Category Archives: Integration

The Rate/Accumulation Question

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AP Type Questions 1

These questions are often in context with a lot of words describing a situation in which some things are changing. There are usually two rates given acting in opposite ways. Students are asked about the change that the rates produce over some time interval either separately or together.

The rates are often fairly complicated functions. If they are on the calculator allowed section, students should store them in the equation editor of their calculator and use their calculator to do any integration or differentiation that may be necessary.

The integral of a rate of change is the net amount of change

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right)

over the time interval [a, b]. If the question asked for an amount, look around for a rate to integrate.

The final accumulated amount is the initial amount plus the accumulated change:

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

where {{x}_{0}} is the initial time, and  f\left( {{x}_{0}} \right) is the initial amount. Since this is one of the main interpretations of the definite integral the concept may come up in a variety of situations.

What students should be able to do:

  • Be ready to read and apply; often these problems contain a lot of writing which needs to be carefully read.
  • Recognize that rate = derivative.
  • Recognize a rate from the units given without the words “rate” or “derivative.”
  • Find the change in an amount by integrating the rate. The integral of a rate of change gives the amount of change (FTC):

\displaystyle \int_{a}^{b}{{f}'\left( t \right)dt}=f\left( b \right)-f\left( a \right).

  • Find the final amount by adding the initial amount to the amount found by integrating the rate. If x={{x}_{0}} is the initial time, and f\left( {{x}_{0}} \right)  is the initial amount, then final accumulated amount is

\displaystyle f\left( x \right)=f\left( {{x}_{0}} \right)+\int_{{{x}_{0}}}^{x}{{f}'\left( t \right)}\,dt,

  • Understand the question. It is often not necessary to as much computation as it seems at first.
  • Use FTC to differentiate a function defined by an integral.
  • Explain the meaning of a derivative or its value in terms of the context of the problem.
  • Explain the meaning of a definite integral or its value in terms of the context of the problem. The explanation should contain (1) what it represents, (2) its units, and (3) how the limits of integration apply in context.
  • Store functions in their calculator recall them to do computations on their calculator.
  • If the rates are given in a table, be ready to approximate an integral using a Riemann sum or by trapezoids.
  • Do a max/min or increasing/decreasing analysis.

Shorter questions on this concept appear in the multiple-choice sections. As always, look over as many questions of this kind from past exams as you can find.

For some previous posts on this subject see January 21, 23, 2013

Interpreting Graphs

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AP Type Questions 1

The long name is “Here’s the graph of the derivative, tell me things about the function.”

Most often students are given the graph identified as the derivative of a function. There is no equation given and it is not expected that students will write the equation (although this may be possible); rather, students are expected to determine important features of the function directly from the graph of the derivative. They may be asked for the location of extreme values, intervals where the function is increasing or decreasing, concavity, etc. They may be asked for function values at points.

The graph may be given in context and student will be asked about that context. The graph may be identified as the velocity of a moving object and questions will be asked about the motion and position. (Motion problems will be discussed as a separate type in a later post.)

Less often the function’s graph may be given and students will be asked about its derivatives.

What students should be able to do:

  • Read information about the function from the graph of the derivative. This may be approached as a derivative techniques or antiderivative techniques.
  • Find where the function is increasing or decreasing.
  • Find and justify extreme values (1st  and 2nd derivative tests, Closed interval test, aka.  Candidates’ test).
  • Find and justify points of inflection.
  • Find slopes (second derivatives, acceleration) from the graph.
  • Write an equation of a tangent line.
  • Evaluate Riemann sums from geometry of the graph only.
  • FTC: Evaluate integral from the area of regions on the graph.
  • FTC: The function, g(x), maybe defined by an integral where the given graph is the graph of  the integrand, f(t), so students should know that if,  \displaystyle g\left( x \right)=g\left( a \right)+\int_{a}^{t}{f\left( t \right)dt} then  {g}'\left( x \right)=f\left( x \right)  and  {{g}'}'\left( x \right)={f}'\left( x \right).

The ideas and concepts that can be tested with this type question are numerous. The type appears on the multiple-choice exams as well as the free-response. They have accounted for almost 25% of the points available on recent test. It is very important that students are familiar with all of the ins and outs of this situation.

As with other questions, the topics tested come from the entire year’s work, not just a single unit. In my opinion many textbooks do not do a good job with these topics.

Study past exams; look them over and see the different things that can be asked.

For some previous posts on this subject see October 15, 17, 19, 24, 26, 2012 January 25, 28, 2013

Logarithms

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Students first encounter logarithms in Algebra 2 or their last pre-calculus course. They are usually defined as exponents and their properties “proven” by reference to the corresponding properties of exponents. This is good enough at the time, but in calculus we are ready to define logarithms properly. The result is the definition in terms of a function defined by an integral.

In fact, the definition flows naturally from the properties of inverse functions (no pun intended, because the flow is quite natural). But is this a definition? Or a proof? The solution of certain differential equations discussed in my last post may be used to derive the logarithm function.

Let’s begin by pretending we know nothing about logarithms, and investigate the inverse of ex.  (See The Derivative of Exponential Functions)

This function ex contains points of the form \left( X,{{e}^{X}} \right) so its inverse will contain points of the form \left( {{e}^{X}},X \right), which, since we like the first coordinate of functions to be x, we may also call \left( x,\ln \left( x \right) \right), where ln(x) will be the name of the inverse of ex. Remember, at the moment ln(x) is just a notation for the inverse of ex, we do not know anything about logarithms (yet).

So \left( {{e}^{X}},X \right) =\left( x,\ln \left( x \right) \right) and X\ne x. For example, (0, 1) is a point on eX, so (1, 0) is a point on the ln(x) function, and so ln(1) = 0.

In a previous post  we defined the number e and the function ex in such a way that \displaystyle \frac{d}{dx}{{e}^{x}}={{e}^{x}}. Now, at \left( X,{{e}^{X}} \right) the derivative is eX, so at the corresponding point on its inverse, \left( {{e}^{X}},X \right), the derivative is the reciprocal of the derivative of eX which is \frac{1}{{{e}^{X}}}=\frac{1}{x}. That is

\displaystyle \frac{d}{dx}\ln \left( x \right)=\frac{1}{x}

We can use this idea to define ln(x) as a function defined by an integral. Solving the differential equation by the method suggested in a recent post we get:

\displaystyle \ln \left( x \right)=\ln \left( a \right)+\int_{a}^{x}{\frac{1}{t}dx}

We can pick any positive number for a and a convenient one is the one we already found a = 1 where  ln(1) = 0, so

\displaystyle \ln \left( x \right)=\int_{1}^{x}{\frac{1}{t}dx},\quad x>0

This then is the “official definition” of the natural logarithm function. The domain is x > 0 (the range of ex) and the range is all real numbers (the domain of ex).

Graphically, ln(x) is the area between the graph of \displaystyle \frac{1}{t} and the t-axis between 1 and x. If 0<x<1, then \ln \left( x \right)<0 and if x>1, \ln \left( x \right)>0.

logarithm

From this definition we can prove all the familiar properties of logarithms \left( \ln \left( ab \right)=\ln \left( a \right)+\ln \left( b \right)\text{ etc}\text{.} \right) and show that this is a function of the kind we called “logarithm” all along. This is in all the textbooks so I will not go into this here.

One of these properties tells us that for a>0\text{ and }a\ne 1, \displaystyle {{\log }_{a}}\left( x \right)=\frac{\ln \left( x \right)}{\ln \left( a \right)} and since ln(a) is a constant

\displaystyle {{\log }_{a}}\left( x \right)=\frac{1}{\ln \left( a \right)}\int_{1}^{x}{\frac{1}{t}dt}

Finally, you will see this antiderivative formula: \displaystyle \int_{{}}^{{}}{\frac{1}{x}dx=\ln \left| x \right|+C}. The absolute value sign is to remind you that the argument of the logarithm function must be positive, since in some situations x itself may be negative.

Revised slightly 8-28-2018

Painting a Point

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Accumulation 7: An application (of paint)

Suppose you started with a point, the origin to be specific, and painted it. You put on layers and layers of paint until your point grows to a sphere with radius r. Let’s stop and admire your work part way through the job; at this point the radius is {{x}_{i}} and 0\le {{x}_{i}}\le r.

How much paint will you need for the next layer?

Easy: you need an amount equal to the surface area of the sphere, 4\pi {{x}_{i}}^{2}, times the thickness of the paint. As everyone knows paint is thin, specifically \Delta x thin. So we add an amount of paint to the sphere equal to 4\pi {{x}^{2}}\left( \Delta x \right).

The volume of the final sphere must be the same as the total amount of paint. The total amount of paint must be the (Riemann) sum of all the layers or \displaystyle \sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)}. As usual \Delta x is very thin, tending to zero as a matter of fact, so the amount of paint must be

\displaystyle \underset{\Delta x\to 0}{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{4\pi {{x}_{i}}^{2}\left( \Delta x \right)=\int_{0}^{r}{4\pi {{x}^{2}}}dx=\left. \tfrac{4}{3}\pi {{x}^{3}} \right|_{0}^{r}=\tfrac{4}{3}\pi {{r}^{3}}}.

A standard related rate problem is to show that the rate of change of the volume of a sphere is proportional to its surface area – the constant of proportionality is dr/dt. So it should not be a surprise that the integral of this rate of change of the surface area is the volume. The integral of a rate of change is the amount of change.

Interestingly this approach works other places as long as you properly define “radius:”

  • A circle centered at the origin with radius x and perimeter of 2\pi x, gains area at a rate equal to its perimeter times the “thickness of the edge” \Delta x: \displaystyle A=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{2\pi x\Delta x}=\int_{0}^{r}{2\pi x\,dx=\pi {{r}^{2}}}
  • A square centered at the origin with “radius” x with sides whose length are s =2 x, gains area at a rate equal to its perimeter (8x) times the “thickness of the edge” \Delta x:  \displaystyle A=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{8x\Delta x=\int_{0}^{x}{8x\,dx}}=4{{x}^{2}}={{(2x)}^{2}}={{s}^{2}}
  • A cube centered at the origin with “radius” x and edges of length 2x, gains volume at a rate equal to its surface area, 6(4{{x}^{2}}), times the “thickness of the face” \Delta x – think paint again: V=\underset{n\to \infty }{\mathop{\text{lim}}}\,\sum\limits_{k=1}^{n}{\text{6(4}{{\text{x}}^{2}})\Delta x}=\int_{0}^{x}{24{{x}^{2}}dx}=8{{x}^{3}}={{(2x)}^{3}}={{s}^{3}}

Accumulation and Differential Equations

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Accumulation 6: Differential equations

When students first learn about antiderivatives, they are given simple initial value problems to solve such as \frac{dy}{dx}=3-2x,\quad y\left( 1 \right)=-5. They are instructed to find the antiderivative, then tack on a +C , then substitute in the initial condition, then solve for C and finally write the particular solution. So we hope to see:

 \frac{dy}{dx}=3-2x

y=3x-{{x}^{2}}+C

-5=3(1)-{{1}^{2}}+C

-7=C

y=3x-{{x}^{2}}-7

But thinking of it as an accumulation problem you can write

\displaystyle y=-5+\int_{1}^{x}{3-2t\,dt}

y=-5+\left. \left( 3t-{{t}^{2}} \right) \right|_{1}^{x}

 y=-5+\left( 3x-{{x}^{2}} \right)-\left( 3(1)-{{1}^{2}} \right)

y=3x-{{x}^{2}}-7

Now I’ll admit there is not too much difference in the amount of work involved there, but once the first line is on the paper there is not much to remember in what to do from there on. Common student mistakes in the first method include entirely forgetting the +C and not using the initial condition correctly.

A differential equation in which dy/dx can be written as a function of x only, \frac{dy}{dx}={f}'\left( x \right), with an initial condition \left( a,f\left( a \right) \right) has the solution

\displaystyle f\left( x \right)=f\left( a \right)+\int_{a}^{x}{{f}'\left( t \right)dt}

Look familiar? Yes, that’s my favorite accumulation equation.

Furthermore, if all you need is some function values, this expression can be entered into a graphing calculator and the values calculated without finding an antiderivative.

In my first post on accumulation, I discussed the AP exam question 2000 AB 4. The solution to part (a) looks like this

\displaystyle \int_{0}^{3}{\sqrt{t+1}dt}=\left. \tfrac{2}{3}{{\left( t+1 \right)}^{3/2}} \right|_{0}^{3}=\tfrac{2}{3}\left( {{4}^{3/2}}-{{0}^{3/2}} \right)=\tfrac{14}{3}\text{ gallons}

The scoring standard included a second method that went this way:

L(t)=\text{ gallons leaked in first }t\text{ minutes}

\frac{dL}{dt}=\sqrt{t+1};\quad L\left( t \right)=\tfrac{2}{3}{{\left( t+1 \right)}^{3/2}}+C

L(0)=0;\quad C=-\tfrac{2}{3}

L(t)=\tfrac{2}{3}{{\left( t+1 \right)}^{3/2}}-\tfrac{2}{3};\quad L(3)=\tfrac{14}{3}

You decide which is easier. If you’re still not sure compare the two methods shown for part (c) on the scoring standard.                                               

Stamp Out Slope-intercept Form!

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Accumulation 5: Lines

Ban Slope Intercept

If you have a function y(x), that has a constant derivative, m, and contains the point \left( {{x}_{0}},{{y}_{0}} \right) then, using the accumulation idea I’ve been discussing in my last few posts, its equation is

\displaystyle y={{y}_{0}}+\int_{{{x}_{0}}}^{x}{m\,dt}

\displaystyle y={{y}_{0}}+\left. mt \right|_{{{x}_{0}}}^{x}

\displaystyle y={{y}_{0}}+m\left( x-{{x}_{0}} \right)

This is why I need your help!

I want to ban all use of the slope-intercept form, y = mx + b, as a method for writing the equation of a line!

The reason is that using the point-slope form to write the equation of a line is much more efficient and quicker. Given a point \left( {{x}_{0}},{{y}_{0}} \right) and the slope, m, it is much easier to substitute into  y={{y}_{0}}+m\left( x-{{x}_{0}} \right) at which point you are done; you have an equation of the line.

Algebra 1 books, for some reason that is beyond my understanding, insist using the slope-intercept method. You begin by substituting the slope into y=mx+b and then substituting the coordinates of the point into the resulting equation, and then solving for b, and then writing the equation all over again, this time with only m and b substituted. It’s an algorithm. Okay, it’s short and easy enough to do, but why bother when you can have the equation in one step?

Where else do you learn the special case (slope-intercept) before, long before, you learn the general case (point-slope)?

Even if you are given the slope and y-intercept, you can write y=b+m\left( x-0 \right).

If for some reason you need the equation in slope-intercept form, you can always “simplify” the point-slope form.

But don’t you need slope-intercept to graph? No, you don’t. Given the point-slope form you can easily identify a point on the line,\left( {{x}_{0}},{{y}_{0}} \right), start there and use the slope to move to another point. That is the same thing you do using the slope-intercept form except you don’t have to keep reminding your kids that the y-intercept, b, is really the point (0, b) and that’s where you start. Then there is the little problem of what do you do if zero is not in the domain of your problem.

Help me. Please talk to your colleagues who teach pre-algebra, Algebra 1, Geometry, Algebra 2 and pre-calculus. Help them get the kids off on the right foot.

Whenever I mention this to AP Calculus teachers they all agree with me. Whenever you grade the AP Calculus exams you see kids starting with y = mx + b and making algebra mistakes finding b.

Graphing with Accumulation 2

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Accumulation 4: Graphing Ideas in Accumulation – Concavity

In the last post we saw how thinking about Riemann sum rectangles, RΣR, moving across the graph of the derivative made it easy to see when the function whose derivative was given increased and decreased and had its local extreme values. Today we will consider concavity.

Suppose a derivative is constant, its graph a horizontal line. In this case each successive RΣR is exactly the same size and adds exactly the same amount to the accumulated function. The function’s graph increases (or decreases) by exactly the same amount – it is linear.

For derivatives that are not constant, the change in the resulting function is not constant and the function’s graph bends up or down. This bending of the function is referred to as its concavity. If it bends up, the function increases faster and its graph is concave up; if it bends down, it is increasing slower (or decreasing faster) and concave down.

RSR 2

The graph above shows pairs of RΣR in different intervals as they move along the graph of a derivative. Consider the dark blue rectangle to be the previous position of the red rectangle.

As the RΣR moves from a to b each red rectangle is larger than the dark blue one. Each move adds more to the accumulated sum than the previous one. The graph of the function increases more with each move – it is concave up.

As the RΣR moves from b to d (there are two pairs drawn) each red RΣR has a smaller value than the previous one. (Remember when they are below the x-axis the longer (red) RΣR has a smaller value.) In the interval [b, d] less is added to the accumulated sum (or more is subtracted) with each move to the right. Therefore, the graph of the function bends down – the function is concave down.

In the last section, from d to f the red rectangle now has a larger value than the dark blue one. (Again, remember that when the function has negative values, the shorter rectangle, has the larger value.) The graph of the function again bends up – is concave up.

Putting these ideas together with those in the last post we can see how the moving RΣR idea can distinguish the four shapes of the graph of the accumulating function:

  • On [a, b] the function’s graph is increasing and concave up; the RΣR are positive and getting more positive (longer).
  • On [b, c] the function’s graph is increasing and concave down; the RΣR are positive and getting less positive (shorter).
  • On [c, d] the function’s graph is decreasing and concave down; the RΣR are negative and getting more negative (longer).
  • On [d, e] the function’s graph is decreasing and concave up; the RΣR are negative and getting less negative (shorter).
  • At the extreme values of the derivative, the concavity of the function changes from up to down or down to up. These are called points of inflection.

Questions in which students are asked about the properties of a function given the graph, but not the equation, of the derivative are very common. Many students (including me) find this approach easier and more intuitive than working strictly with derivative ideas.