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FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

the graph of f is shown below let G of X be equal to the definite integral from 0 to X of F of T DT now at first when you see this you're like wow this is this is strange I have a function that is being defined by an integral a definite integral but one of its bounds are X and you should just say well well this is ok a function can be defined any which way and as we'll see it's actually quite straightforward to evaluate this so G of negative 2 G of negative 2 and I'll do the negative 2 in a different color G of negative 2 well what we do is we take this expression right over here this definite integral and everywhere we see an X we replace it with a negative 2 so this is going to be equal to the integral from 0 to X of and I'll write X in a second F of T DT well X is now negative 2 this is now negative 2 and so how do we figure out what this is now before we even look at this graph you might say ok this is the the region under the area of the region under the graph y equals f of T between negative 2 and 0 but you have to be careful notice our upper bound here is actually a lower number than our lower bound right over here so it will be nice to swap those bounds so we can truly view it as the area of the region under F of T above the T axis between those two bounds and so when you swap the bounds this is going to be equal to negative definite integral from negative to negative 2 to 0 of F of T DT and now what we have right over here when I'm squaring off in magenta this is the area under the curve F between negative 2 and 0 so between negative 2 and 0 so that is this area right over here that we care about now what is that going to be well you could there's a bunch of different ways that you could do this you could split off into a square in a triangle the area of this square right over here is 4 it's 2 by 2 and it's make sure to make sure you look at the unit sometimes each square doesn't represent one square unit so but in this case it does so that's 4 and then up here this is half of 4 right if it was all of this that would be 4 this triangle is half of 4 so this is 2 right over there or you could view this as base times height times 1/2 which is going to be 2 times 2 times 1/2 and so this area right over here is 6 so this part is 6 but we can't forget that negative sign so this is going to be equal to negative negative 6 and so G of negative 2 is negative 6