750 Learners
Students often get anxious about the idea of "calculus". It is a branch of mathematics dealing with the constant change of quantities over a period of time. Differential and Integral Calculus are the two main branches of calculus.
Share Post:
Trustpilot | Rated 4.7
1,292 reviews
Calculus is all about understanding how things change over time. It focuses on derivatives and integrals, which essentially involve adding up tiny changes step by step. Derivatives help us know how one quantity changes in relation to another. At the same time, integrals enable us to measure the total effect of those changes, such as finding the area under a curve or the distance traveled.
Ancient Times
In mathematics, the foundation of calculus dates back thousands of years. In the fifth century BC, Democritus worked with ideas based on infinitesimals during the ancient Greek period. If backed by a geometric proof, the Greeks would consider a theorem accurate.
Greek philosophers also viewed ideas based on infinitesimals as paradoxes, as it is always possible to divide a number again, no matter how small it becomes. In the 3rd century BC, Archimedes developed the method of exhaustion, which is used for calculating the area of circles.
The Middle East and India
Like many other areas of science and math, the growth of calculus slowed down in the Western world during the Middle Ages. At the same time, on the other side of the world, people were discovering and studying both integrals and derivatives. An Arab mathematician, Ibn al-Haytham, used formulas he created to find the volume of a paraboloid. A solid shape made by the spinning part of a parabola (a curve) around a line.
During this time, ideas moved between the Middle East and India, since some of them were also found in the Kerala School of Astronomy and Mathematics. By the start of the fourteenth century, many pieces of calculus were already known, but differentiation and integration had not yet been joined together as one subject.
The Fourteenth Century
In the 14th century, a group of mathematical scholars revived the study of mathematics. The group was known as the Oxford Calculators. They explored philosophical questions through the lens of math.
The Seventeenth Century
The early seventeenth century was the most critical time in the history of calculus. During this period, René Descartes created analytical geometry, and Pierre de Fermat studied the highest and lowest points of curves, as well as their tangents. Some of Fermat’s formulas are similar to the ones we still use today, even after almost 400 years.
The Eighteenth Century and Beyond
The debate surrounding the invention of calculus became more and more heated as time wore on, with Newton’s supporters openly accusing Leibniz of copying. A British firm still claims that calculus was Newton’s discovery.
Based on the change and mathematical analysis, calculus can be divided into four parts.
The integral is the inverse process of the differential. The fundamental theorem of calculus links derivatives and integrals. There are two theorems that are the first and the second fundamental theorems of calculus.
First fundamental theorem of integral calculus
The first part of the calculus theorem is often called the first fundamental theorem of calculus. It states that one of the antiderivatives, which is also known as the indefinite integral of a function \(f\), can be found by integrating \(f\) with a variable as the upper limit of integration. This shows that every continuous function has an antiderivative.
Statement: Let f be a continuous function on the closed interval [a, b] and let A(x) be the area function.
Then \(A'(x)=f(x)\), for all \(x ∈ [a, b]\).
\(\begin{array}{l}F(x) = \int_{a}^{x} f(t) dt\end{array}\)
Then F is uniformly continuous on the closed interval [a, b] and differentiable on the open interval (a, b).
\(F'(x) = f(x) ∀ x ∈(a, b)\)
Here, \(F'(x)\) is a derivative function of \(F(x)\).
Second fundamental theorem of integral calculus
The second fundamental theorem of calculus states that if a function \(f\) is continuous on the closed interval [a, b] and \(F\) is an antiderivative, that is an indefinite integral of \(f\) on [a, b], then it can be expressed as follows:
\(F(b)- F(a) = a∫b f(x) dx\)
On the right-hand side, the integral of \(f(x)\) with respect to \(x\) is shown, where \(f(x)\) is the integrand, \(dx\) is the differential, a is the upper limit, and b is the lower limit.
A definite integral of a function always has a unique value and can be interpreted as the limit of a sum. If the function has an antiderivative \(F\) on the interval [a, b], then the definite integral is given by the difference \(F(b) - F(a)\).
In calculus, for each function, we have different formulas. And students should remember and use the correct formula. For a better understanding of calculus, it's important to remember all the basic formulas.
\( \frac{d}{dx}(x^n) = n x^{n-1} \)
\( \frac{d}{dx}(C) = 0 \quad \text{where $C$ is a constant} \)
\( \frac{d}{dx}(C f) = C \frac{d}{dx}(f) \)
\( \frac{d}{dx}(f \pm g) = \frac{d}{dx}(f) \pm \frac{d}{dx}(g) \)
\( \frac{d}{dx}(f g) = f' g + f g' \)\( \frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g \frac{d}{dx}(f) - f \frac{d}{dx}(g)}{g^2} \)
\( \frac{d}{dx}(\sin x) = \cos x \)
\( \frac{d}{dx}(\cos x) = -\sin x \)
\( \frac{d}{dx}(\tan x) = \sec^2 x \)
\( \frac{d}{dx}(\cot x) = -\csc^2 x \)
\( \frac{d}{dx}(\sec x) = \sec x \tan x \)
\( \frac{d}{dx}(\csc x) = -\csc x \cot x \)
\( \frac{d}{dx}(a^x) = a^x \ln a \)
\( \frac{d}{dx}(e^x) = e^x \)
\( \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \)
\( \begin{aligned} &\int 1 \, dx = x + C \\ &\int a \, dx = a x + C \\ &\int \frac{1}{x} \, dx = \ln|x| + C \\ &\int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1} x + C \\ &\int \frac{1}{\sqrt{1 + x^2}} \, dx = \tan^{-1} x + C \\ &\int \frac{1}{|x| \sqrt{x^2 - 1}} \, dx = \sec^{-1} x + C \\ &\int \sin x \, dx = -\cos x + C \\ &\int \cos x \, dx = \sin x + C \\ &\int \sec x \tan x \, dx = \sec x + C \\ &\int \csc x \cot x \, dx = -\csc x + C \\ &\int \sec^2 x \, dx = \tan x + C \\ &\int \csc^2 x \, dx = -\cot x + C \\ &\int e^x \, dx = e^x + C \\ &\int a^x \, dx = \frac{a^x}{\ln a} + C, \quad a>0, a\neq 1 \\ &\int \log x \, dx = x \ln x - x + C \\ &\int \log_a x \, dx = \frac{x \ln x - x}{\ln a} + C \end{aligned} \)
Calculus is a broad topic that includes limits, derivatives, integrals, and many more. Let’s learn some tips and tricks to make it easier for us to learn.
Solve calculus problems in different ways - Performance in math is often based on solving problems. However, with calculus, it is also important to understand the underlying principles behind each solution.
Seek visual aids - For better understanding of the concept, use simple, compelling, and effective visualizations from other sources like YouTube. These are highly revered for their clear, computer generated visuals.
Keep a dictionary of calculus notations and terms - This is the best way to keep track of all the definitions and notations.
Don’t hesitate to ask for help - When you run into trouble while understanding a part of a problem or concept, reach out to your teacher for help.
Mistakes are common in calculus, moreover, students repeat the same mistakes. To master calculus, let’s learn some common mistakes.
We have talked a lot about calculus. We know that we use it in our daily life. Now let’s see the applications of calculus.
Physics: In Physics, calculus is used to calculate motion, force, change, energy, and many more. The speed of a vehicle, rocket, or even the light can be calculated.
Engineering: From building bridges to developing circuits, we use calculus in different fields of engineering.
Economics: Calculus is used to study the market and its behavior. To understand the minimum cost and maximum profit.
Medicine: To study the growth of bacteria and infectious diseases, calculus is used. This data is used to treat cancer and diagnose patients.
Find the derivatives of 3x² + 2x + 1.
The derivatives of 3x2 + 2x + 1 = 6x + 2.
Apply the differentiation rule to find the derivatives of 3x2 + 2x + 1,
Differentiation rule: \( \frac{d}{dx}(f + g) = \frac{d}{dx}(f) + \frac{d}{dx}(g) \)
\( \frac{d}{dx}(3x^2 + 2x + 1) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(1) \)
Derivatives of x2 = 2x
Derivatives of 2x = 2
Derivatives of 1 = 0
Add the derivatives: \( \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x) + \frac{d}{dx}(1) \)
= 3 × 2x + 2 + 0
= 6x + 2.
Therefore, the derivatives of 3x2 + 2x + 1 = 6x + 2.
Calculate ∫2x dx.
\(∫2x dx = x2 + c\)
The power rule of integration \(= ∫xn dx = x(n + 1) / (n + 1) + c\)
\(∫2x dx = 2(x1+1 / 1 + 1) + c\)
= 2(x2 / 2) + c
= x2 + c
Therefore, \(∫2x dx = x2 + c\)
Solve ∫(3x² - 4x + 5)dx.
\(∫(3x2 - 4x + 5)dx = x3 - 2x2 + x + c\)
According to the properties of integral,
\( \int \big(f(x) \pm g(x)\big) \, dx = \int f(x) \, dx \pm \int g(x) \, dx \)
\( \int (3x^2 - 4x + 5) \, dx = \int 3x^2 \, dx - \int 4x \, dx + \int 5 \, dx \)
Using the power rule, \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1 \)
\( \int 3x^2 \, dx = x^3, \quad \int -4x \, dx = -2x^2, \quad \int 5 \, dx = 5x \)
Adding these values to \( \int 3x^2 \, dx - \int 4x \, dx + \int 5 \, dx \)\( \int (3x^2 - 4x + 5) \, dx = x^3 - 2x^2 + 5x + C \)
where ‘c’ is the constant.
Solve ∫(2x² - 3x - 7)dx.
\( \int (2x^2 + 3x - 7) \, dx = \frac{2x^3}{3} + \frac{3x^2}{2} - 7x + C \)
Here, split the integral using linearity
\( \int 2x^2 \, dx + \int 3x \, dx - \int 7 \, dx \)
Now, integrate each term:
\( \int 2x^2 \, dx = \frac{2x^3}{3}, \quad \int 3x \, dx = \frac{3x^2}{2}, \quad \int 7 \, dx = 7x \)
Then, combine results.
\( \int (2x^2 + 3x - 7) \, dx = \frac{2x^3}{3} + \frac{3x^2}{2} - 7x + C \)
Solve ∫(4x² - 5x - 6)dx.
\( \int (4x^2 + 5x - 6) \, dx = \frac{4x^3}{3} + \frac{5x^2}{2} - 6x + C \)
Split the integral using linearity
\( \int 4x^2 \, dx + \int 5x \, dx - \int 6 \, dx \)
Now, integrate each term
\( \int 4x^2 \, dx = \frac{4x^3}{3}, \quad \int 5x \, dx = \frac{5x^2}{2}, \quad \int 6 \, dx = 6x \)
Now, combine the results,
\( \int (4x^2 + 5x - 6) \, dx = \frac{4x^3}{3} + \frac{5x^2}{2} - 6x + C \)
From Numbers to Geometry and beyond, you can explore all the important Math topics by selecting from the list below:
Numbers | Multiplication Tables |
Geometry | Algebra |
Measurement | Trigonometry |
Commercial Math | Data |
Math Formulas | Math Questions |
Math Calculators | Math Worksheets |
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.