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Part C find the average value of F on the interval negative 1 to 1 so the average value of a function over an interval is just going to be so let's just write average the average value of our function is just going to be the integral over the interval negative 1 to 1 of f of X D of X divided by our change in X so divided by sorry this is from negative 1 to 1 negative 1 to 1 f of X D of X divided by our change in X and our change in X is 1 minus negative 1 so this is going to be equal to 1/2 times the integral and here f of X is piecewise defined so what we can do is break up this integral into two intervals we could say the integral of f of X from negative 1 to 0 of f of X DX plus let me just write it this way so we don't have to keep rewriting the one half let me write I'll do use brackets plus the integral from from 0 to 1 of f of X f of X DX and the reason why I broke it up like this is because this function to is has a different definition or it takes it's a different it's piecewise defined it's different when we're when we're less than or equal to 0 at vs. when we are greater than 0 so that's why I like to break it up this way so then we get this is equal to 1/2 times and in brackets in big brackets like this this first part right over here we can write as the integral from negative 1 to 0 what is f of X between negative 1 & 0 it's 1 minus 2 sine of X 1 minus 2 sine of X D DX and then plus this thing right over here plus the integral from 0 to 1 and what is our function when it's between 0 & 1 it's e to the negative 4x e to the negative 4x DX and now we can do each of these integrals separately and so this is going to be equal to this is going to be equal to 1/2 and once again I'll do a big open bracket right over here my pin is getting loose let me tie on the front a little bit better there you go all right back to work so one half open brackets and now let's take the antiderivative of 1 minus 2 sine of X antiderivative of 1 with respect to X is just X negative 2 sine of X well the derivative of cosine of X is negative sine of X so this is just going to be 2 cosine of X and you can you can verify that derivative cosine of X is negative sine of X you have the 2 out front so it's negative 2 sine of X so we're going to evaluate that at 0 and at negative 1 and to that we're going to add so let's do this definite integral over here the antiderivative of e to the negative 4x the antiderivative is going to be equal to negative negative e to the negative 4x over 4 and the way you realize that is if you had just an 8 if this was just an e to the X then its antiderivative would just be e to the X if you have an e to the negative 4x then you know that whatever its antiderivative is is going to have a it's going to be essentially an e to the negative 4x but when you take its derivative you would have to take the derivative of the negative 4x part because of the chain rule and so you're gonna have to have a negative 4 that comes out but we don't see a negative 4 over here we don't see a negative 4 here so we're obviously going to have to divide by a negative 4 so it cancels out another way to think about it is we could have rewritten we could have rewritten this thing so this is equal to e to the negative 4x DX this is exactly what our problem is what we need to take the definite integral and so that it becomes completely clear so that the derivative of this thing is sitting around here we could put a negative 4 over here but you can't just really nearly throw negative 4 there you could put you would have to put a negative 1/4 outside of it in order for it to be in order to negative 1/4 times negative 4 you're not you're just multiplying by 1 which doesn't change the value and then here you'd clearly see here you'd clearly see that this is this right over here is of e to the negative 4x so you'd have e to the negative 4x the antiderivative this is e to the negative 4x and then you have this negative 1/4 over here so either way either way hopefully that makes sense we go into more detail that earlier in the calculus playlist so then this is from we're going to evaluate it at 1 and at 0 evaluated at 1 and at 0 and then we want to close the brackets and so what do we have here this is equal to 1/2 once again open the brackets if you evaluate all of this at 0 you get 0 plus 2 cosine of 0 cosine of 0 is 1 so you just get a 2 when you evaluate it at the 0 so you get a 2 and then from that we want to subtract whatever we get when we evaluated at negative 1 we want to subtract negative 1 plus 2 cosine of negative 1 plus 2 cosine 2 cosine of negative 1 so that's this whole thing right over here evaluates to this whole thing right over there and then we have plus we want to evaluate this at 1 so this gives us negative e to the negative 4 it's 4 times 1 is just or negative 4x when X is 1 is negative 4 over 4 and from that we need to subtract this thing evaluated at 0 we need to subtract this thing about is ear so that's going to be negative e to the 0 over 4 well that's just e to the 0 is 1 so that's just negative 1 negative 1 over 4 and once again all of this is going to be multiplied by 1/2 and now we just have to simplify it so this is equal to well just still throw out the 1/2 here this is equal to 1/2 times so I'll do this in a new color here let me do it well I don't have too many new colors ok so this is equal to 2 plus 1 so this and this together is going to be equal to 3 and then you have this negative outside so that Y that's why it was a plus 1 and then you have a negative times a negative times a positive 2 so minus 2 cosine of negative one and then you have plus plus a minus minus e to the negative four over four and then you have plus one-fourth plus one-fourth and then we want to close our brackets and then we could do one last step of simplification here three we could add the three to the one-fourth three is the same thing as 12 over 4 12 over 4 plus one-fourth is thirteen fourths see if thirteen fourths minus two cosine two cosine of negative one minus e to the negative four over four and then of course you have your one-half sitting out here and it's not the most beautiful or simple thing but this is this is our answer this is the average value of f of x over that interval

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