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Main content
Current time:0:00Total duration:6:14
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)

Video transcript

so we have an f of X right over here it is defined piecewise for X less than zero f of X is X plus one for X is greater than or equal to zero f of X is cosine of PI X and we want to evaluate the definite integral from negative one to one of f of X DX and you might immediately say well which of these which of these versions of f of X am I going to take the antiderivative from because from negative one to zero I would think about X plus one but then from zero to one I would think about cosine PI X and if you were thinking that you're thinking in the right direction and the way that we can make this a little bit more straightforward is to actually split up this definite integral this is going to be equal to the definite integral from negative one to zero of f of X DX plus plus the integral from zero to one of f of X DX now why was it useful for me to split it up this way in particular to split it up split the interval from negative one to one split it into two intervals from negative one to zero and zero to one well I did that because zero x equals zero is where we switch where f of X switches from being X plus 1/2 cosine PI X so if you look at the interval from negative 1 to 0 f of X is x plus 1 so f of X here is X plus 1 and then when you go from 0 to 1 f of X is cosine PI X so cosine of cosine of PI X and so now we just have to evaluate each of these separately and add them together so let's take the definite integral from negative 1 to 0 of X plus 1 DX well let's see the antiderivative X plus 1 is antiderivative X is x squared over 2 I'm just incrementing the exponent and then dividing by that value and then plus X and you could view this I'm doing the same thing if this is X to the 0 it'll be X to the first X to the first over 1 which is just X and I'm going to evaluate that at 0 and subtract from that it evaluated at 1 sorry it evaluate at negative 1 and so this is going to be equal to if I evaluate it at zero let me do this in another color if I evaluate it at zero it's going to be zero squared over two which is well I'll just write it zero squared over two plus zero well all of that's just going to be equal to 0 minus it evaluated at it evaluated at negative one so minus negative one squared negative one squared over two plus negative one so negative one squared is just one so it's 1/2 plus negative one 1/2 plus negative one is or 1/2 minus one is negative one half so all of that is negative one half but then we're subtracting negative one half zero minus negative one half is going to be equal to positive one half so this is going to be equal to positive one half so this first part right over here is positive one half and now let's evaluate the integral from zero to one of cosine pi I don't need that first parenthesis of cosine of PI X DX what is this equal to now if we were just trying to find the antiderivative of cosine of X it's pretty straightforward we know that the derivative with respect to X of sine of X is equal to cosine of X cosine of X but that's not what we have here we have cosine of PI of PI X so there is a technique here you could call it u substitution you could say u is equal to PI X if you don't know how to do that you could still try to think about think this through where we could say all right well maybe it involves sine of pi X somehow so the derivative with respect to X of sine of PI X would be what well we would use the chain rule it would be the derivative of the outside function with respect to the inside or sine of PI X with respect to PI X which would be cosine of PI X and then times the derivative of the inside function with respect to X so it would be x pi or you could say the derivative of sine PI X is I cosign of PI X now we almost have that here except we just need a PI so what if we were to throw a PI right over here but so we don't change the value we also multiply by 1 over pi so if you divide and multiply by the same number you're just you're not changing its value 1 over pi times pi is just equal to 1 but this is useful this is useful because we now know that PI cosine PI X is the derivative of sine PI X so this is all going to be equal to this is equal to 1 let me let me take that 1 over PI so this is equal to 1 over PI x now we're going to evaluate so the antiderivative here we just said a sine sine of PI X and we're going to evaluate that at 1 and at 0 so this is going to be equal to 1 over PI 1 over PI not pi my hand is not listening to my mouth 1 over pi times sine of PI sine of PI minus sine of minus sine of PI times 0 which is just 0 well sine of PI that's 0 sine of 0 is 0 so you're gonna have 1 over pi times 0 minus 0 so this whole thing is just all going to be equal to 0 so this first part was 1/2 this second part right over here is equal to 0 so the whole definite integral is going to be 1/2 plus 0 which is equal to 1/2 so all of that together is equal to 1/2