MA5131 Summary 3
Definition of Anti-Derivative
The anti-derivative of a function $f(x)$ is just defined as any function $F(x)$ such that $F’(x)=f(x)$.
Note that the anti derivative is unique up to a constant. Consider $2$ functions such that $F’(x)=G’(x)=f(x)$. We have $(F(x)-G(X))’=F’(x)-G’(x)=f(x)-f(x)=0$. The gradient of $F(x)-G(x)$ is $0$ everywhere, and therefore $F(x)-G(x)$ is a constant function. Therefore, $F(x)=G(x)+c$, for some $c \in \mathbb{R}$.
The anti-derivative is also known as the integral.
First Fundamental Theorem
Consider a function $F(b)$ that is the area under some curve from $x=a$ to $x=b$. $F(b+h)-F(b) \approx h \cdot f(b)$, where $h$ is small enough, since the this extra area can be approximated as a rectangle. More formally, we can say $\lim \limits_{h \to 0} \frac{F(b+h)-F(b)}{h}=F’(b)=f(b)$. So, the anti-derivative is related the area under the curve.
To denote the area under some function from $x=a$ to $x=b$, we write $\int_{a}^{b} f(x) \, dx$. Interestingly, it is allowed that $a>b$. In that case, it is defined that $\int_{a}^{b} f(x) \, dx=-\int_{b}^{a} f(x) \, dx$.
Estimating Derivative given Table of Values
We can use Riemann sums or trapezium rule.
Riemann sums use rectangles to approximate area, and usually use either the left or right endpoint as the height of the rectangle. Left Riemann sums underestimate when the function is increasing and right Riemann sums overestimate when the function is increasing.
Trapezium rule is similar to Riemann sums but we use trapeziums instead of rectangles. It is a overestimate when the function is concave upwards.
| $x$ | $0$ | $3$ | $4$ | $6$ |
|---|---|---|---|---|
| $f(x)$ | $6$ | $2$ | $8$ | $3$ |
$R_l=f(0)\cdot (3-0) + f(3) \cdot (4-3) + f(4) \cdot (6-4)=18+2+16=36$
$T=\frac 12 ( (f(0)+f(3))\cdot (3-0) + (f(3)+f(4))\cdot (4-3) +(f(4)+f(6))\cdot (6-4) )=\frac 12 (24+10+22)=28$
Common Integrals
- $ \int 1 \, dx=\int \, dx=x+C$
- $\int x^n \, dx=\int \frac{x^{n+1}}{n+1} +C, n \ne -1$
- $\int \frac{1}{x} \, dx = \ln \lvert x \rvert +C$
- $\int e^x=e^x+C$
- $ \int \sin(x) \, dx= - \cos(x) +C$
- $\int \cos(x) \, dx=\sin(x)+C$
- $\int \sec^2(x)=\tan(x)+C$
- $\int \frac{1}{\sqrt{1-x^2}} \, dx=\arcsin(x)+C$
- $\int \frac{1}{1+x^2} \, dx=\arctan(x)$
- $\int \frac{1}{a^2-x^2} \ dx=\frac{1}{2a} \ln\lvert \frac{a+x}{a-x} \rvert+C$
- $\int \frac{1}{x^2-a^2} \ dx=\frac{1}{2a} \ln \lvert \frac{x-a}{x+a} \rvert +C$
Rules
$\int f(x) \pm g(x) \, dx=\int f(x) \, dx \pm \int g(x) \, dx$ $\int kf(x) \, dx=k \int f(x) \, dx$ (linearity of integration)
$\int e^{4x}+\sin(69x) \, dx=\int e^{4x} \, dx + \int \sin(69x) \, dx=\frac{e^{4x}}{4}-\frac{\cos(69x)}{69}$
\[\begin{align} \int \frac{2x^2+5x+4}{x^3+4x^2+4x+3} \, dx =& \int \frac{x+1}{x^2+x+1} + \frac{1}{x+3} \, dx \\\\ =& \int \frac{x+0.5}{x^2+x+1} \, dx + \int \frac{0.5}{x^2+x+1} \, dx + \int \frac{1}{x+3} \, dx\\\\ =& \frac 12 \ln \lvert x^2+x+1 \rvert + \frac{\arctan(\frac{2x+1}{\sqrt{3}})}{\sqrt 3} + \ln \lvert x+3 \rvert +C \end{align}\]It turns out that all functions that is a polynomial divided by another polynomial can be integrated by partial fractions. This is because the fundamental theorem of algebra which states that any real polynomial can be decomposed into real polynomials with at most degree $2$.
$\int_a^b f(x) \, dx=\int_{u(a)}^{u(b)} f(x) \frac{du}{dx} \, du$ (U-substituition)
$\int x^2 e^{x^3} \, dx=\int \frac 13 e^u \, du=\frac 13 e^{x^3}$, $u=x^3, du=3x^2 \cdot dx$
\[\begin{align} \int \sqrt{\frac{1-x}{1+x}} \, dx = & \int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} \, dx \\\\ =& \int \frac{1-x}{\sqrt{1-x^2}} \, dx \\\\ =& \int \frac{1}{\sqrt{1-x^2}} \, dx - \int \frac{x}{\sqrt{1-x^2}} \, dx \\\\ =& \arcsin(x) + \int \frac12 \frac{1}{u} \, du, u=1-x^2, du=-2x \cdot dx \\\\ =& \arcsin(x) + \sqrt{1-x^2} +C \end{align}\]$\int fg’ \, dx= fg - \int f’g \, dx$ (integration by parts)
$\int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx = x\sin(x) + \cos(x) + C$
Reduction Formula
Sometimes we are not able to find a closed form for a integral, so we have to use reduction formula to find a recursive definition for this integral.
For example, let $I_n = \int x^n e^x \, dx$. We can show that $I_n=x^ne^x - nI_{n-1}$ for $n \geq 1$.
Usually for such problems, we will use integration by parts to prove the reduction.
\[\begin{align} I_n=& \int x^n e^x \, dx \\\\ =& x^n e^x - \int nx^{n-1} e^x \, dx\\\\ =& x^ne^x - nI_{n-1} \end{align}\]Riemann Sum
$\lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{k=0}^\infty f(\frac kn)=\int_0^1 f(x) \, dx$
$\lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{k=0}^\infty (\frac kn)^2=\int_0^1 x^2 \, dx=\frac 13$
Sometimes we need to do some work before we get a form of the equation that can be converted into an integral.
\[\begin{align} \lim\limits_{n \to \infty} \sum\limits_{k=0}^\infty \sqrt{\frac{1}{3n^2+2k^2}} =&\lim\limits_{n \to \infty} \sum\limits_{k=0}^\infty \sqrt{\frac{1}{n^2(3+2(\frac{k}{n})^2)}} \\\\ =&\lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{k=0}^\infty \sqrt{\frac{1}{3+2(\frac{k}{n})^2}} \\\\ =&\int_0^1 \frac{1}{3+2x^2} \, dx \\\\ =& \frac{1}{3} \int_0^1 \frac{1}{1+(\sqrt{\frac{2}{3}}x)^2} \\\\ =& \frac{\sqrt{3}}{3 \sqrt{2}} \big[ \arctan{(\sqrt{\frac{2}{3}} x)} \big]_0^1 \\\\ =& \frac{1}{\sqrt{6}} \arctan{(\sqrt{\frac{2}{3}})} \\\\ \end{align}\]Mean-Value Theorem
Suppose $mn$ and $mx$ are the minimum and maximum values respectively on some closed, bounded and continuous interval $[a,b]$.
Then it follows that $mn(b-a) \leq \int_a^b f(x) \, dx \leq mx(b-a)$. Furthermore, from the Mean-Value Theorem, there exists $c\in [a,b]$ such that $\int_a^b f(x) \, dx=f(c)(b-a)$.
Average Value
The average value of the function on $[a,b]$ is given by $\frac{\int_a^b f(x) \, dx}{b-a}$, it can be thought of as $\frac{\text{area}}{\text{length}}$.
Second Fundamental Theorem
$f(x) = \frac{d}{dx} \int_a^x f(t) \, dt$
Alternatively, we define $F(x)$ as a function such that $F’(x)=f(x)$. Then $\frac{d}{dx} \int_a^b f(t) \, dt=\frac{d}{dx} F(b)-F(a)$.
$\frac{d}{dx} \int_{-1}^{x^3} (t^3+2)^{\frac 13} \, dt$
Define $f(t)=(t^3+2)^{\frac 13}$
$\frac{d}{dx} \int_{-1}^{x^3} f(t) \, dt=\frac{d}{dx} F(x^3)-F(-1)=3x^2 f(x^3)=3x^2 \cdot (x^9+2)^{\frac 13}$
Kinematics
$\frac{dx}{dt}=v$, $\int v \, dt=x$
$\frac{dv}{dt}=a$ $\int a \, dt =v$
Note that the displacement of an object is $P_{final}-P_{initial}$ which is $\int v \, dt$.
However, the distance travelled is the total length of the path taken, which is $\int \lvert v \rvert \, dt$.
Imagine a particle with $v(t)=t^2-t-3$ and $x(0)=-3$.
The particle’s position at time $t$, $x(t)$ is given by $x(t)=x(0)+\int_0^t v(k) \, dk=-3+\big[\frac{k^3}{3}-\frac{k^2}{2}-3k]_0^t=\frac{t^3}{3}-\frac{t^2}{2}-3t-3$.
The particle’s acceleration at time $t$, $a(t)$ is given by $a(t)=\frac{d}{dx} v(t)=\frac{d}{dk} k^2-k-3=2k-1$.
Let’s compare the displacement and distance travelled from $t=1$ to $t=3$.
The displacement is $\int_1^3 v(k) \, dk=-1.333$.
The distance travelled is $\int_1^3 \lvert v(k) \rvert \, dk=3.312$.
We observe that the displacement is in no way related to the distance travelled.
Integration by Geometry
Consider the integral $\int_0^1 \sqrt{1-x^2} \, dx$. It is very hard to integrate by only using arithmetic. Putting it into WolframAlpha, you find that $\int \sqrt{1-x^2} \, dx= \frac{1}{2} (\sqrt{1-x^2} \cdot x + \arcsin{(x)})$, which is probably hard to do by hand.
However, recalling that the equation of a circle is $x^2+y^2=1$, we realize that the above integral is just finding the area of $\frac{1}{4}$ of a circle of radius $1$. Which is simply $\frac{\tau}{8}$.