Frequently used formula

Graph Stuff

Extrema

Critical Point

How to find critical point

Differentiate once, then find the root of f’(x) = 0.

The roots are the critical points.

Increasing/Decreasing(1st derivative test)

By ploting the graph of f and f’, you can find out whether it is increasing or decreasing.

Concavity (2nd derivative test)

By ploting the graph of f’ and f’', you can find out whether it is concave up or concave down.

It is better to find out the inflection point as it helps you to find out the concavity.

Point of Inflection

How to find inflection point

Differentiate twice, then find the root of f’'(x) = 0.

The roots is inflection point.

Asymptotes

Horizontal Asymptotes

find $\lim_{x\rightarrow \infty } f(x) = a$ . If the answer $a$ is $\infty$ , there is no horizontal asymptote.

Otherwise horizontal asymptote = $a$ .

Vertical Asymptotes

find $\lim_{x\rightarrow a } f(x) = \pm \infty$ . If the answer $a$ is $\infty$ , there is no vertical asymptote.

Otherwise vertical asymptote = $a$ .

Inclined Asymptotes

There will be 2 cases. the procedure depends on the exponent of the numerator and denominator.

If exponent of numerator > exponent of denominator :

e.g. $f(x) = \frac{x^2-x-2}{x-2}$

Inclined Asymptote = $y = mx+c$

where $m$ = $\lim_{x\rightarrow \infty } \frac{f(x)}{x}$ , $c$ = $\lim_{x\rightarrow \infty } (f(x)-mx)$

if exponent of denominator > exponent of numerator :

e.g. $f(x) = \frac{x^2+1}{x^3-x^2}$

find $\lim_{x\rightarrow \infty } \frac{\frac{numerator}{highest\space exponent}}{\frac{denominator}{highest\space exponent}} = a$

$\lim_{x\rightarrow \infty } \frac{\frac{x^2+1}{x^3}}{\frac{x^3-x^2}{x^3}} = a$

Inclined asymptote = $a$

Integration Stuff

Integration

Basic Integration Rules

Some good examples to try with :

$\int \frac{4}{x^2+9}dx$
$\int \frac{4x^2}{x^2+9}dx$

Definite integrals

Some important properties of Definite integrals

if $f$ is defined at $x=a$ , then $\int_{a}^{a}f(x)dx = 0$

if $f$ is integrable on [a,b], then $\int_{b}^{a}f(x)dx = -\int_{a}^{b}f(x)dx$

Additive Interval Property

if $f$ is integrabble on the three closed intervals determined by a,b,c, then

$\int_{a}^{b} f(x)dx = \int_{a}^{c} f(x)dx + \int_{c}^{b} f(x)dx$

Some good example to try with :

Given $\int_{0}^{2} f(x)dx =3$ , $\int_{1}^{4} f(x)dx = 5$ , $\int_{1}^{2} f(x)dx = 1$ , $\int_{0}^{4} g(x)dx = -x$ .

Find $\int_{2}^{4} f(x)dx$
Find $\int_{0}^{4} f(x)dx$
Find $\int_{0}^{1} [2f(x)+3g(x)]dx$
Find $\int_{2}^{4} [g(x)-2f(x)]dx$

Fundamental Theorem of Calculus (FTC)

if a function $f$ is continuous on the closed interval [a,b] and $F$ is an antiderivative of $f$ on the interval [a,b], then

$\int_{a}^{b} f(x)dx = F(b) - F(a)$

Taylor Expansion

Linear Approximation (Just to give you an idea)

$f(x) \approx L(x) = f(c) + f'(c)(x-c)$

for linear approximation,

$R_n(x) = f''(z)(x-c)^2$

Taylor Series

Find $n$ , $x$ and $c$ .
then find $z$
then find $f^n(x)$
then find $R_n(x)$

Taylor expension of n degree :

$T_n(x) = f(c) + \frac{f'(c)(x-c)^1}{1!}+ \frac{f''(c)(x-c)^2}{2!}+ \frac{f'''(c)(x-c)^3}{3!} ...+ \frac{f^n(c)(x-c)^n}{n!}$

$R_n(x) = \frac{f^{n+1}(z)(x-c)^{n+1}}{(n+1)!}$

$f(x) = T_n(x) + R_n(x)$

Integration by substitution

Basically, use $u$ to make the thing less complicated to solve.

Geniue Steps:

Sub $u = g(x)$

Then $du = g'(x)dx$

Rewrite the integrals in terms of $u$

Find the resulting integrals

then replace $u$ by $g(x)$ .

Some good examples to try with :

$\int (x^2+1)^2(2x)dx$
$\int \sqrt{2x-1}dx$
$\int^1_0 x(x^2+1)^3dx$

General Power Rule for Integration

$\int x^ndx = \frac{x^{n+1}}{n+1} + C$

where $n \neq -1$

e.g.

$\int x^1 dx = \frac{x^{1+1}}{1+1} = \frac{x^2}{2}$

Some good examples to try with :

$\int (3x^2 - \sqrt{5x} + 2)dx$
$\int (x^3-2x^2)(\frac{1}{x}-5)dx$

Trigonometric substitution

Some good examples to try with :

$\int(2+tan^2x)dx$
$\int (2sint-3cost+t^{\frac{5}{4}})dt$
$\int cos(x)(tan(x)+sec(x))dx$
$\int \frac{x^2-2}{x^2+1}dx$
$\int_2^3 \frac{3x^2-1}{(x^3-x)^3}dx$
$\int \frac{dx}{x^2\sqrt{9-x^2}}$

Integration by parts

I like to call it reverse product rule.

you can pick which to be $f(x)$ and $g(x)$ .

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

$f(x)g(x) = \int f'(x)g(x)dx + \int f(x)g'(x) dx$

$\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx$

so basically :

$(AB)' = A'B + B'A$

$AB = \int A'B dx + \int B' A dx$

$\int A'Bdx = AB - \int B'A dx$

For example, to solve $\int \frac{lnx}{x^2}dx$

$\int \frac{lnx}{x^2}dx$

$= \int lnx \frac{1}{x^2}dx$

Let $f(x) = lnx$ , $g'(x) = \frac{1}{x^2}$

Now you get $\int lnx \frac{1}{x^2}dx = \int f(x)g'(x) dx$ .

$\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx$

Now you need to find out $g(x)$ and $f'(x)$ .

Look at the integration table (or you solve it yourself), you get $f'(x) = \frac{1}{x}$ , $g(x) = \frac{-1}{x}$ .

Sub the variables into the equation.

$\int f(x)g'(x) dx = f(x)g(x) - \int f'(x)g(x) dx$

$\int lnx \frac{1}{x^2}dx = (lnx\cdot\frac{1}{x}) - \int(\frac{1}{x} \cdot \frac{-1}{x})dx$

Then you solve the $(lnx\cdot\frac{1}{x}) - \int(\frac{1}{x} \cdot \frac{-1}{x})dx$ to get the answer.

Note : Sometimes you may need to carry integration by part twice to get the answer.