# Evaluating integrals geometrically, without using the fundamental theorem of calculus

I'm designing a lesson for an Introduction to Integral Calculus class, and I want to encourage students to evaluate integrals without just going straight for the antiderivative and using the fundamental theorem of calculus. I want them to think geometrically about the situation before diving in with computations. Here are a few general topics I've thought to include so far:

• Integration of even and odd functions.

• Integration of functions with graphs that are familiar shapes. Like $\int_{-2}^2 \sqrt{4-x^2}\;\mathrm{d}x$.

• Maybe integrating a function without an nice antiderivative on the domain of integration, like $\int_{-a}^a |x|\;\mathrm{d}x\,$? But I would want the function so be more interesting that just $|x|$.

What are good examples of integrals that illustrate the ideas above and would work well in a lesson? Or better, does anyone have other ideas for topics that I should (or shouldn't) include in this lesson?

• I often do an exercise like "estimate the area under this curve given only this graph carefully drawn on graph paper. Get an over estimate and an under estimate." – Pat Devlin Dec 15 '16 at 12:52
• The version of the fundamental theorem of calculus in which you "calculate" the derivative of $\int_{a}^x f(t)\;\mathrm{d}t\,$ for difficult-or-impossible-to-integrate $f(t)$s should be in your list. – Amir Asghari Dec 15 '16 at 13:45
• Notice, $\frac{d}{dx}x|x| = 2|x|$. Hence $\int |x|dx = \frac{1}{2}x|x| + C$ – James S. Cook Dec 15 '16 at 15:11
• One can give them the value of an integral not easily computed directly, for example the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^{2}}\,dx$ and ask them to use it to compute some internally rescaled integral such as $\int_{-\infty}^{\infty} e^{-x^{2}/2}\,dx$. There are many variations on this theme. – Dan Fox Dec 15 '16 at 15:34
• Could could hand them out a piece of paper with an unknown function printed on it (together with a grid and rulers for convenience). – dtldarek Jan 10 '17 at 16:02

A useful trick is the idea of rearranging values of a function. For example, while $\sin^2 \theta$ and $\cos^2 \theta$ have different graphs on $0 \leq \theta \leq \pi/2$ you can tell from the graphs of $\sin \theta$ and $\cos \theta$ that the values which the squared functions take must be equal. In short, $$\int_{0}^{\pi/2} \cos^2 \theta \, d\theta = \int_{0}^{\pi/2} \sin^2 \theta \, d\theta$$ Of course, you can make the same conclusion for other intervals such as $[0, \pi]$ or any $2\pi$-length interval. Use the Pythagorean identity and the fact $\int_{a}^{b} c \, d\theta = c(b-a)$ to derive: $$\frac{\pi}{2} = \int_{0}^{\pi/2} 1 \, d\theta = \int_{0}^{\pi/2} \left(\cos^2 \theta +\sin^2 \theta\right) \, d\theta = \int_{0}^{\pi/2} \cos^2 \theta \, d\theta + \int_{0}^{\pi/2} \sin^2 \theta \, d\theta$$ Hence, $\displaystyle \int_{0}^{\pi/2} \cos^2 \theta \, d\theta = \int_{0}^{\pi/2} \sin^2 \theta \, d\theta = \pi/4$.

This trick paired with those which exploit the fact that multiples of a period of a sinusoid integrate to zero can save a lot of work on common trigonometric integrals. More to the point, these are relevant to your lesson because they allow integration without a deep understanding of yet-taught techniques

( I assume students are largely ignorant of basic trigonometric identities like $\sin^2 \theta = \frac{1}{2} \left( 1 - \cos 2 \theta \right)$, a belief those outside the USA will likely gasp in disbelief, but those within the US will easily recognize as a realist position given current malpractice in our standards of precalculus etc.)

• Historical note: This idea goes back to Roberval, Traité des Indivisibles (ca. 1634, pub. post. 1693); in Ouvrages de mathematique (1731), see p. 251. Roughly thus: In a quarter circle, the sum of the squares of all the sines and cosines will be equal on the one hand to twice the sum of the squares of all the sines and on the other to the sum of the squares of the corresponding radii (taking a radius for each sine). – user615 Jan 1 '17 at 0:53
• @MichaelE2 fascinating. See, this is why I love the stack exchange. – James S. Cook Jan 1 '17 at 1:00
• You can also compute $\int_0^{\frac{\pi}{2}} \sin^2(\theta) d\theta$ by rewriting $\sin^2(\theta) = \sqrt{1-\cos^2(\theta)}$ and using the substitution $x= \cos(\theta)$ to get $\int_0^1 \sqrt{1-x^2} dx$. Geometrically this corresponds to partitioning the quarter circle using small angled sectors, and drawing in a Riemann sum corresponding to that partition. The height of the rectangles are $\sin(\theta)$ and the widths are $\approx \sin(\theta) d\theta$. This can be "shown" geometrically. – Steven Gubkin Apr 29 '19 at 12:53
• @StevenGubkin that is so backwards to how I think about this in Calculus II. Hilarious. I love it. Thanks. – James S. Cook Apr 30 '19 at 2:16

In his Calculus book, Spivak gives these two exercises before he ever introduces the fundamental theorems of calculus.

Evaluate without doing any computations:

$$\int\limits_{-1}^{1} x^3\sqrt{1-x^2} \,\mathrm{d}x \qquad\qquad\int\limits_{-1}^{1} \left(x^5+3\right)\sqrt{1-x^2} \,\mathrm{d}x$$

• The 2nd one is evil (in a possibly good way). My students would never see it. I wonder. It might be a good one to build up to... – Sue VanHattum Dec 17 '16 at 5:34
• @SueVanHattum How do you do the second one? – Ovi Dec 30 '16 at 14:34
• @Ovi You distribute and break it into two integrals, then use the fact that $x^5$ is odd and the fact that $\sqrt{1-x^2}$ is the upper-half of a circle. – Mike Pierce Dec 30 '16 at 15:37

The integrals of sine and cosine have a nice geometric interpretation, if you and your students are comfortable treating the differential of a circular arc as equivalent to a straight hypotenuse (in the limit as $\Delta\theta \rightarrow 0$). This is perhaps more of an example than a simple exercise for the student. But I think it answers the OP's objective of thinking geometrically about integrals instead of in terms of algebraic antidifferentiation.

For instance, $\int_a^b \sin\theta \; d\theta = \cos a - \cos b$:

One may find the reasoning (which I assume the reader can fill in) a little loose, but it's a true picture that helps give the student a new perspective on sine and cosine. I prove the ratio of the lengths of an arc and its subtending chord approaches $1$ as the lengths approach zero as well as prove the derivatives in a similar geometric fashion, so it feels more comfortable to me.

This idea may be found in Pascal, Traité des sinus du quart de cercle 1659, Prop. I: La somme des sinus d’un arc quelconque du quart de cercle est égale à la portion de la base comprise entre les sinus extrêmes multipliée par le rayon.

Notice through a geometric argument that

$$\int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$

$$\int_0^af(x)\ dx=\frac1b\int_0^{ab}f(x/b)\ dx$$

The second which may follow through a squeezing of the integrals or Riemann sums.

Now compute the following integrals:

$$\int_0^{\pi/2}\frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi(\theta)}\ d\theta$$

$$\int_0^{\pi/2}\ln(\sin(\theta))\ d\theta$$

For the first integral:

\begin{align}\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\int_0^{\pi/2}\frac{\sin ^\pi( \frac\pi2-\theta)}{\sin^\pi(\frac\pi2-\theta) +\cos^\pi(\frac\pi2-\theta)}\ d\theta\\& =\int_0^{\pi/2}\frac{\cos^\pi( \theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta\\ \therefore I+I=\int_0^{\pi/2} \frac{\sin^\pi(\theta) + \cos^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta&=\int_0^{\pi/2}1\ d\theta = \frac\pi2\\\therefore\int_0^{\pi/2} \frac{\sin^\pi(\theta)}{\sin^\pi(\theta)+\cos^\pi( \theta)}\ d\theta &=\frac\pi4\end{align}

For the second integral:

\begin{align} \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta = \int_0^{\pi /2} \ln(\cos(\theta))\ d\theta\end{align} \begin{align} \therefore I+I=\int_0^{\pi /2}\ln(\sin(\theta))+\ln(\cos(\theta)) \ d\theta &= \int_0^{\pi/2} -\ln(2)+\ln(2\sin(\theta)\cos(\theta)) \ d\theta \\&= -\frac\pi2\ln(2) +\frac12\int_0^\pi \ln(\sin(\theta))\ d\theta \\&= -\frac\pi2\ln(2) +I\end{align} \begin{align}\therefore \int_0^{\pi /2} \ln(\sin(\theta))\ d\theta =-\frac\pi2\ln(2)\end{align}

• Wow, that's really nifty. :) And this integral is surprising related to some other problems I've been thinking about. I think you can also evaluate this one by writing it all in terms of $\tan^n(\theta)$ (or $\cot^n(\theta)$), making a substitution $u = \tan(\theta)$, and then using this cheeky trick. – Mike Pierce Mar 29 '17 at 20:32
• @MikePierce :-P Can you evaluate the second integral now though? – Simply Beautiful Art Mar 29 '17 at 20:35
• Aw I would never have figured that second one out. That's really good. – Mike Pierce Mar 29 '17 at 20:56

There are many nice problems involving integrals of the "integer floor" function, which are evaluated as series. Eg questions 2 and 3 here: https://maths.org/step/sites/maths.org.step/files/assignments/assignment3.pdf

Just ask the students to think geometrically about the basic properties of integrals.

Thinking of an integral as "the area under the curve," write a geometric explanation of why each of the following properties of integrals is reasonable to believe. $$\int\limits_a^b cf(x) \,\mathrm{d}x \;=\; c\!\int\limits_a^b f(x) \,\mathrm{d}x \qquad\qquad \int\limits_a^b f(x) \,\mathrm{d}x \;=\; \int\limits_0^b f(x) \,\mathrm{d}x - \!\int\limits_0^a f(x) \,\mathrm{d}x\\ \int\limits_a^b \big(f(x)+g(x)\big) \,\mathrm{d}x \;=\; \int\limits_a^b f(x) \,\mathrm{d}x + \!\int\limits_a^b g(x) \,\mathrm{d}x$$

Or for more of a challenge, ask the same question for the property $$\int\limits_0^a f(x) \,\mathrm{d}x = \int\limits_0^a f(a-x) \,\mathrm{d}x$$ If they've covered $u$-substitution it would be reasonable to ask them to prove this formally.

Consider integrating the hyperbola $y = 1/x$:

If the $x$ coordinates of a partition of the interval from $1$ to $b=2$ are multiplied by $a=3$, the partition is transformed to one of the interval from $a = 3$ to $ab=6$. Furthermore, all the rectangles in a Riemann sum over $[1,2]$ are transformed to equal rectangles in a Riemann sum over $[3,6]$, since the bases are multiplied by $a=3$ and the heights multiplied by $1/a=1/3$.

1. Why does this show $\int_1^2 {dx \over x} = \int_3^6 {dx \over x}$? Argue similarly that $\int_1^b {dx \over x} = \int_a^{ab} {dx \over x}$ for any $a,b>0$).

2. Let $L(a) = \int_1^a {dx \over x}$. Show that $L(ab)=L(a)+L(b)$. Such a function is called a logarithmic function. (First argue that $\int_1^{ab} {dx \over x} = \int_1^{a} {dx \over x} + \int_a^{ab} {dx \over x}$.)

3. Given that $L(a)$ is a logarithm, estimate its base. (First recall what the value of $\log_B(B)$ is.)

• Wow! This is really cool! Thank you. – Mike Pierce Dec 27 '16 at 6:52
• @MikePierce You're welcome. For 3, some students may need a more direct hint, such as "You're looking for a value of $a$ such that $L(a)$ approximately $1$" or even "You're looking for a number $a$ and a RIemann sum for $L(a)$ such that the areas of the rectangles add up to approximately $1$," especially if they're working on their own. – user615 Dec 27 '16 at 15:05

From

we have

$$\int_0^x \sqrt{a^2-t^2}\;dt = {1 \over 2}\,x\,\sqrt{a^2-x^2} + {1\over2}\,a^2 \arcsin\left({x \over a}\right)$$

Exercise: Show that the derivative of the right-hand side is $\sqrt{a^2-x^2}$.

Similarly, from

we have

$$\int_0^t \arcsin x \; dx = t\,a - \int_0^a \sin y \; dy$$

where $a = \arcsin t$.

This one may be best kept for when the students know how to integrate $\sin y$.

I like the explication and just the example in this paper (really makes you think about an accumulation and a rate). Very nice graphics also (for box counting).

http://home.hiroshima-u.ac.jp/yhiraya/er/ZR11_Z_03.html

Note, he ended up underestimating hydrocarbon production but the thought process is still very nice. And I think doing something topical like this makes it relevant. Not just another cannonball.