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# Worked example: Breaking up the integral's interval

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)

## Video transcript

so this right over here is the graph of the function G of T it is a function of T and let's define a new function let's call it capital G of X and it's equal to the definite integral between t is equal to negative 3 and T is equal to X of our lower case G of T G of T DT so given how we have just given how we have just defined capital G of X let's see if we can evaluate some about let's see if we can find some values so let's evaluate the function capital G at X is equal to 4 let's also evaluate the function capital G at X is equal to 8 and I encourage you to pause the video right now and try to think about these on your own and then we can work through them together so let's tackle capital G of 4 first so this is going to be this is going to be well if X is equal to 4 this this top boundary is going to be 4 so this is going to be the definite integral of G of T between t is equal to negative 3 and T is equal to 4 G of T G of T DT now what's that going to be equal to well let's see let's look at this graph here so we have negative 3 so this T is equal to negative 3 which is right over here T is equal to negative 3 and we're going to go all the way to T is equal to 4 let me circle that in orange all the way to T is equal to 4 which is right over here so this is one way to think about this is this is going to be the area above the T axis and below the graph of G so it's going to be this area right over here that is above the T axis and below the graph of G of T but we're not going to add to it this area because this area over here I'll shaded in yellow this area I'm shading in yellow over here this is going to be negative why is it going to be negative because we want the area that is above T and below G this is the other way or round.this is below the t-axis and above the graph of G so one way to think about it is we could split it up so we could and actually let me just clear this out so we have more space so this right over here is going to be equal to the integral I'll do that in this purple color the integral from T equals negative 3 to 0 of G of T DT plus I'll do this in the yellow color plus the integral from 0 T equals 0 to T is equal to 4 of G of T DT now what are each of these going to be well this is just a triangle where the base is 3 the base has length 3 the height is length 3 so it's going to be 3 times 3 which is 9 times 1/2 because 3 times 3 would give us the area of this entire square but this triangle is half of that so let's go this right over here is going to be 4.5 this is going to be 4.5 and then what's this area in yellow well let's see we have a triangle its basis or its width here is 4 its height is 4 4 times 4 is 16 which would be the area of this entire square we take half of that it's 8 now we don't just add it to it because once again this is going to be negative area the graph over here is below below the T axis so this we would say we would this integral is going to evaluate to negative 8 once again why is it negative 8 because the graph is below below the T axis and so what do we get we get G of 4 this which is this area essentially minus this area 4.5 minus 8 is going to be let's see 4 minus 8 is negative 4 at a point 5 that's negative 3 point 5 negative 3 point 5 now let's try to figure out what G of 8 is so G of 8 and if you couldn't figure it out the first time around try to pause the video again and now that we figured out G of 4 try to figure out what G of 8 is well G of a is one way to think about it's going to be that - that area and then we're going to add and then we're going to figure out the area and then we're going to have two more areas to think about we're going to go all the way to eight so actually let me make draw the line there so we're going to have to think about we're going to think about this whole area now this whole area now and then we're going to think about this one so we could write this is going to be equal to the integral between let me do that purple color so it's going to be this this integral which is the integral between negative T equals negative three and zero G of T DT Plus this entire yellow region right now the part that we looked at before plus this part over here so I'll just write that as plus the definite integral between T is equal to zero and six G of T DT and then finally plus the definite integral between T is equal to six and T is equal to eight G of T DT now we already know that this first one evaluates to four point five so that one is four point five now what's this one going to be we have a triangle whose base is length six height is four six times four is 24 times one half is going to get us 12 so this is going to be that over there is 12 and then finally what is this area oh and we have to be careful this is below the T axis and above the graph so this is going to be a negative 12 and then finally we have this area which is once again going to be a positive area it's below the graph and above the T axis and so let's see 2 times 4 is 8 times 1/2 is 4 so we're going to have plus 4 plus 4 and so what do we have we end up with four point five plus four is eight point five - twelve so this is going to be equal to this is going to be equal to eight point eight minus twelve would be negative four plus five it is negative three point five again now why did these two things end up being the same well think about what happened here when we went from capital G of four to capital G of eight we subtracted we subtracted this value right over here and we added this value right over here and you see that these two triangles have the same area so we subtracted and added the same amount to get the same exact value