Index

Class 12 level recap

1. Integral Calculus Recap
2. Basic Integral Formulae
3. Some basic trigonometric identities
4. Methods of integration
5. 1. U-substitution
6. 2. Integration by Parts -- The easy method.
7. 3. Integration by partial fractions -- An easy method

College Stuff (Calculus - 1)

1. Evolutes and Involutes
2. 1. Indefinite vs Definite Integrals
3. 2. Even and Odd Functions (Symmetry Rules)
4. 3. Improper Integrals
5. Applications of definite integrals to evaluate surface areas and volumes of revolutions
6. DEFINITE INTEGRALS — MASTER ALGORITHM
7. Area under Curves
8. Volumes of Revolution
9. 1. Disk Method (When there is no hollow region)
10. 2. Washer Method (When there is a hollow region)
11. Gamma Functions
12. Beta Functions

Integral Calculus Recap

Taken from Integral Calculus

A problem well defined, is a problem half solved -- Charles Kettering

https://www.youtube.com/watch?v=hXOrQ0Ao4UE&list=PLF-vWhgiaXWM7Iri0t_AjBfv51tF28PEy&index=1&pp=gAQBiAQB

The above link points to a full comprehensive recap of class 12 recap. Watch it only if you have time.

Basic Integral Formulae

Some basic trigonometric identities

https://www.youtube.com/watch?app=desktop&v=m1OitPmkydY

https://www.geeksforgeeks.org/trigonometric-identities/

https://www.geeksforgeeks.org/trigonometry-table/

Methods of integration

1. U-substitution

So let's say we have an integral

To solve this using u-substitution, we need to:

1. set $x=g(t)$.
2. Differentiate both sides
3. So, we get : $${dx} = g'(t).dt$$
   And then we replace the value of $x$ and $dx$ in the original equation to get:

Example 1 :

1. $$\int sin(mx)dx$$

Let us set $mx=t$ and differentiate both sides w.r.t $x$, we get :

And thus we replace $dx$ in the original equation to get:

or , $$\frac{-cos(mx)}{m} + C$$

Example 2

Let $$1+x^2 = t$$

or $$\implies 0 + 2x = \frac{dt}{dx}$$
or, $$2x dx = dt$$
Replacing this in the original equation,

or ,

2. Integration by Parts -- The easy method.

https://www.youtube.com/watch?v=2I-_SV8cwsw&list=PLF-vWhgiaXWM7Iri0t_AjBfv51tF28PEy&index=2

This is called the D-I method. Differentiate-Integrate. (also known as the LIATE method)

Example 1.

So let's say we have this integral here :

We need to make a small table.

The D column represents differentiation and I column represents integration.

Now we choose each term to be either differentiated or integrated.

Here the easy choice seems to differentiate $x^2$ and integrate $sin(3x)$

Now we continue the respective operations on both sides for a while.

Each column having alternating signs.

What to differentiate and what to integrate?

• Inverse trigonometric functions (like $ta{n}^{-1}$, $si{n}^{-1}$) — differentiate these if present.
• Logarithmic functions (like ln(x)) — differentiate these if present.
• Algebraic functions (like $x^2$, $x$, constants) — differentiate these if no inverse or logarithmic terms are present.
• Trigonometric functions (like $sin(x)$, $cos(x)$) — differentiate these if there are no inverse, logarithmic, or algebraic terms to use.
• Exponential functions (like $e^x$, $a^x$) — these are typically integrated as a last choice.

When to stop the operations:

1. If the column of D reaches 0.
2. If the products of a row (product of both the value of D and I) can be integrated.
3. If any the products of any row (product of both the value of D and I) result in the original integral.

So here we see that the column of D has reached 0.

So we multiply downwards, diagonally, along with each respective sign, and write them together for the final answer.

Like this.

So the final answer becomes,

Example 2

Let's say we have another integral

Example 3:

Let's say we have another integral, $$e^x.sin(x)dx$$

Remembering our rules, we differentiate $e^x$, since functions like $e^x$ are integrated as a last choice. And the remaining $sin(x)$ will be differentiated.

So ,

So we stop the operations and write diagonal products first.

or $$2\int{e^x.sin(x)dx} = e^{x}.[sin(x)-cos(x)]$$
or, $$\int{e^x.sin(x)dx} = \boxed{\frac{e^{x}.[sin(x)-cos(x)]}{2}}$$

3. Integration by partial fractions -- An easy method

https://www.youtube.com/watch?v=OFEfGtCsVKg&list=PLF-vWhgiaXWM7Iri0t_AjBfv51tF28PEy&index=4&pp=gAQBiAQB

This method can only be applied if the denominator can be factorized into linear factors

or that in denominator power of $x$ is 1, or it is a product of terms in which power of $x$ is 1.

Example 1

So let's say we have an integral :

We can see that it's denominator is in the form of a product of linear factors.

So the way to tackle this, is to take one term at a time.

We first select $(x-1)$.

Equate this term to 0. We get $x=1$.

Now we apply this value back in the original equation, but there's a twist.

We modify this part of the process by removing the selected term from denominator, but instead placing it in the form of a natural log (log or $ln$), and we don't substitute the value of $x$ in that log term.

1. So, for $(x-1)$

or $$\frac{-ln(x-1)}{6}$$
2. For $(x+2)$, we get $x=-2$

or $$\frac{2(-2) - 1}{(-2-1)(-2-3)}ln(x+2)$$

or $$\frac{-5.ln(x+2)}{15} = \frac{-ln(x+2)}{3}$$
3. For $(x-3)$, we get $x=3$.

or $$\frac{2(3) - 1}{(3-1)(3+2)}.ln(x-3)$$
or $$\frac{5.ln(x-3)}{10} = \frac{ln(x-3)}{2}$$
Finally we put all the obtained terms together along with their signs.

as our complete answer.

The given video link has another example on this.

Evolutes and Involutes

Evaluation of definite and improper integrals

1. Indefinite vs Definite Integrals

Indefinite Integral

Indefinite Integral (the integral we are used to by default)

• Reverse of differentiation
• Gives a function
• Always include $+C$

Definite Integral

An integral which has defined specific upper and lower limits.

• Gives a number
• No $+C$ (constants cancel)
• Represents net area

2. Even and Odd Functions (Symmetry Rules)

Odd Function

Examples of such odd functions:

$x^3$, $x$, $\sin (x)$

Reason: Areas cancel.

For dissimilar limits:

Even Function

Examples: $x^2$, $\cos (x)$.

Reason: Graph is symmetric about y-axis.

For dissimilar limits:

3. Improper Integrals

An improper integral is a definite integral that extends the concept of integration to cases where the interval of integration is unbounded (infinite) or the integrand has an infinite discontinuity.

The Case of Infinity

For:

or

The integral:

• Converges if $p>1$
• Diverges if $p\le 1$

Key idea: If $p>1$ then the integral shrinks, but shrinks slowly, enough so that it can be
integrated. If $p\le 1$, then the integral shrinks too fast to be integrated.

Powers even as close to 1 as:

will converge, because that tiny extra power makes the function shrink just fast enough.

However the moment $p=1$, the function will diverge.

Example:

Exhibit A:

Exhibit B:

Near-Zero Case

For:

or

The integral:

• Converges if $p<1$
• Diverges if $p\ge 1$

Key idea: If $p<1$ then the integral blows up, but blows up slowly, enough so that it can be
integrated. If $p\ge 1$, then the integral blows up too fast to be integrated.

Powers even as close as $0.99$, will converge, but the moment $p=1$, the function will diverge.

The boundary is sharp.

Example:

Exhibit A:

or:

or:

Exhibit B:

5. Why Conditions Flip

At $\mathrm{\infty }$ → concerned about shrinking
At $0$ → concerned about explosion

Practice examples for the evaluation of definite and improper integrals.

1. Even and Odd Functions

(A)

Since $x^5$ is odd.

(B)

(C)

2. Improper Integrals -- The Case of Infinity

(A)

Here: $p=3>1$.

So the function will converge.

(B)

Since $p=\frac{1}{2}=0.5<1$,

The function diverges.

The value is:

(C)

Since $p=1.2>1$,

The function converges.

3. Improper Integrals -- Near Zero

(A)

Here,

$p=0.7<1$, so the function will converge.

(B)

Here, $p=2>1$, so the function will diverge.

Value:

(C)

Since $p=0.99<1$, the function will converge.

I am not going to solve this one, since the concept is already clear at this point.

Applications of definite integrals to evaluate surface areas and volumes of revolutions

Before we proceed, let's sum up definite integrals in one master algorithm

DEFINITE INTEGRALS — MASTER ALGORITHM

We separate everything into three layers:

1. Orientation
2. Symmetry (Parity)
3. Sign / Area Type

PART 1 — ORIENTATION RULE (Always Check First)

Rule:

or

(limits were swapped manually here).

When to use:

• Whenever limits appear reversed.
• Whenever you swap limits manually.

When NOT to use:

• If limits are already in increasing order.

This has nothing to do with symmetry or sign.
It is purely about direction.

PART 2 — SYMMETRY / PARITY RULE (Only When 0 is Structurally Present, i.e present among the limits)

Is interval symmetric around 0?
If yes → use parity.
If no → ignore parity.

or,

Trigger condition:

• Interval is $[-a,a]$, or
• You are converting $[-b,-a]$ to $[a,b]$

Odd Function

Examples of such odd functions:

$x^3$, $x$, $\sin (x)$

Even Function

Examples: $x^2$, $\cos (x)$.

PART 3 — SIGN ANALYSIS (Determines Area Sign)

This determines whether the integral is:

• Positive
• Negative
• Zero

Ask:

Is function positive, negative, or changing sign?

This determines:

• Net area sign
• Whether splitting is needed
• Whether absolute value is needed

This is done is as follows:

Suppose we have three range brackets (or zeroes):

Let's say from $-2$ to $2$.

And we get the zeroes as:

The rule to test whether the function $f(x)$ is positive or negative between each bracket, is:

• Always check with values between the ends of the bracket.

In the case of zeroes where you cannot find any middle value, you can use either end, like for example $-2$ in $[-2,-1]$ or $2$ in $[1,2]$.

However in the case of the bracket of $[-1,1]$, we can find a middle value, which is $0$.

So, we can test if the function $f(x)$ is positive or negative at $x=0$, which will tell us whether the function is positive or negative in $[-1,1]$.

Area under Curves

Are we asked for:

• Net area? → use integral normally.
• Total/geometric area? → integrate absolute value.

1. Net Area (Standard Definite Integral)

This calculates:

Signed area.

• Area above x-axis → positive.
• Area below x-axis → negative.
• Opposite sides cancel.

Odd symmetry cancellation happens here.

2. Geometric Area (Total Area)

This calculates:

Total physical area.

Everything becomes positive.

No cancellation.

🔵 When To Split the Integral

Split when:

• The function changes sign in the interval.
• You are calculating geometric area.
• The interval crosses 0 and you want symmetry benefits.

Example:

Quick Summary Table

Example

Let's say we have:

We have to find the:

• Net Area
• Geometric Area

For Net Area

The limits go from -3 to 3, so the integral has to be split:

Now we solve:

This could also have been directly solved by just:

For Geometric Area

For the integrand,

Set $x^2=0$

Thus $x^2=0$.

$x=0$.

So we going from $-3$ to $3$, we get three range brackets:

The function would be: $x^2=0$

So, for any $x\ne 0$, $x^2>0$

For the first bracket, $x^2=9$
For the second bracket, $x^2=0$
For the third bracket: $x^2=0$

So the function is positive for all three range brackets

So the geometric area would be:

We can technically ignore the middle bracket of $[0,0]$ since the end output would be just zero. Adding or subtracting that won't make any difference.

Solving these integrals,

Thus, the net and geometric area are the same.

Practice Problems

1. Find net and geometric area of

Net area = $0$.

For geometric area,

Setting, $x^3=0$, $x=0$.

Then we split the integral:

At:

• $x=-2$, $x^3=-8$, $[-2,0]$
• $x=0$, $x^3=0$, $[0,0]$
• $x=0,x^3=0$., $[0,2]$

So, the function is negative between $[-2,0]$ and positive for $[0,2]$

So, the geometric area:

2. Find the net and geometric area of

First, we split the integral:

Both are even functions, so by rule of parity:

Now, for the geometric area:

The limits go from $-2$ to $2$.

Now, setting $x^2-1=0$,

$x=\pm 1$.

So the zeroes are:

At:

• $x=-2$, $x^2-1=3$
• $x=0$, $x^2-1=-1$
• $x=1$, $x^2-1=0$

So the function is positive for the range brackets of $[-2,-1],[1,2]$, but negative for $[-1,1],$

Now, we will split the integral accordingly:

Volumes of Revolution

This is where definite integrals stop being just “area under a curve” and start becoming 3D geometry.

We revolve a curve around an axis → it sweeps out a solid → we compute its volume using integration.

There are two primary methods you must master:

1. Disk / Washer Method
2. Shell Method

1. Disk Method (When there is no hollow region)

Situation:

You rotate a curve around an axis and it creates a solid disk (like stacking coins).

If a function $y=f(x)$, is rotated about the x-axis then,

Why is function squared?

Because:

• At each x, the radius = $f(x)$
• Area of circular cross-section = $\pi r^2$.

So, a small volume element will equal:

Then we can integrate it to find the volume.

Example 1.

Find volume formed by rotating:

from $x=0$ to $x=2$ about the $x$-axis.

Step 1: Find radius.

At each $x$, the radius is $f(x)$ which is $y=x$. (just $x$).

Step 2: Apply formula:

2. Washer Method (When there is a hollow region)

Now suppose you rotate the region between two curves:

and

with $f(x)>g(x)$.

You get a hollow solid, like a washer.

The formula for the volume is:

where:

• $R(x)$ = outer radius
• $r(x)$ = inner radius

WASHER METHOD — PURE ALGORITHM

We assume:

• Region between two curves
• Rotated about a horizontal or vertical line

STEP 1 — Identify Axis of Rotation

Is it:

• y = 0 (x-axis)?
• x = 0 (y-axis)?
• y = k ?
• x = k ?

Write it down clearly.

Everything depends on this.

STEP 2 — Decide Variable of Integration

If axis is horizontal (like y = something):

→ Use vertical slices
→ Integrate with respect to dx

If axis is vertical (like x = something):

→ Use horizontal slices
→ Integrate with respect to dy

(Exam shortcut: washers use slices perpendicular to axis.)

STEP 3 — Identify Outer and Inner Radius

Radius = distance from axis.

Always compute distance using:

Then:

• Outer radius = farther curve from axis
• Inner radius = closer curve to axis

Do NOT think “which function is bigger.”

Think “which is farther from axis.”

STEP 4 -- Apply Washer Formula

Where:

• R = outer radius
• r = inner radius

Gamma Functions

What is it?

The Gamma function is an extension of the factorial to all real numbers, including fractions.

Formally defined as:

• The $e^{-t}$ term forces convergence despite the infinite upper limit — it kills the function fast enough as $t\to \mathrm{\infty }$

• Defined for all n > 0

The Reduction Formula

This is the core property. Every Gamma value connects to the previous one by simple multiplication — same unwinding pattern as a factorial.

So Gamma is the factorial function, shifted by 1:

• Γ(5) = 4! = 24
• Γ(6) = 5! = 120

⚠️ Watch the shift — Γ(5) = 4!, not 5!

Base Values

Half-Integer Values

Derived using the reduction formula starting from $\Gamma (\frac{1}{2})=\sqrt{\pi }$ .

Pattern:

• Numerator: 1, 1, 3, 15 → each multiplies by the next odd number
• Denominator: 1, 2, 4, 8 → powers of 2

💡 If you blank in the exam, just re-derive on the spot using the reduction formula. Takes 20 seconds.

Beta Functions

What is it?

A function of two inputs that measures a weighted interaction between two exponents.

• Limits are 0 to 1 (unlike Gamma which goes 0 to ∞)
• Two parameters m and n

Key Properties

Symmetry:

Bridge Formula:

This connects Beta back to Gamma. Almost every exam question uses this.

⚠️ Denominator is Γ(m+n) — add the arguments FIRST, then take Gamma. Not Γ(m) + Γ(n).

Trigonometric Form

Which means:

since:

• $\frac{p+1}{2}=m$
• $\frac{q+1}{2}=n$

Algorithm for Trig Integrals

1. Read off p and q from the powers of sin and cos
2. Compute m = (p+1)/2 and n = (q+1)/2
3. Extend the half-integer Gamma table as far as needed using Γ(n+1) = n·Γ(n)
4. Plug into bridge formula
5. Let √π cancel — keep everything as pure fractions, never decimals
6. Simplify fraction to final answer

Worked Example

• p = 5, q = 4
• m = $\frac{5+1}{2}=3$, n = $\frac{4+1}{2}=\frac{5}{2}$
• Γ(3) = $2!=2$
• Γ(5/2) = $\frac{3\pi }{4}$
• Γ(11/2) = 9/2 · Γ(9/2) = 9/2 · $\frac{105\pi }{16}$ = $\frac{945\pi }{32}$