative three so plus negative nine and so 14 minus nine is equal to five. This is fun. Let's do more of these. All right. Okay, so here this first integral the integral from zero to five of f(x)dx, so this is pretty straightforward. This right over here, we're gonna go from zero to five. Zero to five and of f(x)dx, so we're talking about this area there which they tell us is four so that was pretty easy to evaluate and now we're gonna subtract, we're gonna subtract
going from negative eight to negative four times 2f(x), well let's just take this two outside. So if we just take this two outside, then this just becomes the
integral from negative eight to negative four of f(x) and so this thing, this thing right over here evaluates to five. It's this area they're talking about. So this is all going to simplify to four minus that two that we brought out, minus two times five. Times five, which is equal to let's see four minus 10 which is equal to negative six. All right. Let's do another one of these. So here I have the integral
from negative seven to negative five so I'm going from negative seven to negative five which is, it's gonna be right around there so I want to find, I want to find this area right over here. So I want to find that area. That's that and then I'm gonna go from
negative five to zero. So then this is going to be going from negative five to zero so it's gonna be all of that. Now there's a couple of
ways you can think about doing it. You can assume I have some symmetry here and they don't tell it for sure but it looks very symmetric
around x equals negative five so you could assume that this eight is split between these two regions but an easier way to do it is just to realize look, I went from negative seven to negative five and then from negative five to zero and I'm integrating the same thing. F(x)dx so this integral I can rewrite as the integral from negative seven so negative seven all the way to zero of f(x), f(x)dx and so really that's just going to be the net area between negative seven and zero and so we have the positive eight there so this is going to be
equal to the positive eight and then we have the negative one there so minus one which is equal to seven.