Good Question 13

Posted on December 12, 2017 by Lin McMullin

Let’s end the year with this problem that I came across a while ago in a review book:

Integrate $\int{x\sqrt{x+1}dx}$

It was a multiple-choice question and had four choices for the answer. The author intended it to be done with a u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.

1. Find the antiderivative using a u-substitution.
2. Find the antiderivative using integration by parts.
3. Find the antiderivative using a different u-substitution.
4. Find the antiderivative by adding zero in a convenient form.

Your answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by

“Simplifying” your answer to 2 and get a third form of the answer.
“Simplifying” your answer to 1, 3, 4 and get that third form again.

Give it a try before reading on. The solutions are below the picture.

Method 1: u-substitution

Integrate $\int{x\sqrt{x+1}dx}$

$u=x+1,x=u-1,dx=du$

$\int{x\sqrt{x+1}dx=}\int{\left( u-1 \right)\sqrt{u}}du=\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\tfrac{5}{2}{{u}^{5/2}}-\tfrac{3}{2}{{u}^{3/2}}$

$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$

Method 2: By Parts

Integrate $\int{x\sqrt{x+1}dx}$

$u=x,du=dx$

$dv=\sqrt{x+1}dx,v=\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}$

$\int{x\sqrt{x+1}dx}=\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\int{\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}dx}=$

$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C$

Method 3: A different u-substitution

Integrate $\int{x\sqrt{x+1}dx}$

$u=\sqrt{x+1},x={{u}^{2}}-1,$

$du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu$

$2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=$

$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$

This gives the same answer as Method 1.

Method 4: Add zero in a convenient form.

Integrate $\int{x\sqrt{x+1}dx}$

$\int{x\sqrt{x+1}}dx=\int{x\sqrt{x+1}+\sqrt{x+1}-\sqrt{x+1} dx=}$

$\int{\left( x+1 \right)\sqrt{x+1}-\sqrt{x+1}}dx=$

$\int{{{\left( x+1 \right)}^{3/2}}-{{\left( x+1 \right)}^{1/2}}dx}=$

$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$

This, also, gives the same answer as Methods 1 and 3.

So, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?

No, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of ${{\left( x+1 \right)}^{3/2}}$ . (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)

Simplify the answer for Method 2:

$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C=$

$\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( x-\tfrac{2}{5}\left( x+1 \right) \right)+C=$

$\displaystyle \tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( \frac{5x-2x-2}{5} \right)+C=$

$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$

Simplify the answer for Methods 1, 3, and 4:

$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C=$

$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3\left( x+1 \right)-5 \right)+C=$

$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$

So, same answer and same constant.

Is this a good question? No and yes.

As a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.

From another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students’ perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (… maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.

My next post will be after the holidays.

Happy Holidays to Everyone!

Applications of integrals, part 2: Volume problems

Posted on December 5, 2017 by Lin McMullin

One of the major applications of integration is to find the volumes of various solid figures.

Volume of Solids with Regular Cross-sections This is where to start with volume problems. After all, solids of revolution are just a special case of solids with regular cross-sections.

Volumes of Revolution

Subtract the Hole from the Whole and Does Simplifying Make Things Simpler?

Visualizing Solid Figures

Visualizing Solid Figures 1 Making models et al.
Visualizing Solid Figures 2 Using Winplot to visualize solids with regular cross-sections.
Visualizing Solid Figures 3 Using Winplot to visualize the Disk and Washer methods.
Visu alizing Solid Figures 4 Using Winplot to visualize finding volumes by cylindrical shells. While no problem on the AP Calculus exams will require a solution by the Cylindrical Shell method, students may use the method and be eligible for full credit. So this is “enrichment.”
Challenge – the half-full problem comparing the disk and shell methods and Visualizing Solid Figures 5 the detailed solution

Area and Volume (Type 4) also Area and Volume Question review notes.

Why you Never Need Cylindrical Shells

Painting a Point

Applications of integrals, part 1: Areas & Average Value

Posted on November 28, 2017 by Lin McMullin

Usually the first application of integration is to find the area bounded by a function and the x-axis, followed by finding the area between two functions. We begin with these problems

First some calculator hints

Graphing Integrals using a graphing calculator to graph functions defined by integrals

Graphing Calculator Use and Definition Integrals – Exam considerations Suggestions for using a calculator efficiently in area/volume problems

Area Problems

Area Between Curves

Under is a Long Way Down How to avoid “negative area.”

Density Functions Not often asked on the AP exams, but a good application related to area, nevertheless.

Who’d a thunk it? Some more complicated area problems for CAS solution.

Improper Integrals and Proper Areas – a BC topic

Average Value

Average Value of a Function

What’s a Mean Old Average Anyway – Discusses the different “average” in calculus

Half-full and Half Empty – Average Value

Average Value Activity to help students discover the Average Value formula

Starting Integration

Posted on November 21, 2017 by Lin McMullin

Behind every definite integral is a Riemann sums. Students need to know about Riemann sums so that they can understand definite integrals (a shorthand notation for the limit if a Riemann sun) and the Fundamental theorem of Calculus. Theses posts help prepare students for Riemann sums.

The Old Pump Where I start Integration
Flying into Integrationland Continues the investigation in the Old Pump – the airplane problem
Working Towards Riemann Sums
Definition of the Definite Integral and the FTC – a more exact demonstration from last Friday’s post and The Fundamental Theorem of Calculus – an older demonstration
More about the FTC The derivative of a function defined by an integral – the other half of the FTC.
Good Question 11 Riemann Reversed – How to find the integral, given the Riemann sum
Properties of Integrals
Variation on a Theme – 2 Comparing Riemann sums
Trapezoids – Ancient and Modern – some history

The Definite Integral and the FTC

Posted on November 17, 2017 by Lin McMullin

The Definition of the Definite Integral.

The definition of the definite integrals is: If f is a function continuous on the closed interval [a, b], and $a={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<\cdots <{{x}_{{n-1}}}<{{x}_{n}}=b$ is a partition of that interval, and $x_{i}^{*}\in [{{x}_{{i-1}}},{{x}_{i}}]$ , then

$\displaystyle \underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=\int\limits_{a}^{b}{{f\left( x \right)dx}}$

The left side of the definition is, of course, any Riemann sum for the function f on the interval [a, b]. In addition to being shorter, the right side also tells you about the interval on which the definite integral is computed. The expression $\left\| {\Delta x} \right\|$ is called the “norm of the partition” and is the longest subinterval in the partition. Usually, all the subintervals are the same length, $\frac{{b-a}}{n}$ , and this is the last you will hear of the norm. With all the subdivisions of the same length this can be written as

$\displaystyle \underset{{n\to \infty }}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{f\left( {x_{i}^{*}} \right)}}\frac{{b-a}}{n}=\int\limits_{a}^{b}{{f\left( x \right)dx}}$

Other than that, there is not much more to the definition. It is simply a quicker and more efficient notation for the sum.

The Fundamental Theorem of Calculus (FTC).

First recall the Mean Value Theorem (MVT) which says: If a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there exist a number, c, in the open interval (a, b) such that ${f}'\left( c \right)\left( {b-a} \right)=f\left( b \right)-f\left( a \right)$ .

Next, let’s rewrite the definition above with a few changes. The reason for this will become clear.

$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{{f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)}}$

Since every function is the derivative of another function (even though we may not know that function or be able to write a closed-form expression for it), I’ve expressed the function as a derivative, I’ve also chosen the point in each subinterval, ${{c}_{i}}$ , to be the number in each subinterval guaranteed by the MVT for that subinterval.

Then, $\displaystyle {f}'\left( {{{c}_{i}}} \right)\left( {{{x}_{i}}-{{x}_{{i-i}}}} \right)=f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)$ . Making this substitution, we have

$\int\limits_{a}^{b}{{{f}'\left( x \right)dx}}=\underset{{\left| {\left| {\Delta x} \right|} \right|\to 0}}{\mathop{{\lim }}}\,\sum\limits_{{i=0}}^{n}{{\left( {f\left( {{{x}_{i}}} \right)-f\left( {{{x}_{{i-1}}}} \right)} \right)}}$

$\displaystyle =f\left( {{{x}_{1}}} \right)-f\left( {{{x}_{0}}} \right)+f\left( {{{x}_{2}}} \right)-f\left( {{{x}_{1}}} \right)+f\left( {{{x}_{3}}} \right)-f\left( {{{x}_{2}}} \right)+\cdots +f\left( {{{x}_{n}}} \right)-f\left( {{{x}_{{n-1}}}} \right)$

$\displaystyle =f\left( {{{n}_{n}}} \right)-f\left( {{{x}_{0}}} \right)$

And since ${{x}_{0}}=a$ and ${{x}_{n}}=b$ ,

$\displaystyle \int_{a}^{b}{{{f}'\left( x \right)dx}}=f\left( b \right)-f\left( a \right)$

This equation is called the Fundamental Theorem of Calculus. In words, it says that the integral of a function can be found by evaluating the function of which the integrand is the derivative at the endpoints of the interval and subtracting the values. This is a number that may be positive, negative, or zero depending on the function and the interval. The function of which the integrand is the derivative, is called the antiderivative of the integrand.

The real meaning and use of the FTC is twofold:

It says that the integral of a rate of change (i.e. a derivative) is the net amount of change. Thus, when you want to find the amount of change – and you will want to do this with every application of the derivative – integrate the rate of change.
It also gives us an easy way to evaluate a Riemann sum without going to all the trouble that is necessary with a Riemann sum; simply evaluate the antiderivative at the endpoints and subtract.

At this point I suggest two quick questions to emphasize the second point:

Find $\int_{3}^{7}{{2xdx}}$ .

Ask if anyone knows a function whose derivative is 2x? Your students will know this one. The answer is x², so

$\displaystyle \int_{3}^{7}{{2xdx}}={{7}^{2}}-{{3}^{2}}=40$ .

Much easier than setting up and evaluating a Riemann sum!

2. Then ask your students to find the area enclosed by the coordinate axes and the graph of cos(x) from zero to $\frac{\pi }{2}$ . With a little help they should arrive at

$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}$ .

Then ask if anyone knows a function whose derivative is cos(x). it’s sin(x), so

$\displaystyle \int_{0}^{{\pi /2}}{{\cos \left( x \right)dx}}=\sin \left( {\frac{\pi }{2}} \right)-\sin \left( 0 \right)=1-0=1$ .

At this point they should be convinced that the FTC is a good thing to know.

There is another form of the FTC that is discussed in More About the FTC.

Antidifferentiation

Posted on November 14, 2017 by Lin McMullin

We now turn to integration. The first thing to decide is when to teach antidifferentiation. Many books do this at the end of the last differentiation chapter or the first thing in the first integration chapter. Some teachers, myself included, prefer to wait until after presenting the Fundamental Theorem of Calculus. Still others wait until after teaching all the applications. The reasons for this are discussed in more detail in the first post below, Integration Itinerary.

If you teach this later, come back and look at it then.

Integration itinerary – a discussion of when to teach antidifferentiation.

The following posts are on different antidifferentiation techniques.

Antidifferentiation u-substitution

Why Muss with the “+C”?

Good Question 12 – Parts with a Constant?

Integration by Parts I (BC only)

Integration by Parts II (BC only)

Parts and More Parts (BC only) More on the tabular method and on reduction formulas

Modified Tabular Integration (BC only) With this you don’t need to make a table.