Calculus Tutorial

Calculus is the study of the changes taking place in a system over time. Mathematicians, scientists, and engineers use calculus in their work to make predictions based on the information they have. Calculus allows us to calculate change, even when it's not constant, which is usually the case in real-life situations. It is a very practical branch of mathematics and finds application in fields as diverse as science and technology, economics and statistics, and medicine and demography. There are two branches of calculus- differential calculus and integral calculus.

 Basics of Calculus

Calculus includes a lot of material from algebra and trigonometry, which is why calculus tutors insist on students having a thorough understanding of these subjects. Calculus is useful in problems where just algebraic knowledge will not suffice. Any situation which involves a change occurring at a non-constant rate employs calculus principles to find the answer.

Solved Example

Question: Solve $\int_{1}^{2}x^5dx$
$\int_{1}^{2}x^5dx$ = [$\frac{x^6}{6}]_{1}^{2}$

[ $\int x^n$ = $\frac{x^{n +1}}{n + 1}$]

= $\frac{1}{6}$ [$2^6 - 1^6$]

= $\frac{1}{6}$ [64 - 1]

= $\frac{1}{6}$ * 63

= 10.5

Calculus Help

Learning calculus is simple; most of the problems can be solved using a particular set of steps. Students who are serious about pursuing science, math, or technology related courses need a good working knowledge of calculus and the best way to learn the subject is to be thorough with all the formulae and equations, right from day one.

Solved Examples

Question 1: Evaluate $\int_{0}^{1}$ $\frac{(tan^{-1}z)^2}{1 + z^2}$ dz
Given:  $\int_{0}^{1}$ $\frac{(tan^{-1}z)^2}{1 + z^2}$ dz

Substitution Method:

Put $tan^{-1}z$ = t

By differentiating, we get

$\frac{1}{1 + z^2}$ dz = dt

When z = 0, t = $tan^{-1}$0 = 0

When z = 1, t = $tan^{-1}$1 = $\frac{\pi}{4}$


$\int_{0}^{\frac{\pi}{4}}$ $t^2 dt$ = [$\frac{t^3}{3}]_{0}^{\frac{\pi}{4}}$

= $\frac{1}{3}$[($\frac{\pi}{4})^3$]

= $\frac{\pi^3}{192}$

Question 2: Solve the differential equation $\frac{dy}{dx}$ + $\frac{y}{x}$ = sin x
Given differential equation is $\frac{dy}{dx}$ + $\frac{y}{x}$ = sin x

Step 1:
The general differential equation $\frac{dy}{dx}$ + Py = Q

Integrating factor (IF) = $e^{\int Pdx}$

Then the solution is y * IF = $\int Q * IF$ $dx$

By comparing given and general equation, we have

P = $\frac{1}{x}$ and Q = sin x

Step 2: Find IF

IF = $e^{\int (\frac{1}{x})dx}$ = $e^{log\ x}$ = x

Step 3

The solution for the given differential equation is

y * x = $\int sin x * x$ $dx$

= -x cos x + sin x + c  (Integrating by parts)

=> y = - cos x + $\frac{sin\ x}{x}$ + $\frac{c}{x}$.