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

Worked examples: Definite integral properties 2

Video transcript

so what we're going to do in this video is several examples where we evaluate expressions with definite integrals and so right over here we have the definite goal from negative two to three if two f of X DX plus the definite go from three to seven of three f of X DX and all we know about f of X is the graph of y equals f of X from negative from x equals negative six to x equals seven they also give us the areas between f of X y equals f of 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 of X I can relate that to the area's over here and I know I can do that and this is a very common integration property it applies to definite and indefinite integrals but if I'm taking the integral of K f of X DX this is the same thing as K times the integral of f of X DX so let's just apply that property there which is really you're taking the scalar 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 of X DX plus three times the integral from three to seven of f of 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 of X DX well we can view that is the area between the curves y equals f of X and the x-axis between x equals negative 2 and x equals 3 so between x equals 2 and x equals 3 they give us the area between y equals f of 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 of X so we're going to go from three to seven and once again well this this is going to evaluate to a negative value because f of X is below the x-axis there and it's going to evaluate to negative 3 so this is all going to be this is going to be 2 times 7 so 14 plus 14 plus 3 times negative 3 so plus negative 9 and so 14 minus 9 is equal to 5 this is fun let's do more of these all right okay so here this first integral the integral from 0 to 5 of f of X D of X so this is pretty straightforward this right over here we're going to go from 0 to 5 0 to 5 and of f of X D of X DX so we're talking about this area there which they tell us is 4 so that was pretty easy to evaluate and now we're going to subtract we're going to subtract going from negative 8 to negative 4 times 2 f of X well let's just take this 2 outside so if we just take this 2 outside then this just becomes the integral from negative 8 to negative 4 of f of X and so this thing this thing right over here evaluates to 5 it's this area they're talking about so this is all going to simplify to 4 - that - that we brought out - 2 times 5 times 5 which is equal to let's see 4 minus 10 which is equal to negative 6 all right let's do another one of these so here I have the integral from negative 7 to negative 5 so I'm going from negative 7 to negative 5 which is it's going to 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 going to go from negative 5 to 0 so then this is going to be going from negative five to zero so it's going to be all of that now there's a couple of ways you could think about doing it you could assume I have some symmetry here and we 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'm going from negative seven to negative five and then from negative five to zero and I'm integrating the same thing f of X DX so this integral I can rewrite as the integral from negative seven so negative seven all the way to zero of f of X f of 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