Definite Integrals are used to find areas of the complex curve, volumes of irregular shapes, and other things. Definite Integrals are defined by, let us take p(x) to be the antiderivative of a continuous function f(x) defined on [a, b] then, the definite integral of f(x) over [a, b] is denoted by\int\limits_{a}^{b}f(x)dx and is equal to [p(b) - p(a)].
\bold{\int\limits_{a}^{b}f(x)dx} = p(b) - p(a)
The numbers a and b are called the limits of integration where a is called the lower limit and b is called the upper limit. The interval [a, b] is called the interval of the integration.
Note
- Constant of Integration is not included in the evaluation of the definite integral.
- \bold{\int\limits_{a}^{b}f(x)dx} is read as "integral of f(x) from a to b"
How to Evaluate Definite Integrals?
To find the definite integral of f(x) over interval [a, b] i.e.,\int\limits_{a}^{b}f(x)dx we have following steps:
Step 1: Find the indefinite integral ∫f(x) dx.
Step 2: Evaluate p(a) and p(b) where, p(x) is the antiderivative of f(x), p(a) is the value of antiderivative at x = a, and p(b) is the value of antiderivative at x = b.
Step 3: Calculate p(b) - p(a).
The value obtained in Step 3 is the desired value of the definite integral.
Properties of Definite Integral
Various Properties of Definite Integral have been discussed below:
Property 1)\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(z)dz}
Proof:
Let p(x) be an antiderivative of f(x). Then,
\frac{d}{dx} {p(x)} = f(x)
\frac{d}{dz} {p(z)} = f(z)
\int\limits_{a}^b f(x)dx = \big[p(x)\big]_a^b = p(b) - p(a) -----------------(i)
\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}
and\int\limits_{a}^b f(z)dz = \big[p(z)\big]_a^b = p(b) - p(a) -----------------(ii)
From (i) and (ii)
\bold{\int\limits_{a}^{b}f(x)dx= \int\limits_{a}^{b}f(z)dz}
Property 2)\bold{\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx}
If the limits of the definite integral are interchanged then, its value changes by a minus sign only.
Proof:
Let p(x) be an antiderivative of f(x). Then,
\int\limits_{a}^b f(x)dx = p(b) - p(a)
and-\int\limits_{b}^a f(x)dx = -[p(a) - p(b)] = p(b) - p(a)
\bold{\int\limits_{a}^{b}f(x)dx=-\int\limits_{b}^{a}f(x)dx}
Property 3)\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx} where, a<c<b
Proof:
Let p(x) be the antiderivative of f(x). Then,
\int\limits_{a}^b f(x)dx = p(b) - p(a) ---------------(i)
\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx = [p(c) - p(a)] + [p(b) - p(c)] = p(b) - p(a) ---------------(ii)
From (i) and (ii)
\bold{\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{c}f(x)dx+\int\limits_{c}^{b}f(x)dx}
Property 4)\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}
Proof:
Let x = a - t. Then, dx = d(a - t) ⇒ dx = -dt
When x=0 ⇒ t = a and x = a ⇒ t = 0
\int\limits_{0}^{a}f(x)dx=-\int\limits_{a}^{0}f(a - t)dt
⇒\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a - t)dt [Using second property]
⇒\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a - x)dx [Using first property]
\bold{\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx}
Property 5)\bold{\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}
Proof:
Using third property
\int\limits_{-a}^{a}f(x)dx=\int\limits_{-a}^{0}f(x)dx+\int\limits_{0}^{a}f(x)dx ----------(i)
Let x = -t, dx = -dt
Limits: x = -a ⇒ t = a and x = 0 ⇒ t=0
\int\limits_{-a}^{0}f(x)dx=\int\limits_{a}^{0}f(-t)(-dt) =-\int\limits_{a}^{0}f(-t)dt=\int\limits_{0}^{a}f(-t)dt [By second property]
⇒\int\limits_{-a}^{0}f(x)dx=\int\limits_{0}^{a}f(-x)dx [By first property] -------------(ii)
From (i) and (ii)
\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}f(-x)dx+\int\limits_{0}^{a}f(x)dx
⇒\int\limits_{-a}^{a}f(x)dx=\int\limits_{0}^{a}[f(-x)+f(x)]dx
⇒\int\limits_{-a}^{a}f(x)dx = \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(-x) = f(x) \\ 0 & , if f(-x) = -f(x) \end{cases}
\bold{\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}}
Property 6) If f(x) is a continuous function defined on [0, 2a]\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}
Proof:
Using third property
\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{a}^{2a}f(x)dx -----------(i)
Consider
Let x = 2a - t, dx = -d(2a - t) ⇒ dx = -dt
Limits: x = a ⇒ t = a and x = 2a ⇒ t=0
\int\limits_{a}^{2a}f(x)dx=-\int\limits_{a}^{0}f(2a - t)dt
⇒\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - t)dt [Using second property]
⇒\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx [Using first property]
Substituting\int\limits_{a}^{2a}f(x)dx=\int\limits_{0}^{a}f(2a - x)dx in (i)
\int\limits_{0}^{2a}f(x)dx=\int\limits_{0}^{a}f(x)dx+\int\limits_{0}^{a}f(2a - x)dx = \int\limits_{0}^{a}[f(x) + f(2a - x)]dx
\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}
Property 7)\bold{\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx}
Proof:
Let t = a + b - x ⇒ dt = -dx
Limits: x = a, y=b and x = b, y =a
After putting value and limit of t in\int\limits_{a}^{b}f(a + b - x)dx
⇒\int\limits_{a}^{b}f(a + b - x)dx =-\int\limits_{b}^{a}f(t)dt
⇒\int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(t)dt [Using second property]
⇒\int\limits_{a}^{b}f(a + b - x)dx =\int\limits_{a}^{b}f(x)dx [Using first property]
\bold{\int\limits_{a}^{b}f(x)dx= \int\limits_{a}^{b}f(a + b - x)dx}
Applications of Definite Integrals :
There are numerous definite integral applications across different fields. Some of them include the following : Area under Curves : Definite integrals allow one to calculate the area between a curve and the axis of x within some specified interval. It is used in geometry and physics in determining the size of a region. Total Distance Covered : Integrating the absolute value of the velocity function over an interval would yield a value of total distance traveled by an object. This is very important in Physics and Engineering in the analysis of motion. Work Done by a Force : In physics, the work done by a variable force is calculated by integration of the function of force along a path. This explains the energy transfer and mechanical systems. Total Revenue and Cost Functions : Definite integrals are very important in economic theory for finding the total revenue and total cost functions. Integration of the revenue and cost function within certain intervals is pretty useful in studying profit margins and business decisions. Modeling of Population Growth and Decay : In problems involving biology and ecology, definite integrals are used to study the growth and decay of a population. If the functions of growth or decay in a population are integrated over time, insights into the dynamics and sustainability of the population can be gained. Evaluating Probabilities using Probability Density Functions : In statistics, definite integrals are used for evaluating probabilities, where a probability density function is given. The integration of such functions along certain intervals gives the probability of events happening within that range.
Definite Integral as Limit of Sum
The definite integral of f(x) over the interval [a, b], denoted by\int\limits_a^bf(x)dx , is defined as the limit of a sum given by:
\int\limits_a^bf(x)dx = \displaystyle \lim_{n\to\infty}\displaystyle \sum_{r=1}^n hf(a + rh) \\ or \\ \displaystyle\lim_{n\to\infty}\displaystyle \sum_{r=0}^{n-1} hf(a + rh)
where nh = b - a
f(x) is said to be integrable over [a, b] if the above two limits exist and are equal.
Challenging Definite Integral
Problem1: Evaluate the definite integral:\int\limits^2_0 x^2[x]dx
Solution:
The integral contains the greatest integer.
For limit 0 to 1 greatest integer function [x] = 0
For limit 1 to 2 greatest integer function [x] = 1
So, we split the above integral into two parts using the following definite integral property.
\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx
\int\limits_0^2x^2[x]dx = \int\limits_0^1x^2[x]dx+\int\limits_1^2x^2[x]dx
=\int\limits_0^1x^2.0dx+\int\limits_1^2x^2.1dx
\int\limits_0^2x^2[x]dx = 0+\int\limits_1^2x^2dx
\int\limits_0^2x^2[x]dx = \int\limits_1^2x^2dx
\int\limits_0^2x^2[x]dx = \big[\frac{x^3}{3}\big]_1^2
= [8/3] - [1/3]
\int\limits_0^2x^2[x]dx = 7/3
Problem 2: Evaluate:\int\limits^3_1|x^2-2x| dx
Solution:
The integral contains a mod function.
For limit 1 to 2 let x=1, the function f(x) = x2 - 2x = 1-2 = -1 is negative.
For limit 2 to 3 let x=3, the function f(x) = x2 - 2x = 9-6 = 3 is positive.
So, we split the above limit of integral into two parts using the following definite integral property.
\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx
\int\limits^3_1|x^2-2x| dx =\int\limits^2_1|x^2-2x| dx +\int\limits^3_2|x^2-2x| dx
\int\limits^3_1|x^2-2x| dx =\int\limits^2_1-(x^2-2x) dx +\int\limits^3_2(x^2-2x) dx
\int\limits^3_1|x^2-2x| dx =\int\limits^2_1(-x^2+2x) dx +\int\limits^3_2(x^2-2x) dx
\int\limits^3_1|x^2-2x| dx =\big[x^2-\frac{x^3}{3}\big]^2_1 +\big[\frac{x^3}{3}-x^2\big]^3_2
\int\limits^3_1|x^2-2x| dx =\big[4-\frac{8}{3}\big]-\big[1-\frac{1}{3}\big] +\big[9-9\big]-\big[\frac{8}{3}-4\big]
\int\limits^3_1|x^2-2x| dx = 2/3 + 4/3
\int\limits^3_1|x^2-2x| dx = 2
Problem 3: Evaluate:\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx
Solution:
\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx
Let √x = t
1/ (2√x) dx = dt
dx/√x = 2dt
Limits: If x=0, t=0 and x=π2/4, t=π/2
\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2\int\limits_0^{\pi/2}costdt
\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2\big[sint\big]_0^{\pi/2}
\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2(1 - 0)
\int\limits_0^{\pi^2/4}\frac{ cos \sqrt x}{\sqrt x}dx = 2
Problem 4: Evaluate:\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\sqrt {cotx}}dx
Solution:
I =\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\sqrt {cotx}}dx
I =\int\limits^{\pi /3}_{\pi/6}\frac{1}{1+\frac{\sqrt {sinx}}{\sqrt{cosx}}}dx
I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx ----------(1)
Using property\int\limits^b_af(x)dx = \int\limits^b_af(a+b-x)dx
I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sin(\frac{\pi}{2}-x)}}{\sqrt{cos(\frac{\pi}{2}-x)}+\sqrt {sin(\frac{\pi}{2}-x)}}dx
I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx ----------(2)
Adding (1) and (2)
2I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}}{\sqrt{cosx}+\sqrt {sinx}}dx+\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx
2I =\int\limits^{\pi /3}_{\pi/6}\frac{\sqrt{sinx}+\sqrt{cosx}}{\sqrt{sinx}+\sqrt {cosx}}dx
2I =\int\limits^{\pi /3}_{\pi/6}1.dx
2I =\big[x\big]^{\pi/3}_{\pi/6}
2I = (π/3) - (π/6)
2I = π/6
I = π/12
Problem 5: Evaluate:\int\limits_0^{\pi/2} |sinx - cosx|dx
Solution:
The integral contains mod function.
sin x - cos x = 0
sin x = cos x
tan x = 1
x = π/4
Between limit 0 to π/4, |sinx - cosx| is negative and between π/4 to π/2, |sinx - cosx| is positive
So, we divide the above limits of integral using the following formula
\int\limits_a^b f(x)dx = \int\limits^c_af(x)dx + \int\limits^b_cf(x)dx
\int\limits_0^{\pi/2} |sinx - cosx|dx = \int\limits_{0}^{\pi/4} -(sinx - cosx)dx+\int\limits_{\pi/4}^{\pi/2} (sinx - cosx)dx
\int\limits_0^{\pi/2} |sinx - cosx|dx = \int\limits_{0}^{\pi/4} (cosx - sinx)dx+\int\limits_{\pi/4}^{\pi/2} (sinx - cosx)dx
\int\limits_0^{\pi/2} |sinx - cosx|dx = \big[sinx+cosx\big]_{0}^{\pi/4}+ \big[- cosx-sinx\big]_{\pi/4}^{\pi/2}
\int\limits_0^{\pi/2} |sinx - cosx|dx = \bigg(\frac{2}{\sqrt2}-1\bigg)+\bigg(\frac{2}{\sqrt2}-1\bigg)
\int\limits_0^{\pi/2} |sinx - cosx|dx = 2√2-2
\int\limits_0^{\pi/2} |sinx - cosx|dx = 2(√2-1)
Problem 6: Prove that:\int\limits_0^{\pi/2}\text{sin2x log tanx dx}=0
Solution:
Let I =\int\limits_0^{\pi/2}\text{sin2x log tanx dx} ---------(i)
⇒ I =\int\limits_0^{\pi/2}{sin2(\frac{\pi}{2}-x)} \text{log tan}(\frac{\pi}{2}-x) dx
Using property:\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx
⇒ I =\int\limits_0^{\pi/2}\text{sin2x log cotx dx} ---------(ii)
Adding (i) and (ii)
2I =\int\limits_0^{\pi/2}\text{sin2x (log tanx + log cotx)dx}
2I =\int\limits_0^{\pi/2}\text{sin2x (log tanx cotx)dx}
2I =\int\limits_0^{\pi/2}\text{sin2x (log 1)dx} =0
I = 0
Problem 7: Evaluate:\int\limits_{-\pi}^{\pi}|cosx|dx
Solution:
I =\int\limits_{-\pi}^{\pi}|cosx|dx
|cos x | is an even function
Using property:\int\limits_{-a}^{a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & \text{, if $n$ is even} \\ 0 & \text{, if $n$ is odd} \end{cases}
I = 2\int\limits_{0}^{\pi}|cosx|dx
I = 2 {\int\limits_{0}^{\pi/2}|cosx|dx+\int\limits_{\pi/2}^{\pi}|cosx|dx }
cos x is negative if π/2< x ≤ π
I = 2{\int\limits_{0}^{\pi/2}cosxdx+\int\limits_{\pi/2}^{\pi}-cosxdx }
I = 2{\big[sinx\big]_0^{\pi/2}-\big[sinx\big]_{\pi/2}^\pi }
I = 2 + 2 = 4
Problem 8: Evaluate:\int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx
Solution:
I =\int\limits_1^2\frac{\sqrt x}{\sqrt{3-x}\hspace{0.1cm}+\sqrt{x}}dx ----------(i)
Using property\int\limits_{a}^{b}f(x)dx = \int\limits_{a}^{b}f(a + b - x)dx
I =\int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{3-(3-x)}\hspace{0.1cm}+\sqrt{3-x}}dx
I =\int\limits_1^2\frac{\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx ---------(ii)
Adding (i) and (ii)
2I =\int\limits_1^2\frac{\sqrt {x}+\sqrt {3-x}}{\sqrt{x}\hspace{0.1cm}+\sqrt{3-x}}dx
2I =\int\limits_1^21.dx = \big[x\big]_1^2
2I = 2 - 1
2I = 1
I = 1/2
Problem 9: Show that:\int\limits_0^{\pi/2}f(sin2x)\ sinx\ dx = \sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx
Solution:
I =\int\limits_0^{\pi/2}f(sin2x)\ sinx\ dx
Using property\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx
I =\int\limits_0^{\pi/2}f\{sin 2(\frac{\pi}{2}-x)\}\ sin(\frac{\pi}{2}-x)\ dx
I =\int\limits_0^{\pi/2}f\{sin(\pi-2x)\}\ cosx\ dx
I =\int\limits_0^{\pi/2}f(sin2x)\ cosx\ dx
Adding (i) and (ii)
2I =\int\limits_0^{\pi/2}f(sin2x)\ (sinx+cos x)\ dx
Using property\bold{\int\limits_{0}^{2a}f(x)dx= \begin{cases} 2\int\limits_{0}^{a}f(x)dx & , if f(2a - x) = f(x) \\ 0 & , if f(2a - x) = -f(x)\end{cases}}
2I =2\int\limits_0^{\pi/4}f(sin2x)\ (sinx+cos x)\ dx
2I =2\sqrt{2}\int\limits_0^{\pi/4}f(sin 2x)\big( \frac{1}{\sqrt{2}}sinx + \frac{1}{\sqrt{2}}cosx\big)\ dx
2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin 2x\}\ sin(x+\frac{\pi}{4})\ dx
Using property\int\limits_{a}^{b}f(x)dx=\int\limits_{a}^{b}f(a+b-x)dx
2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin 2(\frac{\pi}{4}-x)\}\ sin(\frac{\pi}{4}-x+\frac{\pi}{4})\ dx
2I =2\sqrt{2}\int\limits_0^{\pi/4}f\{sin (\frac{\pi}{2}-2x)\}\ sin(\frac{\pi}{2}-x)\ dx
2I =2 \sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx
I =\sqrt{2}\int\limits_0^{\pi/4}f(cos 2x)\ cosx\ dx
Problem 10: Evaluate:\int\limits_0^{\pi/2}\frac{x\ sinx\ cosx}{sin^4x+cos^4x}dx
Solution:
I =\int\limits_0^{\pi/2}\frac{x\ sinx\ cosx}{sin^4x+cos^4x}dx
Using property\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx
I =\int\limits_0^{\pi/2}\frac{(\frac{\pi}{2}-x)\ sin(\frac{\pi}{2}-x)\ cos(\frac{\pi}{2}-x)}{sin^4(\frac{\pi}{2}-x)+cos^4(\frac{\pi}{2}-x)}dx
I =\int\limits_0^{\pi/2}\frac{(\frac{\pi}{2}-x)\ cosx\ sinx }{cos^4x+sin^4x}dx
Adding (i) and (ii)
2I =\frac{\pi}{2}\int\limits_0^{\pi/2}\frac{sinx\ cosx}{sin^4x+cos^4x}dx
Let t = sin2x ⇒ dt = 2sinx cosx dx
Limits: x= 0, t = 0 and x = π/2, t =1
2I =\frac{\pi}{4}\int\limits_0^{\pi/2}\frac{1}{(1-t)^2+t^2}dt
2I =\frac{\pi}{8}\int\limits_0^{\pi/2}\frac{1}{\big(t-\frac{1}{2}\big)^2+\big(\frac{1}{2}\big)^2}dt
2I =\frac{\pi}{8}\times2\big[tan^{-1}(2t-1)\big]^1_0
I =\frac{\pi}{8}\big(\frac{\pi}{4}+\frac{\pi}{4}\big)
I =\frac{{\pi}^2}{16}
Similar Reads
Maths Mathematics, often referred to as "math" for short. It is the study of numbers, quantities, shapes, structures, patterns, and relationships. It is a fundamental subject that explores the logical reasoning and systematic approach to solving problems. Mathematics is used extensively in various fields
5 min read
Basic Arithmetic
What are Numbers?Numbers are symbols we use to count, measure, and describe things. They are everywhere in our daily lives and help us understand and organize the world.Numbers are like tools that help us:Count how many things there are (e.g., 1 apple, 3 pencils).Measure things (e.g., 5 meters, 10 kilograms).Show or
15+ min read
Arithmetic OperationsArithmetic Operations are the basic mathematical operations—Addition, Subtraction, Multiplication, and Division—used for calculations. These operations form the foundation of mathematics and are essential in daily life, such as sharing items, calculating bills, solving time and work problems, and in
9 min read
Fractions - Definition, Types and ExamplesFractions are numerical expressions used to represent parts of a whole or ratios between quantities. They consist of a numerator (the top number), indicating how many parts are considered, and a denominator (the bottom number), showing the total number of equal parts the whole is divided into. For E
7 min read
What are Decimals?Decimals are numbers that use a decimal point to separate the whole number part from the fractional part. This system helps represent values between whole numbers, making it easier to express and measure smaller quantities. Each digit after the decimal point represents a specific place value, like t
10 min read
ExponentsExponents are a way to show that a number (base) is multiplied by itself many times. It's written as a small number (called the exponent) to the top right of the base number.Think of exponents as a shortcut for repeated multiplication:23 means 2 x 2 x 2 = 8 52 means 5 x 5 = 25So instead of writing t
9 min read
PercentageIn mathematics, a percentage is a figure or ratio that signifies a fraction out of 100, i.e., A fraction whose denominator is 100 is called a Percent. In all the fractions where the denominator is 100, we can remove the denominator and put the % sign.For example, the fraction 23/100 can be written a
5 min read
Algebra
Variable in MathsA variable is like a placeholder or a box that can hold different values. In math, it's often represented by a letter, like x or y. The value of a variable can change depending on the situation. For example, if you have the equation y = 2x + 3, the value of y depends on the value of x. So, if you ch
5 min read
Polynomials| Degree | Types | Properties and ExamplesPolynomials are mathematical expressions made up of variables (often represented by letters like x, y, etc.), constants (like numbers), and exponents (which are non-negative integers). These expressions are combined using addition, subtraction, and multiplication operations.A polynomial can have one
9 min read
CoefficientA coefficient is a number that multiplies a variable in a mathematical expression. It tells you how much of that variable you have. For example, in the term 5x, the coefficient is 5 — it means 5 times the variable x.Coefficients can be positive, negative, or zero. Algebraic EquationA coefficient is
8 min read
Algebraic IdentitiesAlgebraic Identities are fundamental equations in algebra where the left-hand side of the equation is always equal to the right-hand side, regardless of the values of the variables involved. These identities play a crucial role in simplifying algebraic computations and are essential for solving vari
14 min read
Properties of Algebraic OperationsAlgebraic operations are mathematical processes that involve the manipulation of numbers, variables, and symbols to produce new results or expressions. The basic algebraic operations are:Addition ( + ): The process of combining two or more numbers to get a sum. For example, 3 + 5 = 8.Subtraction (−)
3 min read
Geometry
Lines and AnglesLines and Angles are the basic terms used in geometry. They provide a base for understanding all the concepts of geometry. We define a line as a 1-D figure that can be extended to infinity in opposite directions, whereas an angle is defined as the opening created by joining two or more lines. An ang
9 min read
Geometric Shapes in MathsGeometric shapes are mathematical figures that represent the forms of objects in the real world. These shapes have defined boundaries, angles, and surfaces, and are fundamental to understanding geometry. Geometric shapes can be categorized into two main types based on their dimensions:2D Shapes (Two
2 min read
Area and Perimeter of Shapes | Formula and ExamplesArea and Perimeter are the two fundamental properties related to 2-dimensional shapes. Defining the size of the shape and the length of its boundary. By learning about the areas of 2D shapes, we can easily determine the surface areas of 3D bodies and the perimeter helps us to calculate the length of
10 min read
Surface Areas and VolumesSurface Area and Volume are two fundamental properties of a three-dimensional (3D) shape that help us understand and measure the space they occupy and their outer surfaces.Knowing how to determine surface area and volumes can be incredibly practical and handy in cases where you want to calculate the
10 min read
Points, Lines and PlanesPoints, Lines, and Planes are basic terms used in Geometry that have a specific meaning and are used to define the basis of geometry. We define a point as a location in 3-D or 2-D space that is represented using coordinates. We define a line as a geometrical figure that is extended in both direction
14 min read
Coordinate Axes and Coordinate Planes in 3D spaceIn a plane, we know that we need two mutually perpendicular lines to locate the position of a point. These lines are called coordinate axes of the plane and the plane is usually called the Cartesian plane. But in real life, we do not have such a plane. In real life, we need some extra information su
6 min read
Trigonometry & Vector Algebra
Trigonometric RatiosThere are three sides of a triangle Hypotenuse, Adjacent, and Opposite. The ratios between these sides based on the angle between them is called Trigonometric Ratio. The six trigonometric ratios are: sine (sin), cosine (cos), tangent (tan), cotangent (cot), cosecant (cosec), and secant (sec).As give
4 min read
Trigonometric Equations | Definition, Examples & How to SolveTrigonometric equations are mathematical expressions that involve trigonometric functions (such as sine, cosine, tangent, etc.) and are set equal to a value. The goal is to find the values of the variable (usually an angle) that satisfy the equation.For example, a simple trigonometric equation might
9 min read
Trigonometric IdentitiesTrigonometric identities play an important role in simplifying expressions and solving equations involving trigonometric functions. These identities, which include relationships between angles and sides of triangles, are widely used in fields like geometry, engineering, and physics. Some important t
10 min read
Trigonometric FunctionsTrigonometric Functions, often simply called trig functions, are mathematical functions that relate the angles of a right triangle to the ratios of the lengths of its sides.Trigonometric functions are the basic functions used in trigonometry and they are used for solving various types of problems in
6 min read
Inverse Trigonometric Functions | Definition, Formula, Types and Examples Inverse trigonometric functions are the inverse functions of basic trigonometric functions. In mathematics, inverse trigonometric functions are also known as arcus functions or anti-trigonometric functions. The inverse trigonometric functions are the inverse functions of basic trigonometric function
11 min read
Inverse Trigonometric IdentitiesInverse trigonometric functions are also known as arcus functions or anti-trigonometric functions. These functions are the inverse functions of basic trigonometric functions, i.e., sine, cosine, tangent, cosecant, secant, and cotangent. It is used to find the angles with any trigonometric ratio. Inv
9 min read
Calculus
Introduction to Differential CalculusDifferential calculus is a branch of calculus that deals with the study of rates of change of functions and the behaviour of these functions in response to infinitesimal changes in their independent variables.Some of the prerequisites for Differential Calculus include:Independent and Dependent Varia
6 min read
Limits in CalculusIn mathematics, a limit is a fundamental concept that describes the behaviour of a function or sequence as its input approaches a particular value. Limits are used in calculus to define derivatives, continuity, and integrals, and they are defined as the approaching value of the function with the inp
12 min read
Continuity of FunctionsContinuity of functions is an important unit of Calculus as it forms the base and it helps us further to prove whether a function is differentiable or not. A continuous function is a function which when drawn on a paper does not have a break. The continuity can also be proved using the concept of li
13 min read
DifferentiationDifferentiation in mathematics refers to the process of finding the derivative of a function, which involves determining the rate of change of a function with respect to its variables.In simple terms, it is a way of finding how things change. Imagine you're driving a car and looking at how your spee
2 min read
Differentiability of a Function | Class 12 MathsContinuity or continuous which means, "a function is continuous at its domain if its graph is a curve without breaks or jumps". A function is continuous at a point in its domain if its graph does not have breaks or jumps in the immediate neighborhood of the point. Continuity at a Point: A function f
11 min read
IntegrationIntegration, in simple terms, is a way to add up small pieces to find the total of something, especially when those pieces are changing or not uniform.Imagine you have a car driving along a road, and its speed changes over time. At some moments, it's going faster; at other moments, it's slower. If y
3 min read
Probability and Statistics
Basic Concepts of ProbabilityProbability is defined as the likelihood of the occurrence of any event. It is expressed as a number between 0 and 1, where 0 is the probability of an impossible event and 1 is the probability of a sure event.Concepts of Probability are used in various real life scenarios : Stock Market : Investors
7 min read
Bayes' TheoremBayes' Theorem is a mathematical formula used to determine the conditional probability of an event based on prior knowledge and new evidence. It adjusts probabilities when new information comes in and helps make better decisions in uncertain situations.Bayes' Theorem helps us update probabilities ba
13 min read
Probability Distribution - Function, Formula, TableA probability distribution is a mathematical function or rule that describes how the probabilities of different outcomes are assigned to the possible values of a random variable. It provides a way of modeling the likelihood of each outcome in a random experiment.While a Frequency Distribution shows
13 min read
Descriptive StatisticStatistics is the foundation of data science. Descriptive statistics are simple tools that help us understand and summarize data. They show the basic features of a dataset, like the average, highest and lowest values and how spread out the numbers are. It's the first step in making sense of informat
5 min read
What is Inferential Statistics?Inferential statistics is an important tool that allows us to make predictions and conclusions about a population based on sample data. Unlike descriptive statistics, which only summarize data, inferential statistics let us test hypotheses, make estimates, and measure the uncertainty about our predi
7 min read
Measures of Central Tendency in StatisticsCentral tendencies in statistics are numerical values that represent the middle or typical value of a dataset. Also known as averages, they provide a summary of the entire data, making it easier to understand the overall pattern or behavior. These values are useful because they capture the essence o
11 min read
Set TheorySet theory is a branch of mathematics that deals with collections of objects, called sets. A set is simply a collection of distinct elements, such as numbers, letters, or even everyday objects, that share a common property or rule.Example of SetsSome examples of sets include:A set of fruits: {apple,
3 min read
Practice