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# 4.4: Integration

Difficulty Level: At Grade Created by: CK-12

## Indefinite Integrals Calculus

CONTENT

This content is essentially presented as a kind of game. Students should be presented with the material as such too, so that it seems more fun. Given a function, can you imagine whose derivative it is?

PROCESS

This material is nice because students can quickly check their own work. Put a problem on the board, like:

\begin{align*}\int\limits e^{3x} dx\end{align*}

and then tell students to guess at the answer. They will know the answer is correct when taking its derivative gives back \begin{align*}e^{3x}\end{align*}. As students become more familiar with simple examples like these, more complicated ones like:

\begin{align*}\int\limits x e^x dx\end{align*}

can be presented as a sort of riddle. It may even be fun to have students divide into small groups and think about this problem. Remember, the goal is to come up with some function \begin{align*}f(x)\end{align*} whose derivative is equal to \begin{align*}f'(x)=x\ e^x\end{align*}. Focusing heavily on the additive constant will only frustrate students and take the fun out of the game.

For the problem above, students should be encouraged to literally just guess, try their guess, and then try to fix it. If they were to try \begin{align*}f(x)=x e^x\end{align*} they would find that:

\begin{align*}f'(x)=e^x+x \ e^x\end{align*}

So the derivative is almost right, except for that pesky \begin{align*}e^x\end{align*}. Maybe if they subtracted the integral of that. But \begin{align*}\int\limits e^x dx = e^x\end{align*} so they could just try instead \begin{align*}f(x)=x e^x-e^x\end{align*}. This function has derivative:

\begin{align*}f'(x)=e^x+x e^x-e^x=x e^x\end{align*}

Bingo! You can tell students that what they’ve actually just done is called integration by parts, an extremely important technique for integration.

PRODUCTS

Students should be tested in this material with simple questions like

\begin{align*}\int\limits \left (x^2 + \frac{3}{x} \right )dx \\ \int\limits (x^{17}-17)dx\end{align*}

Solve the following equations for\begin{align*}x\end{align*}assuming that the additive constant \begin{align*}C\end{align*} is zero:

\begin{align*}2 + x & = \int\limits x \ dx \\ x^4-x^3 & = \int\limits (4x-5)(x^2+2x-2) \ dx\end{align*}

## Initial Value Problem

CONTENT

This is a short but important lesson introducing the way in which the additive constant for differential equations or integrals are fixed. After completing the integration we are left with some function \begin{align*}y = f(x)+C\end{align*}, so it should be clear that by giving some point \begin{align*}(x,y)\end{align*} we will have everything in this equation except for \begin{align*}C\end{align*}. This equation, however, can always be solved algebraically since \begin{align*}C\end{align*} is just added linearly to \begin{align*}f(x)\end{align*}.

PROCESS

This material is effectively taught with a variety of examples. It is best to go through at least three or four in class with much detail. For example, write the standard population problem on the board:

A “forest” is planted with one tree in it, and this tree can spawn \begin{align*}10\end{align*} trees per year. Then if the next year there are \begin{align*}11\end{align*} trees in the forest, the population can grow at a rate of \begin{align*}110\end{align*} trees per year. The next year if there are \begin{align*}121\end{align*} trees, then the population can grow at a rate of \begin{align*}1210\end{align*} trees per year. So the population growth rate \begin{align*}p'(t)=10 p(t)\end{align*} where \begin{align*}p(t)\end{align*} is the population at time \begin{align*}t\end{align*}. This equation is written more simply as:

\begin{align*}p'=10p\end{align*}

and is a standard differential equation. The initial population is given as \begin{align*}p(0)=1\end{align*} trees, then we can solve this by thinking about what function gives itself times a constant back. Well the function \begin{align*}e^t\end{align*} gives itself back and we can modify it slightly to see that:

\begin{align*}\frac{d}{dt} (e^{10t}) = 10 (e^{10t})\end{align*}

However, this is true for any multiplicative constant in front since for any \begin{align*}C\end{align*} whatsoever we have:

\begin{align*}\frac{d}{dt} (Ce^{10t}) = 10 (Ce^{10t})\end{align*}

so it would seem that our differential equation is solved by the function:

\begin{align*}p=Ce^{10t}\end{align*}

and we can use the fact that \begin{align*}p(0)=1\end{align*} to find \begin{align*}C\end{align*}:

\begin{align*}\text{at}\ t=0: 1=Ce^{(10)(0)} =C\end{align*}

so \begin{align*}C=1\end{align*} and the population after \begin{align*}t\end{align*} years is given by:

\begin{align*}p(t)=e^{10t}\end{align*}

Presenting examples of this kind, and perhaps having students solve them again on their own after with different numbers are a great way of familiarizing them with differential equations. This will also make the material down the road less surprising when it comes and will give students all-important intuition.

PRODUCTS

Students will gain a lot here by simply recalling the steps you have performed. A nice technique for testing them is to have them clear their desks (or tables), and then present carefully a problem. They should be told to ask lots of questions since they will have to completely understand everything you do. Then, you can cover or erase your work and give them a similar problem with different numbers. In attempting to recall what you did they will have to go through the same logical progression.

## The Area Problem

CONTENT

Students may naturally be confused about why area is being discussed, so it is important to say that they have actually been finding areas by solving these differential equations! The CONTENT here is best introduced with as many pictures as possible. Notation can quickly get out of hand and students are likely to become lost trying to keep track of it all. A better idea is to draw progressive pictures with more and more rectangles. Then they can develop the notation on their own!

PROCESS

The best technique for introducing this material in a way that will be remembered is to have students “discover” it for themselves. Have students get out graph paper and have them draw a close up picture of the function \begin{align*}y=x^2\end{align*} between the points \begin{align*}1\end{align*} and \begin{align*}3\end{align*}:

Then, students should think about how to find the area between the curve and the \begin{align*}x-\end{align*}axis. They should be guided gently to choosing some rectangles to approximate this, and then choosing more and more rectangles to get a better and better approximation.

PRODUCTS

Students will have a good time coming up with their own notation and techniques for forming better and better approximations to the area in question. This is probably the best way to test this material since it will give students an opportunity to understand how this all works without having conventions shoved into their brains. They may even grow to actually enjoy math.

## Definite Integrals

CONTENT

This topic should be thought of for the time-being as a kind of definition. We can see that in the limit that \begin{align*} n \rightarrow \infty \end{align*} the Riemann sums approach the area under the curve. So we simply define the symbol:

\begin{align*} \int\limits_{b}^{a} f(x) dx\end{align*} : = Area under \begin{align*}f(x)\end{align*} between \begin{align*}x = a\end{align*} and \begin{align*}x = b\end{align*}.

Then we can see that we have the definition given in the book since the area on the right here is the limit as \begin{align*} n \rightarrow \infty \end{align*} of the sum of \begin{align*}n\end{align*} Riemann rectangle areas.

PROCESS To teach this material it is best to work a number of detailed examples for the students beforehand. One nice technique would be to have a fairly complicated problem like:

Then the teacher could have the students put all of their materials away and work the problem in detail. The students should be told to ask careful questions and make sure that they understand every step. Then afterwards they can be given the same problem or a similar problem like:

PRODUCTS

Students should be able to effectively find areas for simple functions, although this may take a long time even for just one problem. Therefore it is a good idea to give students very few problems and just have them show all the work.

## Evaluating Definite Integrals

CONTENT

This section is the payoff for calculus. It is among the most interesting and beautiful results in all of math and anybody teaching it should not make light of this. The connection between two seemingly distinct geometric concepts in the fundamental theorem of calculus should surprise and fascinate students.

On the one hand, the derivative of a function gives its rate of change or steepness. More precisely, \begin{align*} f'(x) \end{align*} describes how the function \begin{align*}f(x)\end{align*} is changing at each point \begin{align*}x\end{align*} by giving the slope of the tangent line to the curve \begin{align*}y=f(x)\end{align*}. On the other hand the integral \begin{align*}\int\limits_{b}^{a} f'(x) dx\end{align*} gives the area under the graph of \begin{align*}f'(x)\end{align*} between the points \begin{align*}x = a\end{align*} and \begin{align*}x = b\end{align*}. That these two operations should be inverses of one-another is somewhat profound.

On the other hand, we can make sense of this result geometrically by thinking about constructing a brick wall. Suppose a wall is \begin{align*}10\end{align*} bricks tall at one point. Then extending the wall to the right will require \begin{align*}10\end{align*} bricks for each step to the right. If the wall were \begin{align*}20\end{align*} bricks tall instead, then we’d require \begin{align*}20\end{align*} bricks for each step to the right. Another way to say this is that when the height is \begin{align*}f\end{align*}, the area of the wall increases as we move to the right by an amount of \begin{align*}f\end{align*} per step to the right. Now imagine that the top of the wall follows along a nice function \begin{align*}f(x)\end{align*} like below:

If we let the function \begin{align*}F(z)\end{align*} be the area of the wall from the left edge (the y-axis) to the point \begin{align*}z\end{align*} then for each step to the right \begin{align*}F(z)\end{align*} changes by an amount equal to the height at that point. Another way to say that is that: \begin{align*} F'(z) = f(z) && (1) \end{align*}

On the other hand, we can see that by its definition:

\begin{align*}F(Z) = \int\limits_{z}^{0} f(x) dx \end{align*}

and so the equation \begin{align*}(1)\end{align*} above is the sought-after fundamental theorem of calculus (FTC).

PROCESS

To teach this material, it is a great project to have students all put their books away and focus. Tell them they will have to figure out everything you do on their own so they should concentrate and ask questions if anything is unclear. Then you can go through the proof of the FTC slowly explaining each step, and when complete have the students form small groups. They should discuss each step on the board and the reasoning behind it, to make sure each member of the group understands. Then you should erase the board and have the group write up a thorough proof of the theorem on their own, recalling where to go from each point.

PRODUCTS

This material is effectively tested by having students evaluate a number of definite integrals by finding some antiderivative and applying the FTC.

## The Fundamental Theorem of Calculus

CONTENT

This lesson essentially covers the converse of the theorem discussed previously. Earlier we learned that if \begin{align*}F(x)\end{align*} is any antiderivative of \begin{align*}f(x)\end{align*}, then \begin{align*}\int\limits_{a}^{x}f(x)dx = F(x)\end{align*} is the area between \begin{align*}a\end{align*} and \begin{align*}x\end{align*} under \begin{align*}f\end{align*}. Now we prove the converse, that if \begin{align*}F(x)\end{align*} is some function that gives the area under \begin{align*}f(x)\end{align*} then we must have that \begin{align*}F(x)\end{align*} is some antiderivative of \begin{align*}f(x):F'(x) = f(x)\end{align*}. The intuition here is the same as before, given by the diagram below:

The rate of change for the area \begin{align*}F(z)\end{align*} as \begin{align*}z\end{align*} increases is equal to the height, of \begin{align*}f(z)\end{align*}.

PROCESS

To teach this material, it is another great project to have students all put their books away and focus. Tell them they will again have to figure out everything you do on their own so they should concentrate and ask questions if anything is unclear. This time go through the proof of the converse to the theorem above slowly explaining each step, and when complete have the students form small groups. They should discuss each step on the board and the reasoning behind it, to make sure each member of the group understands. Then you should erase the board and have the group write up a thorough proof of the theorem on their own, recalling where to go from each point.

PRODUCTS

In giving students problems to prove their understanding it is a good idea to have them plot every function they are evaluating. If they are looking for a particular area under or between given functions, they should plot the functions and shade in the area. This way all of their work is constantly connected to the underlying geometric idea.

## Integration by Substitution

CONTENT

This lesson describes two important tricks for coming up with antiderivatives. The first could be explained simply as a means of reducing complex looking integrals to simpler more friendly ones. Or, even on a more basic level, substitutions can be used to make complicated looking functions in general look simpler. Consider the function \begin{align*}f(x) = e ^ {3x + 2}\end{align*}. By setting \begin{align*}u = 3x + 2\end{align*} this function can equivalently be written as \begin{align*}g(u)=e^u\end{align*}. If there is a “differential” \begin{align*}dx\end{align*} multiplying the function \begin{align*}f(x)\end{align*}, then we have that:

\begin{align*}\frac{du}{dx} = 3 \Rightarrow du = 3 dx \end{align*}

So that:

\begin{align*}e ^ {3x + 2} dx = 3 e ^ u dx \end{align*}

which looks a lot simpler. The second trick is integration by parts, and should be tried whenever an integral is a product of two functions that each have known antiderivatives.

PROCESS

To teach this it is nice to do a problem in detail and then have students work individually on a similar problem trying to recall your steps as they go. To teach integration by parts put a problem on the board, like:

\begin{align*}\int x e ^ x dx \end{align*}

and present this as a sort of riddle. It may even be fun to have students divide into small groups and think about this problem. Remember, the goal is to come up with some function \begin{align*}f(x)\end{align*} whose derivative is equal to \begin{align*}f'(x) = x e ^ x\end{align*} . Focusing heavily on the additive constant will only frustrate students and take the fun out of the game.

For the problem above, students should be encouraged to literally just guess, try their guess, and then try to fix it. If they were to try \begin{align*}f(x)=x e^x\end{align*} they would find that:

\begin{align*}f'(x) = e ^ x + x e ^ x \end{align*}

So the derivative is almost right, except for that pesky \begin{align*}e^x\end{align*}. Maybe if they subtracted the integral of that. But \begin{align*}\int e ^ x dx = e ^ x\end{align*} so they could just try instead \begin{align*}f(x) = x e ^ x - e ^ x\end{align*}. This function has derivative:

\begin{align*}f'(x) = e ^ x + x e ^ x - e ^ x = x e ^ x\end{align*}

After this the formal equations can be introduced and they are much more likely to be understood than if they are discussed without a concrete example.

PRODUCTS

Substitutions should be tested by having students perform integrals that are simplified with a substitution. Since the integration itself is not the topic here, it is worthwhile to simply provide a number of functions calling for substitution or integration by parts and have the students think about what to do. For example, the following are good questions:

1. Show that \begin{align*}\frac{\mathrm {Sin}(x)}{\mathrm {Cos}(x)} dx = \frac{du}{u}\end{align*} using an appropriate \begin{align*}u-\end{align*}substitution.

2. Show that \begin{align*}\frac{x dx}{\sqrt{1 - x ^ 2}} = u ^ {-\frac{1}{2}}\end{align*} using an appropriate \begin{align*}u-\end{align*}substitution.

3. Show that \begin{align*}\frac{dx}{\sqrt{4 - x ^ 2}} = du \end{align*} using an appropriate substitution. (hint: This problem will involve the identity that \begin{align*}1 - \mathrm {Sin} ^ 2 (x) = \mathrm {Cos} ^ 2 (x))\end{align*}

4. Show that \begin{align*} x ^ 2 e ^ x dx = \frac{d}{dx} ( x ^ 2 e ^ x) - \frac{d}{dx} ( 2x e ^ x ) dx + 2e ^ x dx\end{align*} and that therefore:

\begin{align*} \int x ^ 2 e ^ x dx = x ^ 2 e ^ x - 2x e ^ x + 2e ^ x\end{align*}

5. Show that \begin{align*}\mathrm {log}(x) dx = \frac{d}{dx} ( x \mathrm {log} (x) ) dx - 1\end{align*} and that therefore:

\begin{align*} \int \text {log} (x) dx = x \text {log} x - x \end{align*}

## Numerical Integration

CONTENT

This lesson discusses the important topic of numerical integration. Almost every problem in physics can be formulated as an integral problem; however, the integrals are never easy to solve. In fact, they can almost never be done exactly. So we are left with trying to make better and better approximations, and this is done by choosing a fine mesh and breaking the area into pieces that we can sum.

Here we see two basic ways of breaking up the region under the curve. It should be noted that no matter what kind of shape we use (rectangle, trapezoid, or parabola), we will always get the exact area in the limit that \begin{align*}n \rightarrow \infty\end{align*} or equivalently as the mesh becomes finer and finer. The different approximating techniques are good because they may allow us to get closer to the true value with a more course mesh, that is, a lower value of \begin{align*}n\end{align*}. A lower value of \begin{align*}n\end{align*} means fewer terms in the summation and a potentially faster computing speed. On the other hand, more complicated shapes require more computing power to determine each term individually and this will eventually overtake the gain made by having fewer terms. In fact, the optimal shape for approximation is a very interesting optimization problem in computation theory and involves a great deal of calculus.

PROCESS/PRODUCTS

A great project here would be to have students write a program in their TI calculator that calculates areas based upon different approximation techniques. The program could take a function \begin{align*}y = f(x)\end{align*}, endpoints \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, and some number of partitions \begin{align*}N\end{align*} as its inputs and from this calculate the function’s value at the \begin{align*}N\end{align*} points and use this to formulate an approximation.

Alternatively, students could simply be asked to find numerical approximations for integrals that they are not likely to know how to solve such as:

1. Approximate \begin{align*}\int\limits_{1}^{0} x ^ 2 dx \end{align*} in a variety of ways and compare your answers with the exact answer.

2. Approximate \begin{align*}\int\limits_{\pi}^{0}\mathrm {Sin}(x) dx\end{align*} in a variety of ways and compare your answers with the exact answer.

3. Approximate \begin{align*}\int\limits_{1}^{0} e ^ { - x ^ 2} dx\end{align*} in a variety of ways.

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