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Current time:0:00Total duration:2:19

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

so we want to evaluate the definite integral from 3 to 3 of f of X DX and we're given the graph of f of X and of y equals f of X and the area between f of X and the x-axis over different intervals well when you look at this you actually don't even have to look at this graph over here because in general if I have the definite integral of any function f of X DX from let's say a to the same value from one value to the same value this is always going to be equal to 0 we're going from 3 to 3 we could be going from negative pi to negative pi it's always going to be zero one way to think about we're starting and stopping here at 3 so we're not capturing any area let's do another one so here we want to find the definite integral from 7 to 4 of f of X DX so we want to go from 7 to 4 so we want to go from 7 to 4 now you might be tempted to say okay well look the area between f of X and X is 2 so maybe this thing is 2 but the key realization is this area only applies when you have the lower bound as the lower bound and the and the higher value as the higher bound so the integral from 4 to 7 of f of X DX this thing this thing is equal to 2 this thing is depicting that area right over there so what about this where we've switched it instead of going from 4 to 7 we're going from 7 to 4 well the key realization is is if you switch the bounds and this is a key definite integral property that's going to make that's going to give you the negative value so this is going to be equal to the negative of the integral from 4 to 7 of f of X DX and so this is going to be negative and we just figured out the integral from 4 to 7 of f of X DX well now that is this area f of X is above the x-axis it's a positive area so it is going to be so this thing right over here is going to evaluate to positive 2 but we have that negative out front so our original expression would evaluate to negative two

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