# Question about domain under a curve with derivative?

##### New member
So I have this question: The second part I'm not sure at all how to find the equation for the derivative of the curve.. when I don't have the equations for h(x), how would I find the derivative equation? I do know that you put (5/2) in for x once you have the equation.

Also, the x value that gives the maximum h(x) would have to be 4 right? Any help would be great!

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#### Deveno

##### Well-known member
MHB Math Scholar
It looks like you are confusing domain with range.

f is only defined on [0,6], so we can only integrate when the upper limit is between these values.

If $0 \leq 2x - 1 \leq 6$, what does that say about $x$?

For part (b) you are expected to use the Fundamental Theorem of Calculus. What does that say?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So I have this question:

The second part I'm not sure at all how to find the equation for the derivative of the curve.. when I don't have the equations for h(x), how would I find the derivative equation? I do know that you put (5/2) in for x once you have the equation.

Also, the x value that gives the maximum h(x) would have to be 4 right? Any help would be great!
Welcome to MHB, pleasehelpme! I think you did not read the question properly for (A).
It did not ask to integrate f over its domain (which you calculated just fine).

What is the lowest value of x for which h(x) is defined?
And the highest value?

As for (B).
Suppose F(t) is the anti-derivative of f(t).
Then:
$$h(x) = \int_0^{2x-1} f(t) dt = F(2x-1) - F(0)$$
Take the derivative:
$$h'(x) = F'(2x-1)\cdot 2 - 0$$

Care to finish it?

##### New member
It looks like you are confusing domain with range.

f is only defined on [0,6], so we can only integrate when the upper limit is between these values.

If $0 \leq 2x - 1 \leq 6$, what does that say about $x$?

For part (b) you are expected to use the Fundamental Theorem of Calculus. What does that say?
Okay, correct me if I'm off.. Does this mean .5<(or equal to) x <(or equal to) 3.5? Or am I not supposed to calculate each side through adding one and dividing by 2?.. Maybe I'm wrong.

##### New member
Welcome to MHB, pleasehelpme! I think you did not read the question properly for (A).
It did not ask to integrate f over its domain (which you calculated just fine).

What is the lowest value of x for which h(x) is defined?
And the highest value?

As for (B).
Suppose F(t) is the anti-derivative of f(t).
Then:
$$h(x) = \int_0^{2x-1} f(t) dt = F(2x-1) - F(0)$$
Take the derivative:
$$h'(x) = F'(2x-1)\cdot 2 - 0$$

Care to finish it?
I'm still working on part A. Do you calculate it like an algebraic equation?

So to finish it...(for part B) h'(x)=F'(4x-2)?..which is 8?

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#### Deveno

##### Well-known member
MHB Math Scholar
Okay, correct me if I'm off.. Does this mean .5<(or equal to) x <(or equal to) 3.5? Or am I not supposed to calculate each side through adding one and dividing by 2?.. Maybe I'm wrong.

Yes, option 2: "solve" each inequality as if they were equations...you will get an interval that $x$ must lie in.

Again, I ask you....what does the Fundamental Theorem of Calculus say? (It's called "fundamental" for a reason....)

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I'm still working on part A. Do you calculate it like an algebraic equation?
Yes. You already figured it out.

So to finish it...(for part B) h'(x)=F'(4x-2)?..which is 8?
Erm... I'm afraid not.
F'(2x-1) means the derivative of F applied to 2x-1 (instead of t).
Since F(t) is the anti-derivative of f(t), that means that F'(t) = f(t).
So:
$$h'(x) = F'(2x-1) \cdot 2 = f(2x-1) \cdot 2 = 2f(2x-1)$$
That leaves substituting $x=\frac 5 2$.

Alternatively, h'(5/2) is the derivative for x=5/2.
That is the rate of change of h(x) when x=5/2.
At that point we have:
$$h\left(\frac 5 2\right) = \int_0^{2\cdot 5/2 - 1} f(t) dt = \int_0^{4} f(t) dt$$
The right hand side represents the area below the graph of f between x=0 and x=4.
What might be the rate of change at x=4?

##### New member
Yes. You already figured it out.

Erm... I'm afraid not.
F'(2x-1) means the derivative of F applied to 2x-1 (instead of t).
Since F(t) is the anti-derivative of f(t), that means that F'(t) = f(t).
So:
$$h'(x) = F'(2x-1) \cdot 2 = f(2x-1) \cdot 2 = 2f(2x-1)$$
That leaves substituting $x=\frac 5 2$.

Alternatively, h'(5/2) is the derivative for x=5/2.
That is the rate of change of h(x) when x=5/2.
At that point we have:
$$h\left(\frac 5 2\right) = \int_0^{2\cdot 5/2 - 1} f(t) dt = \int_0^{4} f(t) dt$$
The right hand side represents the area below the graph of f between x=0 and x=4.
What might be the rate of change at x=4?
Oh okay! But I'm not sure how to calculate the rate of change at 4. I feel stupid. h(x) is at it's maximum at 4 isn't it? And then the area starts reducing, but I don't think that's what you're talking about.

- - - Updated - - -

Yes, option 2: "solve" each inequality as if they were equations...you will get an interval that $x$ must lie in.

Again, I ask you....what does the Fundamental Theorem of Calculus say? (It's called "fundamental" for a reason....)
If I remember correctly it is F'(x)=f(x) when F(x) is any antiderivative of f?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Oh okay! But I'm not sure how to calculate the rate of change at 4. I feel stupid. h(x) is at it's maximum at 4 isn't it? And then the area starts reducing, but I don't think that's what you're talking about.
That is exactly it.
Since h(x) is at its maximum, its rate of change is 0.

Btw, can you substitute $x=\frac 5 2$ in $h′(x)=2f(2x−1)$ and evaluate its value?

#### Deveno

##### Well-known member
MHB Math Scholar
The Fundamental Theorem of Calculus states:

If $f$ is continuous on $[a,b]$ with:

$\displaystyle F(x) = \int_a^x f(t)\ dt$

then on $(a,b),\ F'(x) = f(x)$.

Now, here, $h$ is not quite of that form, we have:

$\displaystyle h(x) = \int_0^{2x-1} f(t)\ dt$

where:

$f(t)$ has the graph shown.

So to mimic the formula in the fundamental theorem, write:

$\displaystyle F(u) = \int_0^u f(t)\ dt$

where $u(x) = 2x - 1$.

We know that $h(x) = F(u) = F(u(x)) = (F\circ u)(x)$

so: $h'(x) = (F'(u(x))(u'(x))$ by the *chain rule*,

and we know that $F'(u) = f(u)$.

That means if: $x = \frac{5}{2}$, then:

$f(u) = f(2\frac{5}{2} - 1) = f(5 - 1) = f(4)$.

So $F'(u(x)) = f(u(x)) = f(4)$ for $x = \frac{5}{2}$.

But that is just one factor of $h'(x)$, we also need to compute the other factor:

$u'(x) = 2$.

So $h'(\frac{5}{2}) = 2\ast f(4)$.

It should be obvious from the graph of $f$ that "near" 4 we are adding very little area to the total, since the "slice" of either of the two semi-circles has almost no height.

##### New member
That is exactly it.
Since h(x) is at its maximum, its rate of change is 0.

Btw, can you substitute $x=\frac 5 2$ in $h′(x)=2f(2x−1)$ and evaluate its value?
Alright, so that's 2*f(4), is there a way to simplify that more or is that my answer? Also if the rate of change is 0 how does that affect h'(5/2)? Does it make my answer 0, or 2*f(4)?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Alright, so that's 2*f(4), is there a way to simplify that more or is that my answer? Also if the rate of change is 0 how does that affect h'(5/2)? Does it make my answer 0, or 2*f(4)?
You have the graph of f.
What is its value at 4?

##### New member
You have the graph of f.
What is its value at 4?
The value of y at 4 is 0. But is that what it's asking for with h'(5/2)? or would it want 2f(4)?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
The value of y at 4 is 0. But is that what it's asking for with h'(5/2)? or would it want 2f(4)?
$$h'(5/2) = 2f(4) = 2 \cdot 0 = 0$$

##### New member
$$h'(5/2) = 2f(4) = 2 \cdot 0 = 0$$
Oh! That makes so much sense! I feel ridiculous for asking now. Thanks so much!

#### Deveno

##### Well-known member
MHB Math Scholar
I'm a little troubled by this entire thread. You haven't given any indication that you even understand the fundamental theorem of calculus at all.

You also don't seem to recognize why you need to use the chain rule to calculate $h'(x)$.

And what I mean is, so, OK, we know what $h'(\frac{5}{2})$ is for THIS problem. What will you do when you encounter a similar problem?

For example, can you find $k(1)$ where:

$\displaystyle k(x) = \int_0^{e^{2x}} t^2\sqrt{1-t^2}\ dt$?

If it takes you longer than 5 minutes to write down the answer, then I'm afraid the larger issues remain unresolved.

##### New member
I'm a little troubled by this entire thread. You haven't given any indication that you even understand the fundamental theorem of calculus at all.

You also don't seem to recognize why you need to use the chain rule to calculate $h'(x)$.

And what I mean is, so, OK, we know what $h'(\frac{5}{2})$ is for THIS problem. What will you do when you encounter a similar problem?

For example, can you find $k(1)$ where:

$\displaystyle k(x) = \int_0^{e^{2x}} t^2\sqrt{1-t^2}\ dt$?

If it takes you longer than 5 minutes to write down the answer, then I'm afraid the larger issues remain unresolved.
I'm doing my best. I'll try to solve that, but I can't tell if for the upper limit it says e^2x or e^2(pi). I will try it. This is an advanced work sheet and I'm just trying to see if I can get through the steps. I know that later on I will need to understand the Fundamental Theorems better, but for now we're just supposed to see if we can slowly use the steps to figure out the answers. (What our professor said, although I get that we should understand them better and not worry about the answers). And I have tried so far to understand the two theorems, but I just know some strict formulas for them, and I'm having trouble using those when something looks a little different in the actual problem.

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##### New member
I'm doing my best. I'll try to solve that, but I can't tell if for the upper limit it says e^2x or e^2(pi). I will try it. This is an advanced work sheet and I'm just trying to see if I can get through the steps. I know that later on I will need to understand the Fundamental Theorems better, but for now we're just supposed to see if we can slowly use the steps to figure out the answers. (What our professor said, although I get that we should understand them better and not worry about the answers). And I have tried so far to understand the two theorems, but I just know some strict formulas for them, and I'm having trouble using those when something looks a little different in the actual problem.
So when I tried this I ended up with 296*7.3i.

I don't think this is right. I like using the first fundamental theorem because I understand it better and that's what I used when I think I was supposed to use the second. I'm going to ask my professor to explain the second one and the chain rule because I'm just not grasping it. When I did your example I set it up as d/dx(your example) so that I could use the first rule (f(h(x))*h'(x)-f(g(x))*g'(x)) which I'm comfortable with, instead of the second and I didn't realize I had done that until I was done! Is the second rule more confusing to others too, or am I just missing something?

#### Deveno

##### Well-known member
MHB Math Scholar
I'm doing my best. I'll try to solve that, but I can't tell if for the upper limit it says e^2x or e^2(pi). I will try it. This is an advanced work sheet and I'm just trying to see if I can get through the steps. I know that later on I will need to understand the Fundamental Theorems better, but for now we're just supposed to see if we can slowly use the steps to figure out the answers. (What our professor said, although I get that we should understand them better and not worry about the answers). And I have tried so far to understand the two theorems, but I just know some strict formulas for them, and I'm having trouble using those when something looks a little different in the actual problem.
I believe you are doing your best. What I am saying underscores the difference between "using a formula" and "knowing what it means". If you can do the second, you can do the first, but often not the other way around. You see, when using things in some other context than in the classroom (for example, trying to find an actual area of some actual curve, given by some real function, because, maybe you need to know to determine if your new manufacturing process will bankrupt you) things usually don't come in "the exact form" we need them in, so one has to get better at "manipulating things".

As a side note, the upper limit is $e^{2x}$. Your answer shouldn't be an approximation, but rather "an exact form" (so you can use square root symbols, and irrational numbers in general, don't try to simplify to "decimals").

I'm not exactly sure what you mean by "the first rule" and "the second rule".

The Fundamental Theorem of Calculus comes in "two flavors".

One is:

Let $f:[a,b] \to \Bbb R$ be a continuous function (this means continuous on all of the "inside" $(a,b)$, with "one-sided continuity" at the endpoints).

Then, for $x \in (a,b)$ (an "inside" point), if we define:

$\displaystyle F(x) = \int_a^x f(t)\ dt$

the function $F$ is differentiable on $(a,b)$ with:

$F'(x) = f(x)$.

(a function $G$ such that $G'(x) = f(x)$ is called an "anti-derivative" for $f$. One caution: anti-deriviatives are not unique. For example, $x^2 + 2$ and $x^2$ are both anti-derivatives of $2x$).

Like any worth-while theorem, it can be proven, but like any worth-while theorem, the proof is harder than using it (that's why we have theorems, to avoid "proving" every little thing we do from scratch). Wikipedia has a good "pictoral" explanation of why it works:

File:FTC geometric2.png - Wikipedia, the free encyclopedia

The "other flavor" is more useful for definite integrals:

If $F(x)$ is an anti-derivative for $f(x)$ on the interval $[a,b]$, then:

$\displaystyle \int_a^b f(x)\ dx = F(b) - F(a)$.

The problem here (in both your original problem, and the second problem I posed for you) is that the upper limit isn't "just $x$", it's a function of $x$.

So if we call that function $u(x)$, the Fundamental Theorem (the first one, not the second) tells us:

$F'(u) = f(u)$

However, $h$ is a function of $x$, not of $u$. If we want to measure how fast $h$ changes with respect to $x$, we have to account for how fast $u$ changes with respect to $x$ (it changes twice as fast, the -1 just tracks "which values of $x$ it starts from").

In your particular problem, you "get lucky" because $f(4) = 0$, so the multiplying factor doesn't "trip you up". I designed MY problem so no such coincidence would occur.

********

I suspect what you mean by "the first rule" is just using what I am calling "the second theorem" and just substituting in $x$ for $b$. Let me see if I can make clear the difference.

In the first version, $F(x)$ is a function, and we are talking about the FUNCTION we get when we differentiate the function:

$\displaystyle F(x) = \int_a^x f(t)\ dt$

that is: "$x$ is a variable".

In the second version, $F(b) - F(a)$ is a NUMBER, which represents the area under the curve $f(t)$ from $t = a$ to $t = b$. Here, $b$ is a CONSTANT.

To prove the first from the second, you would need to show that:

1) the function $F(x)$ is indeed a function, as defined.
2) that $F$ is continuous, and differentiable on $(a,b)$.

The trouble is, you need to do this for every single possible anti-derivative of $f$ (and there are infinitely many), and every single point in the interval $(a,b)$ (of which there are ALSO infinitely many). That's a lot harder than it might seem, at first glance. And you can't "mix up different anti-derivatives" because you lose continuity.

It may be that "your first rule" is for deriving values for:

$\displaystyle k(x) = \int_{g(x)}^{h(x)} f(t)\ dt$

in which case the formula you give:

$k'(x) = f'(h(x))(h'(x)) - f'(g(x))(g'(x))$ is indeed correct, and has the chain rule built into it.

Note that if $g(x) = a$ (a constant function), the second term you're subtracting goes away since it has 0 derivative.

*************

I can also understand that your professor may well take the approach "learn how to use the formulas first" and that "you'll learn the (proof) mechanics later" (in a more advanced class, perhaps, or perhaps later in the course). It is perhaps the case that he/she is teaching non-math-majors as well as math majors, and wants to impart some useful information without getting bogged down in theory. I take this as a concession of failure on the professor's part, in the interests of expediency.

It is perhaps understandable, a course has only so long to spend on any given topic, and not all students in the class need to know the "theory" to use the benefits (you don't have to know how a calculator works to use it, either). But in "real life" (in math, as in other things) theory pwns examples. Knowing how to fish, and having once caught a fish, are two different things.