23 Area of Circle, Volume of Sphere, Indefinite Integrals, Partial Fraction Decomposition

Lecture from: 22.05.2024 | Video: Video ETHZ

Review: Our Integration Toolkit So Far

Quick Recap: Key Integration Techniques

1. Integration by Parts: This is our “reverse product rule” for integrals. ${\int }_a^{b}f(x)g^{\prime }(x)dx={\left[f(x)g(x)\right]}_a^b-{\int }_a^b{f}^{\prime }(x)g(x)dx$ It’s super useful when you have a product and differentiating one part while integrating the other makes the new integral simpler.

2. Substitution Rule (Change of Variables): This is our “reverse chain rule.” ${\int }_a^{b}g(\varphi (x)){\varphi }^{\prime }(x)dx={\int }_{\varphi (a)}^{\varphi (b)}g(t)dt$ If you spot an “inside function” $\varphi (x)$ and its derivative ${\varphi }^{\prime }(x)$ in your integrand, substitution can often transform the integral into an easier one. (Here, we’d let $t=\varphi (x)$, so $dt={\varphi }^{\prime }(x)dx$).

Clicker Question: Recognizing Patterns for Substitution

Let’s test our pattern recognition for the substitution rule. Suppose $f$ is some integrable function.

Today’s Menu: Areas, Antiderivatives, and Rational Functions

• Calculating the area of a circle (a classic application!).
• Understanding the indefinite integral (the family of all antiderivatives).
• A systematic way to integrate rational functions (ratios of polynomials) using partial fraction decomposition.

Application: Area of a Circle

A Famous Area Problem We all know the formula for the area of a circle is $\pi r^2$. But can we derive this using calculus? Yes! We’ll find the area of a semicircle and then double it.

A semicircle with radius $r>0$, centered at the origin and above the x-axis, can be described by the function $f(x)=\sqrt{r^2-x^2}$ for $x\in [-r,r]$.

(This comes from the equation of a circle $x^2+y^2=r^2$, so $y=\pm \sqrt{r^2-x^2}$. We take the positive root for the upper semicircle).

The area of this semicircle is the area between the x-axis and the graph of $f(x)$, which is given by the integral: ${\text{Area}}_{\text{semicircle}}={\int }_{-r}^r\sqrt{r^2-x^2}dx$

To calculate this integral, we’ll use a trigonometric substitution, which is a clever application of the substitution rule.

Let $x=\varphi (t)=r\sin (t)$.

We need to choose the limits for $t$ such that as $t$ varies, $x=r\sin (t)$ covers the interval $[-r,r]$ exactly once.

If $x=-r$, then $r\sin (t)=-r⟹\sin (t)=-1⟹t=-\pi /2$. If $x=r$, then $r\sin (t)=r⟹\sin (t)=1⟹t=\pi /2$. So, we’ll let $t$ run from $-\pi /2$ to $\pi /2$.

Our substitution is $\varphi :[-\pi /2,\pi /2]\to [-r,r]$, $t\mapsto r\sin (t)$.

Now, we need $dx={\varphi }^{\prime }(t)dt=(r\sin (t){)}^{\prime }dt=r\cos (t)dt$.

And the term $\sqrt{r^2-x^2}$ becomes: $\sqrt{r^2-(r\sin (t){)}^2}=\sqrt{r^2-r^2{\sin }^2(t)}=\sqrt{r^2(1-{\sin }^2(t))}$.

Using the identity ${\cos }^2(t)+{\sin }^2(t)=1$, we have $1-{\sin }^2(t)={\cos }^2(t)$. So, $\sqrt{r^2{\cos }^2(t)}=\sqrt{(r\cos (t){)}^2}=|r\cos (t)|$.

Since $t\in [-\pi /2,\pi /2]$, $\cos (t)\ge 0$. Also $r>0$. So, $|r\cos (t)|=r\cos (t)$.

Substituting everything into the integral: The limits $x=-r$ to $x=r$ become $t=-\pi /2$ to $t=\pi /2$. ${\int }_{-r}^r\sqrt{r^2-x^2}dx={\int }_{-\pi /2}^{\pi /2}(r\cos (t))\cdot (r\cos (t)dt)={\int }_{-\pi /2}^{\pi /2}{r}^2{\cos }^2(t)dt=r^2{\int }_{-\pi /2}^{\pi /2}{\cos }^2(t)dt$

Now we need to evaluate ${\int }_{-\pi /2}^{\pi /2}{\cos }^2(t)dt$.

We use integration by parts. Recall $\int udv=uv-\int vdu$.

Let $u=\cos (t)$ and $dv=\cos (t)dt$. Then $du=-\sin (t)dt$ and $v=\sin (t)$.

So, $\int {\cos }^2(t)dt=\int \cos (t)\cdot \cos (t)dt=\cos (t)\sin (t)-\int \sin (t)(-\sin (t))dt$ $=\cos (t)\sin (t)+\int {\sin }^2(t)dt$.

Now use ${\sin }^2(t)=1-{\cos }^2(t)$: $\int {\cos }^2(t)dt=\cos (t)\sin (t)+\int (1-{\cos }^2(t))dt=\cos (t)\sin (t)+\int 1dt-\int {\cos }^2(t)dt$ $\int {\cos }^2(t)dt=\cos (t)\sin (t)+t-\int {\cos }^2(t)dt$.

Bring the $\int {\cos }^2(t)dt$ term to the left side: $2\int {\cos }^2(t)dt=\cos (t)\sin (t)+t$.

So, $\int {\cos }^2(t)dt=\frac{1}{2}(\cos (t)\sin (t)+t)+C$.

Now, the definite integral: ${\int }_{-\pi /2}^{\pi /2}{\cos }^2(t)dt={\left[\frac{1}{2}(\cos (t)\sin (t)+t)\right]}_{-\pi /2}^{\pi /2}$

Evaluate at the upper limit $t=\pi /2$: $\frac{1}{2}(\cos (\pi /2)\sin (\pi /2)+\pi /2)=\frac{1}{2}(0\cdot 1+\pi /2)=\frac{1}{2}(\pi /2)=\pi /4$.

Evaluate at the lower limit $t=-\pi /2$: $\frac{1}{2}(\cos (-\pi /2)\sin (-\pi /2)-\pi /2)=\frac{1}{2}(0\cdot (-1)-\pi /2)=\frac{1}{2}(-\pi /2)=-\pi /4$.

The definite integral is $(\pi /4)-(-\pi /4)=\pi /4+\pi /4=\pi /2$.

So, the area of the semicircle is $r^2{\int }_{-\pi /2}^{\pi /2}{\cos }^2(t)dt=r^2\cdot \frac{\pi }{2}=\frac{\pi r^2}{2}$. The area of the full circle is twice this: $\pi r^2$. Success!

Can We Calculate the Volume of a Sphere using Integration?

Yes! Imagine a sphere of radius $r$. We can think of slicing it into thin circular disks perpendicular to, say, the x-axis.

The equation of the circle (in the xy-plane, forming the sphere by revolution) is $x^2+y^2=r^2$, so $y=f(x)=\sqrt{r^2-x^2}$ gives the radius of a disk at position $x$.

The area of such a circular disk at position $x$ is $A(x)=\pi (\text{radius at }x{)}^2=\pi (f(x){)}^2=\pi (\sqrt{r^2-x^2}{)}^2=\pi (r^2-x^2)$.

To find the volume, we “sum up” the volumes of these infinitesimally thin disks (Area $\times dx$) from $x=-r$ to $x=r$: $V_{\text{sphere}}={\int }_{-r}^{r}A(x)dx={\int }_{-r}^r\pi (r^2-x^2)dx$ $=\pi {\int }_{-r}^r{r}^{2}dx-\pi {\int }_{-r}^r{x}^{2}dx$ The first integral: $\pi {\left[r^{2}x\right]}_{-r}^r=\pi (r^2(r)-r^2(-r))=\pi (r^3-(-r^3))=\pi (2r^3)=2\pi r^3$. The second integral: $\pi {\left[\frac{x^3}{3}\right]}_{-r}^r=\pi \left(\frac{r^3}{3}-\frac{(-r{)}^3}{3}\right)=\pi \left(\frac{r^3}{3}-\frac{-r^3}{3}\right)=\pi \left(\frac{r^3}{3}+\frac{r^3}{3}\right)=\pi \frac{2r^3}{3}$. So, $V_{\text{sphere}}=2\pi r^3-\frac{2\pi r^3}{3}=\frac{6\pi r^3-2\pi r^3}{3}=\frac{4}{3}\pi r^3$.

This is indeed the familiar formula for the volume of a sphere!

The Indefinite Integral: The Family of Antiderivatives

Let $I$ be an interval and $f:I\to \mathbb{R}$ be continuous.

The Fundamental Theorem of Calculus (Part 1) tells us that an antiderivative $F$ of $f$ exists. For example, $F(x)={\int }_a^{x}f(t)dt$ (for some $a\in I$) is one such antiderivative.

We know that if $F(x)$ is one antiderivative, then any other antiderivative $G(x)$ must differ from $F(x)$ by only a constant: $G(x)=F(x)+C$.

We write the indefinite integral of $f(x)$ as: $\int f(x)dx=F(x)+C$ where $F(x)$ is any particular antiderivative of $f(x)$, and $C$ is an arbitrary constant of integration. This notation represents the entire family of functions whose derivative is $f(x)$.

“Integration is the inverse operation of differentiation.”

Many derivatives are already known to us, which directly gives us a list of common antiderivatives:

• Power Rule (for $s\ne -1$): $\int x^{s}dx=\frac{1}{s+1}{x}^{s+1}+C$
• Logarithmic Case (for $s=-1$): $\int x^{-1}dx=\int \frac{1}{x}dx=\ln |x|+C$
• Exponential: $\int e^{x}dx=e^x+C$
• Trigonometric: $\int \cos (x)dx=\sin (x)+C$ $\int \sin (x)dx=-\cos (x)+C$
• Inverse Trigonometric related: $\int \frac{1}{\sqrt{1-x^2}}dx=\arcsin (x)+C$ (for $x\in (-1,1)$) $\int \frac{1}{1+x^2}dx=\arctan (x)+C$
• Hyperbolic Functions: (Recall $\sinh (x)=\frac{e^x-e^{-x}}{2}$, $\cosh (x)=\frac{e^x+e^{-x}}{2}$) $\int \sinh (x)dx=\cosh (x)+C$ $\int \cosh (x)dx=\sinh (x)+C$
• Inverse Hyperbolic related: $\int \frac{1}{\sqrt{x^2+1}}dx=\mathrm{arsinh}(x)+C$ (where $\mathrm{arsinh}:\mathbb{R}\to \mathbb{R}$ is the inverse of $\sinh :\mathbb{R}\to \mathbb{R}$) $\int \frac{1}{\sqrt{x^2-1}}dx=\{\begin{array}{ll}\mathrm{arcosh}(x)+C& \text{for }x>1\\ -\mathrm{arcosh}(-x)+C& \text{for }x<-1\end{array}$ (where $\mathrm{arcosh}:[1,\infty )\to [0,\infty )$ is the inverse of $\cosh :[0,\infty )\to [1,\infty )$)

Techniques for Indefinite Integrals

The rules for definite integrals (integration by parts, substitution) have analogous versions for indefinite integrals.

Integration by Parts (Indefinite Version)

$\int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx$ (Remember the $+C$ is implicitly handled if we think of $\int$ as “an antiderivative of”).

Examples:

1. $\int \ln (x)dx=?$ Let $f(x)=\ln (x)$ and $g^{\prime }(x)=1$. Then $f^{\prime }(x)=1/x$ and $g(x)=x$. $\int \ln (x)\cdot 1dx=\ln (x)\cdot x-\int \frac{1}{x}\cdot xdx$ $=x\ln (x)-\int 1dx=x\ln (x)-x+C$.

2. $\int x\ln (x)dx=?$ Let $f(x)=\ln (x)$ and $g^{\prime }(x)=x$. Then $f^{\prime }(x)=1/x$ and $g(x)=x^2/2$. $\int \ln (x)\cdot xdx=\ln (x)\cdot \frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx$ $=\frac{x^2}{2}\ln (x)-\int \frac{x}{2}dx=\frac{x^2}{2}\ln (x)-\frac{x^2}{4}+C$.

3. $\int x\sin (x)dx=?$ Let $f(x)=x$ and $g^{\prime }(x)=\sin (x)$. Then $f^{\prime }(x)=1$ and $g(x)=-\cos (x)$. $\int x\sin (x)dx=x(-\cos (x))-\int 1\cdot (-\cos (x))dx$ $=-x\cos (x)+\int \cos (x)dx=-x\cos (x)+\sin (x)+C$.

4. (Skipped, see script for other examples like $\int e^x\sin (x)dx$ which requires parts twice).

5. Reduction Formula for $J_n=\int \frac{1}{(1+x^2{)}^n}dx$, for $n\ge 1$. This can be used to derive a recurrence relation. Let’s look at $J_1=\int \frac{1}{1+x^2}dx=\arctan (x)+C$.

   Consider $J_n$ for $n\ne 1$. We can try parts on $J_{n-1}$ to relate it to $J_n$. For $J_1=\int 1\cdot \frac{1}{1+x^2}dx$.

   Let $f(x)=\frac{1}{1+x^2}$ and $g^{\prime }(x)=1$. Then $f^{\prime }(x)=\frac{-2x}{(1+x^2{)}^2}$ and $g(x)=x$. $J_1=\frac{x}{1+x^2}-\int x\frac{-2x}{(1+x^2{)}^2}dx=\frac{x}{1+x^2}+\int \frac{2x^2}{(1+x^2{)}^2}dx$.

   We can write $2x^2=2(1+x^2)-2$. So, $\int \frac{2x^2}{(1+x^2{)}^2}dx=\int \frac{2(1+x^2)-2}{(1+x^2{)}^2}dx=\int \left(\frac{2}{1+x^2}-\frac{2}{(1+x^2{)}^2}\right)dx$ $=2\int \frac{1}{1+x^2}dx-2\int \frac{1}{(1+x^2{)}^2}dx=2J_1-2J_2$. So, $J_1=\frac{x}{1+x^2}+2J_1-2J_2$. $2J_2=\frac{x}{1+x^2}+J_1$. $J_2=\int \frac{1}{(1+x^2{)}^2}dx=\frac{1}{2}\frac{x}{1+x^2}+\frac{1}{2}{J}_1=\frac{x}{2(1+x^2)}+\frac{1}{2}\arctan (x)+C$.

Substitution (Indefinite Version)

To calculate $\int f(x)dx$: We replace $x=\varphi (u)$ (so $u$ is the new variable). Then $dx={\varphi }^{\prime }(u)du$.

The integral becomes $\int f(\varphi (u)){\varphi }^{\prime }(u)du$. After integrating with respect to $u$, we must substitute back $u={\varphi }^{-1}(x)$ to express the result in terms of $x$.

Example (like the clicker):

• If $e^x=u$, then $x=\ln (u)$, so $dx=\frac{1}{u}du$.
• If $1/x=u$, then $x=1/u$, so $dx=-\frac{1}{u^2}du$.

Example: $\int \frac{1}{a^2+x^2}dx$

Let $x=a\cdot u$, so $u=x/a$. Then $dx=a\cdot du$. $\int \frac{1}{a^2+(au{)}^2}(adu)=\int \frac{a}{a^2(1+u^2)}du=\frac{1}{a}\int \frac{1}{1+u^2}du$ $=\frac{1}{a}\arctan (u)+C$.

Substitute back $u=x/a$: $=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C$.

(Examples (2)-(4) skipped, see script).

Partial Fraction Decomposition

Integrating Ratios of Polynomials: A Systematic Approach How do we find antiderivatives of rational functions, i.e., functions of the form $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials? The method of partial fractions breaks these down into simpler pieces that we do know how to integrate.

To calculate antiderivatives of rational functions, $f(x)=\frac{P(x)}{Q(x)}$, where $P,Q$ are real polynomials.

Step 1: Ensure Degree of Numerator is Less than Degree of Denominator

If $\mathrm{deg}(P)\ge \mathrm{deg}(Q)$, we first perform polynomial long division to write $\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}$, where $S(x)$ is a polynomial (easy to integrate) and $R(x)$ is a polynomial with $\mathrm{deg}(R)<\mathrm{deg}(Q)$.

We then focus on integrating $\frac{R(x)}{Q(x)}$. From now on, assume $\mathrm{deg}(P)<\mathrm{deg}(Q)$.

Step 2: Normalize and Factorize the Denominator $Q(x)$

Assume $Q(x)$ is “normiert” (monic), meaning its leading coefficient (coefficient of the highest power of $x$) is 1. If not, divide both $P(x)$ and $Q(x)$ by this leading coefficient.

E.g., $Q(x)=x^n+\dots$

Now, theoretically, any real polynomial $Q(x)$ can be factored over the real numbers into a product of linear factors $(x-x_j)$ and irreducible quadratic factors $(x^2+px+q)$ (where $p^2-4q<0$).

For simplicity in these notes, we’ll first consider the case where all roots of $Q(x)$ are real.

Case: “All roots of $Q$ are real”

Then $Q(x)$ can be factored as $Q(x)={\prod }_{j=1}^k(x-x_j{)}^{n_j}$, where $x_j$ are the distinct real roots of $Q$ and $n_j$ is the multiplicity of the root $x_j$.

The sum of multiplicities $\sum n_j=\mathrm{deg}(Q)$.

Step 3: The Partial Fraction Setup

The theory of partial fractions states that $\frac{P(x)}{Q(x)}$ (with $\mathrm{deg}(P)<\mathrm{deg}(Q)$) can be written as a sum of simpler fractions: $\frac{P(x)}{Q(x)}={\sum }_{j=1}^k{\sum }_{l=1}^{n_j}\frac{C_{jl}}{(x-x_j{)}^l}$ where $C_{jl}$ are real constants that we need to determine.

For each distinct root $x_j$ of multiplicity $n_j$, we have terms with denominators $(x-x_j{)}^1,(x-x_j{)}^2,\dots ,(x-x_j{)}^{n_j}$.

Step 4: Determine the Coefficients $C_{jl}$

To find the coefficients $C_{jl}$, we multiply both sides of the equation by $Q(x)$ (clearing denominators). This results in an identity between two polynomials (the original $P(x)$ on one side, and a combination of $C_{jl}$ terms multiplied by parts of $Q(x)$ on the other).

We then equate the coefficients of corresponding powers of $x$ on both sides. This leads to a system of linear equations for the unknown $C_{jl}$.

Step 5: Integrate the Simpler Fractions

The right-hand side (the sum of partial fractions) can then be integrated term by term. We need to know how to integrate terms like $\frac{1}{x-x_j}$ and $\frac{1}{(x-x_j{)}^l}$ for $l\ge 2$.

• $\int \frac{1}{x-x_j}dx=\ln |x-x_j|+C$
• For $l\ge 2$: $\int \frac{1}{(x-x_j{)}^l}dx=\int (x-x_j{)}^{-l}dx=\frac{1}{-l+1}(x-x_j{)}^{-l+1}+C=-\frac{1}{(l-1)(x-x_j{)}^{l-1}}+C$.

Example

Integrate $\int \frac{1-x}{x^2+x}dx=\int \frac{1-x}{x(x+1)}dx$. Here $P(x)=1-x$, $Q(x)=x(x+1)$. $\mathrm{deg}(P)=1<\mathrm{deg}(Q)=2$. Roots of $Q(x)$ are $x_1=0$ (multiplicity $n_1=1$) and $x_2=-1$ (multiplicity $n_2=1$).

Equation: $\frac{1-x}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$ Multiply by $x(x+1)$: $1-x=A(x+1)+Bx$ $1-x=Ax+A+Bx=(A+B)x+A$.

Compare coefficients:

• Coefficient of $x^0$ (constant term): $A=1$.
• Coefficient of $x^1$: $A+B=-1$. Since $A=1$, $1+B=-1⟹B=-2$.

So, $\frac{1-x}{x(x+1)}=\frac{1}{x}-\frac{2}{x+1}$. Then $\int \frac{1-x}{x^2+x}dx=\int \frac{1}{x}dx-\int \frac{2}{x+1}dx=\ln |x|-2\ln |x+1|+C$.

Remark: In general, to find the coefficients $C_{jl}$, one must solve a system of linear equations. The Gauss elimination method from linear algebra can be used for this.

(If $Q(x)$ has irreducible quadratic factors, the equation includes terms like $\frac{Ax+B}{x^2+px+q}$, which integrate to logarithms and arctangents).

Continue here: 24 Integration of Convergent Series and Swapping Limit and Integrals, Improper Integrals