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Worked examples: Finding definite integrals using algebraic properties

FUN‑6 (EU)
FUN‑6.A (LO)
FUN‑6.A.1 (EK)
FUN‑6.A.2 (EK)

Video transcript

- Do you want to evaluate the definite integral from three to three, of F of X, D X. 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, D X, from let's say, A to the same value, from one value to the same value, this is always going to be equal to zero. We're going from three to three. We could be going from negative pi to negative pi. It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one. So here, we want to find the definite integral from seven to four, of F of X, D X. So we want to go from seven to four. So we want to go from seven to four. Now you might be tempted to say, okay well look, the area between F of X and X, is two, so maybe this thing is two. But the key realization is, this area only applies when you have the lower bound as the lower bound, and the higher value as the higher bound. So the integral from four to seven of F of X, D X, this thing. This thing is equal to two. This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four. Well the key realization is, is if you switch the bounds, this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven, of F of X, D X. And so this is going to be negative, and we just figured out the integral from four to seven of F of X, D X, 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 two, but we have that negative out front, so our original expression would evaluate to negative two.
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