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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition) > Unit 12

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 #4a

Taking derivatives and integrals of strangely defined functions. Created by Sal Khan.

## Want to join the conversation?

- At3:33we try to derive the integral evaluated from 0 to x. Would we not want to subtract the lower limit of f(0) from f(x)? Why is g'(x)=2+f(x) and not g'(x)=2+(f(x)-f(0))? This changes the answer since f(0) has a value.(7 votes)
- When you integrate, you get something like F(x) - F(0). F(x) is a function of x, while F(0) is an evaluation of F(x) at x = 0, which is some constant number. When you derive F(x) - F(0), d[F(x)]/dx = f(x), while d[F(0)]/dx = 0, since the derivative of a constant is zero. This equates to f(x) + 0 = f(x). Therefore, g'(x) = 2 + [f(x) - 0] = 2 + f(x).(13 votes)

- Why does the derivative of f(t) is f(x)?(2 votes)
- because of the first FTC. When you have an integral from a to x of some function with a different variable, in this case f(t) you simply replace it with x to make f(x) when you take the antiderivative.(3 votes)

- Why did Sal put -3 in f(x) when g'(-3) and f(x) are two different functions?(2 votes)
- The derivative of an integral from any number c to x (with respect to x) is just going to be that function at x. Basically, the derivative and integral simply cancel each other out, and since c is simply a constant it will disappear leaving just f(x), or in this case f(-3) since x is -3.(2 votes)

- Was anyone else tempted to find the equation of the circle (y^2 + x^2 = 9) and integrate it?

I always seem to miss the simple solution.(1 vote)

## Video transcript

The continuous
function f is defined on the interval
negative 4 is less than or equal to x is less
than or equal to 3. The graph of f consists of two
quarter circles and one line segment. So this right here is
one quarter circle, then we have another
quarter circle, and then it has this
line segment over here, as shown in the figure above. I put the figure on
the side, so that we have some screen real
estate down here. Let g of x be equal to 2x plus
the definite integral from 0 to x of f of t dt. All right, that's a
little bit interesting. Now, let's do Part a. Part a. Find g of negative 3. So we want to find
g of negative 3. So first of all, it
tells us what g of x is. So g of negative 3,
every time we see an x, we just put a negative 3 there. So it's going to
be 2 times negative 3 plus the definite integral
from 0 to negative 3, f of t dt. So this first part is
pretty straightforward. 2 times negative
3 is negative 6. And then this part
right over here is the definite integral from
0 to negative 3 of f of t dt. So this is essentially
the area under f of t-- or under f of x, if you want to
view it that way-- from 0 to 3. And this is pretty
straightforward to figure out but we
have to be careful here. Because this area
right over here, this would be the integral--
let me do this in a color that you can see. That area right
over there would be the integral from
negative 3 to 0 f of t, or we could even say f of t
dt or f of x dx, either way. That would be this
area right over here. They've swapped it. They have the larger
number as the lower bound. They have 0 as the lower bound. So what we could do,
we could rewrite this. This is the same thing,
this is equal to negative 6 minus-- if you swap the
bounds of integration, you swap the sign
on the integral. So we could say minus from
negative 3 to 0 of f of t dt. And now this thing right over
here-- this expression right over here is the area
under this quarter circle. And the area under
this quarter circle, we can just use a little bit
of geometry to figure that out. We know its radius. The radius here is 3. Radius is equal to 3. The entire circle
is pi r squared. So the entire circle--
the area if this was an entire circle right over
here-- that area would be pi r squared. So pi times 3 squared. So 9 pi. So it would be 9 pi, but this is
only 1/4 of the entire circle. So we want to divide it by 4. So this right over here we'll
evaluate to 9 pi over 4. So Part a, we get it's
equal to negative 6, that's that right over there,
minus 9 pi over 4. And that's at least the
first part of Part a. Then they say find g prime of x. So let's find g prime of x. We did the first part of
Part a right over there. Then we need to
find g prime of x. Well, it's just going to be
the derivative of g of x. The derivative of 2x
is just going to be 2. And then the derivative-- and
this is the fundamental theorem of calculus right over
here-- the derivative of the definite integral
from 0 to x of f of t dt, that's just going to be f of x. So that is g prime of x. We did this second
part right over there. That is g prime of x. And then we need to
evaluate it at negative 3. So g prime of
negative 3 is going to be equal to 2
plus f of negative 3, which is equal to 2 plus-- and
let's look at our-- let's see what f of negative 3 is. This is our function definition. When x is equal to negative
3, our function is at 0. So f of negative 3 is just 0. So it's just two plus
0, which is equal to 2. So g of negative
3 is equal to 2. And we're done with Part a. And the trickiest
part of this is just realizing the bounds
of integration. That you might be tempted to
say, oh, the integral between 0 and negative 3, that's
just this area over here because we're going
between 0 and negative 3. But you've got to
realize that we actually have to swap them and make
them negative if you really want to consider this
area right over here. That this integral is going to
be the negative of this area. And that's what we did when we
went through this step right over there.