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## AP®︎/College Calculus AB

### Course: AP®︎/College Calculus AB > Unit 9

Lesson 3: AP Calculus AB 2011 free response- 2011 Calculus AB free response #1a
- 2011 Calculus AB Free Response #1 (b, c, & d)
- 2011 Calculus AB free response #2 (a & b)
- 2011 Calculus AB free response #2 (c & d)
- 2011 Calculus AB free response #3 (a & b)
- 2011 Calculus AB free response #3 (c)
- 2011 Calculus AB free response #4a
- 2011 Calculus AB free response #4b
- 2011 Calculus AB free response #4c
- 2011 Calculus AB free response #4d
- 2011 Calculus AB free response #5a
- 2011 Calculus AB free response #5b
- 2011 Calculus AB free response #5c
- 2011 Calculus AB free response #6a
- 2011 Calculus AB free response #6b
- 2011 Calculus AB free response #6c

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# 2011 Calculus AB free response #6c

Average value of a piecewise-defined function on an interval. Created by Sal Khan.

## Want to join the conversation?

- At the end, shouldn't we distribute the
`1/2`

over the expression in the brackets to get`(13/8)-cos(-1)-((e^-4)/8)?`

The solution feels unfinished to me without taking this final step.(7 votes) - Why didn't Sal simplify cos(-1)? That's zero and that term drops. Distribute the 1/2 and = 13/8-(e^-4)/8(0 votes)
- Hi blahdee327. I hope you don't mind that I changed your comment to a question and did some editing to it.

Are you sure that cos(-1) = 0? If you graph the function y=cos(x), you'll see that that is not the case. Indeed, the function y = cos(x) is equal to zero at - π/2 which does not equal -1.

You may have gotten confused because cos(0) = 1.(11 votes)

- the exam of AP calculus the word "BC" stands off for integral calculus right¿(2 votes)
- BC is a different course and comprises of a harder level of Calculus. Both of the exams contain integrals.(1 vote)

- At1:07, Sal takes the definite integral from 0 to 1 of the function f(x). The function f(x)at x=0 is defined as 1-2sin(x). Everywhere else on the interval, the function f(x) is defined as e^-4x. Even though we determined in part (a) that the function f(x) is continuous at x=0, how can we take a definite integral of a function over an interval that contains the dividing line between two parts of a piecemeal function. Is this action defined in such a case?(2 votes)
- It would be ok even if the function wasn't continuous (as long as it's defined for every x in the integration interval).

Suppose we have a piecewise defined function f(x): g(x) when x ≥ c, h(x) otherwise. Suppose we now want to integrate f(x) from a to b, and that c is in the interval. We simply have to split the interval in two: first integrate f from a to c, then from x to b.

∫[a,b] f(x) dx = ∫[a,c] f(x) dx + [c,b] f(x) dx = ∫[a,c] h(x) dx + ∫[c,b] g(x) dx

(Being technical, since f(x) is defined to be h(x) only when x < c but not x = c, the integral ∫[a,c] h(x) dx should actually be the limit of ∫[a,t] h(x) dx as t → c-. But this doesn't actually effect the end result.)(1 vote)

## Video transcript

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.