Integration: area under a curve, and end points?

[dogleg]

I don't understand how you get an accurate value for area under a curve using antiderivatives for just the end points. Could someone give me a simple explanation please. (If there is one!)

 

[pka]

I don't understand how you get an accurate value for area under a curve using antiderivatives for just the end points. Could someone give me a simple explanation please. (If there is one!)

There is no simple explanation.
But you can study this.

 

[JeffM]

I don't understand how you get an accurate value for area under a curve using antiderivatives for just the end points. Could someone give me a simple explanation please. (If there is one!)

I do not know whether this is simple enough. If you take the limit of a Riemann Sum for an integrable function between the endpoints of an interval, that is intuitively equal to the area under the curve of that function over that interval, and it also equals the difference between the anti-derivatives of that function computed at the endpoints of the interval. This is an informal explanation. Your text should have examples.

 

[Bob Brown MSEE]

Think about it when driving

I don't understand how you get an accurate value for area under a curve using antiderivatives for just the end points. Could someone give me a simple explanation please. (If there is one!)

A satisfying observation is your car speedometer.

Suppose that you hang your watch on your dashboard so that you can video all three with your phone cam.
1) speedometer = S(t)
2) odometer = D(t)
3) watch = t
When you get back from a long trip you know that you can plot S(t) and D(t) from that video.
It is easy to verify that S(t) = D'(t) graphically, especially if there is a lot of variation.

If you want to know the distance between time t₀ and t_f then naturally
D = D(t_f)-D(t₀)

But think about what you just did! Think about the Avg Speed
( Area under S(t) ) / (t_f - t₀) = ( D(t_f)-D(t₀) ) / (t_f - t₀)
Therefore:
( Area under S(t) ) = D(t_f)-D(t₀)

 

[dogleg]

Thank you both. I am revisiting calculus after studying it some years ago. I used my prof's text which I still have. My concept is the difference of the anti-derivatives at the endpoints of the area predicts or projects (includes) all the intermediate values lying under the function between these points. This may be too crude a way of looking at it. (or wrong) I follow the explanation of Riemann Sums but struggle to put in words how to explain and justify it when only the end points are used in the determination of the area. Obviously more work and focus is needed on my part.

 

[dogleg]

Mr. Brown: I posted my reply before I saw your explanation. I will need some time to think about it. Thank you for your help.

 

[dogleg]

Bob, thanks for your helpful example.

 

[dogleg]

Bob: I realize I was asking the wrong question. I was mixed up about using the anti-derivative to find the area under the curve. For some dumb reason I thought we were just finding a sliver of area at each end point and then using those values to get the 'fill-in' area. Anyway what I need to do is get a clear understanding of how the anti-derivative yields the entire area under the curve (if that is what we want to measure). Thanks again for your help.

 

[Bob Brown MSEE]

Thinking backwards

But think about what you just did! Think about the Avg Speed
( Area under S(t) ) / (t_f - t₀) = ( D(t_f)-D(t₀) ) / (t_f - t₀)
Therefore:
( Area under S(t) ) = D(t_f)-D(t₀)

Assume you actually draw the graphs S(t) and D(t).
Assume you want the area under S(t) starting at time t₀ and ending at time t_f.
Note that D(t) is the anti-derivative of S(t) because S(t) = D'(t).

The reason that you want that area is ...
( Area under S(t) ) = D(t_f)-D(t₀)
You want to know the distance that you drove starting at time t₀ and ending at time t_f.

You will pay no attention to the data on the S(t) graph.
You will pay no attention to the data on the D(t) graph except the two endpoints D(t_f) and D(t₀).

========================

In the above I have invited you to think backwards.
Start with something that you do on every trip, when you want to measure distance.
1) Note odometer setting at the beginning of a trip t₀.
2) Note odometer setting at the end of that trip t_f.
3) Take the difference.
You would think it silly to record any other odometer readings.

-------------------

Now think about what you did and how that relates to the speedometer graph S(t).
The result D(t_f)-D(t₀) is also the area under S(t).

Why? Two reasons.
1) Definition of average speed when using integral of S(t).
2) Fundamental Theorem of Calculus for S(t) = D'(t).

Summary: The Fundamental Theorem of Calculus can be thought of as providing the DEFINITION of average over time for S(t). It is very convenient that the average speed for a trip is simply distance traveled divided by the elapsed time.

 

[Bob Brown MSEE]

Average of a function

In your first class in calculus you learned that the average rate of change on (t_f, t₀) for a function D(t) is defined by ...

                    D(t[SUB]f[/SUB]) - D(t[SUB]0[/SUB])
Avgerage D'(x) =  -----------------
                        t[SUB]f[/SUB] - t[SUB]0[/SUB]

This is good if we know D(t). We do! It is the odometer graph.
But how does that relate to the area under the speedometer's graph S(t)?

The following video shows how we can calculate average rate of change if we only have a graph of S(t) = D'(t).

In the video, f(x) = S(t) when x=t.
Also: b=t_f and a=t_{0
Picture this image in your mind as you watch}
Video

Summary:
The average of a function on an interval is a constant function that has the same area.
In other words: What constant speed could I have driven between t₀ and t_f​ and still cover the same distance?

 

[dogleg]

Bob: Again thank you for your help and the excellent examples and the way you explain them. You are a big help. Now could I ask you to take me through the logic that get us from the function ,say f of x = x squared(parabola) to the anti-derivative x to the third over 3.

 

[Bob Brown MSEE]

Anti-derivitive

The best logic is Anti-derivitive.

f(x) = x³/3 + C
f'(x) = x²​ + 0

That means...
1) Guess an Anti-derivative.
2) Test using differentiation.

 

[dogleg]

Bob: Thanks again. I should never have burdened you with my last question. I went back to my old calculus notes from USF and reviewed the proof for producing the anti-derivative. It tortured me with a binomial expansion but at least I feel comfortable using it. If you are like me I can't use a formula unless I know how to derive it. Deriving the power rule is simple by comparison- at least for me. Again, thank you for your kind help. Loveland CO. must be a nice place. I expect you have plenty of snow now.

 

[Bob Brown MSEE]

This is an EXTREMELY important point.

If you are like me I can't use a formula unless I know how to derive it.

You will do well with that attitude!

School encourages us in our youth to miss-understand intelligence.
Most courses require memorization and testing that the information was memorized.
This is not testing intelligence, this is testing memory.
(some exceptions are, compositions, essays and open book math/science problems)

The most important capacity of the human mind is not memory.
This is becoming more obvious as we out-source ever more of our facts to Google and memory to on-line tools for future searches.

Mastery of facts is NOT the pride of the human mind.
Giving meaning to facts IS the pride of the human mind.

 

[Bob Brown MSEE]

Product rule verses a binomial expansion

I went back to my old calculus notes from USF and reviewed the proof for producing the anti-derivative. It tortured me with a binomial expansion but at least I feel comfortable using it.

Usually, the proof of the power rule is presented.
then d/dx(xⁿ) = nx^n-1 is applied to the
Anti-derivative = xⁿ⁺¹/(n+1)
to prove the latter.

The proof of d/dx(xⁿ) = nx by using the limit definition of the derivative does require evaluation of the first three terms of (x+h)ⁿ. However, there is a proof by induction that does not rely on a binomial expansion. It uses the product rule. I like the product rule because many proofs can follow from it.
1) Constant multiplier
2) Quotient rule
3) Power rule
Learning how to prove the product rule is worthwhile.
Proving the product rule is long, but not hard to understand.
A really intuitive proof is written out on Wikipedia and uses the following figure...

I will post how to use the product rule to get the power rule without a binomial expansion.

 

[Bob Brown MSEE]

Power rule proof without a binomial expansion

I will post how to use the product rule to get the power rule without a binomial expansion.

Given: Product Rule
$\displaystyle \frac{d}{dx}$(f(x)* g(x)) = f'g + g'f

Deriving the induction step
by applying the Product Rule to x*xⁿ
$\displaystyle \frac{d}{dx}\left[x\cdot x^n\right]=1\cdot x^n+x\frac{d}{dx}\left[x^n\right]$
$\displaystyle \frac{d}{dx}\left[x^{n+1}\right]=x^n+x\frac{d}{dx}\left[x^n\right]\text{ }\blacksquare$

Demonstrating the power rule base case
A constant function has zero slope.
$\displaystyle \frac{d}{dx}x^0=0$

Stating the induction hypothesis for x^n-1
$\displaystyle \frac{d}{dx}\left[x^{n-1}\right]=(n-1) x^{n-2}$

Applying the induction step to the induction hypothesis
If:
$\displaystyle \frac{d}{dx}\left[x^{n-1}\right]=(n-1) x^{n-2}$
Then:
$\displaystyle \frac{d}{dx}\left[x^n\right]=x^{n-1}+x\frac{d}{dx}\left[x^{n-1}\right]$
$\displaystyle \frac{d}{dx}\left[x^n\right]=x^{n-1}+ x(n-1) x^{n-2}$
$\displaystyle \frac{d}{dx}\left[x^n\right]=n\cdot x^{n-1}$ QED.