0 energipoint
Studying for a test? Prepare with these 3 lessons on AP Calculus AB solved exams.
See 3 lessons

# 2011 Calculus AB frit svar #6c

Video transskription
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. Sorry, this is from negative 1 to 1. 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 can 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 1/2. I'll use brackets-- plus the integral from 0 to 1 of f of x dx. And the reason why I broke it up like this is because this function has a different definition, or it's piecewise-defined. It's different when we're less than or equal to 0, versus 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 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 and 0? It's 1 minus 2 sine of x 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 and 1? It's 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 1/2-- and once again, I'll do a big open bracket right over here. My pen is getting loose, let me tie on the front a little bit better. There you go. All right, back to work. So 1/2, 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 verify that. Derivative of 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 are going to add-- so let's do this definite integral over here. The antiderivative of e to the negative 4x is going to be equal to negative e to the negative 4x over 4. And the way you realize that is, if this was just an e to the x, then it's 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 be, essentially, an e to the negative 4x. But when you take its derivative, you're going to have to take the derivative of the negative 4x part because of the chain rule. And so you're going to have a negative 4 that comes out. But we don't see a negative 4 over here, so we're obviously going to have to divide by negative 4 so it cancels out. Another way to think about it is, we could have rewritten we 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. So then, 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 willy-nilly throw a negative 4 there. You would have to put a negative 1/4 outside of it in order for it to be negative 1/4 times negative 4. You're just multiplying by 1, which doesn't change the value. And then here, you'd clearly see that this right over here is the derivative of e to the negative 4x. So that you'd have e to the negative 4x. The antiderivative of this is e to the negative 4x. And then you have this negative 1/4 over here. So either way, hopefully that makes sense. We go into more detail of that earlier in the calculus playlist. So then this is from-- we're going to evaluate it 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 evaluate it at negative 1. We want to subtract negative 1 plus 2 cosine of negative 1. So 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. Because negative 4x, when x is 1, is negative 4. Over 4. And from that, we need to subtract this thing evaluated at 0. So that's going to be negative e to the 0 over 4. e to the 0 is 1. So that's just 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, I'll 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's why it was a plus 1. And then you have a negative times a positive 2. So minus 2 cosine of negative 1. And then you have plus-- or maybe I should say minus-- minus e to the negative 4 over 4. And then you have plus 1/4. And then we want to close our brackets. And then we could do one last step of simplification here. 3, we could add the 3 to the 1/4. 3 is the same thing as 12 over 4. 12 over 4 plus 1/4 is 13/4. So you have 13/4 minus 2 cosine of negative 1 minus e to the negative 4 over 4. And then, of course, you have your 1/2 sitting out here. And it's not the most beautiful or simple thing. But this is our answer. This is the average value of f of x over that interval.