Calculus Formula: Limits, Differentials & Integrals

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Shwetha S

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Calculus is a branch of mathematics that deals with the study of the “rate of change”. It is applied in solving equations. Calculus is also referred to as infinitesimal calculus or infinite calculus. The two major branches of calculus are differential calculus and integral calculus. Differential calculus is concerned with rates of change and slopes of curves, whereas Integral calculus is the branch dealing with the accumulation of quantities and the areas under and between curves. Using differential calculus, it is possible to determine velocity and acceleration in mechanics from the position function. 

Read More: Integration 

Key Terms: Calculus, Calculus Formula, Differential Calculus, Integral Calculus, Limits, Derivatives, Integration, Differentiation, Function, Integration


What is Calculus Formula?

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Basic calculus consists of rules and differentiation formulas. This is the method that is used to calculate the derivative of a function. Integration is the method used to calculate the anti-derivative of a function. Calculus formulas describe the rate of change of a function for the given input value using the derivative of a function or differentiation formula. The process of finding the derivative of any given function is called differentiation.

Calculus

Calculus

Read More: Double Integral

Integrals Detailed Video Explanation


Limits Formula

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Limits of functions at a point are the common and coincidence value of the left and right-handed limits. It is denoted as:

\(\lim_{x \to a} f(x) = A\)
where f(x) is a function of x.  

This formula means that,

If at a point x = a the function f(x) takes an indeterminate form, then the values of f(x) which is close to a are considered.  If the value tends to a definite value as x tends to a, then the value obtained is called the Minit of f(x) at x=a.

  • The expected value of the function f(x) shown by the points to the left of point ‘a’ is the left-hand limit of the function at that point. It is denoted as:
\(\lim_{x \to a_-} f(x) = A\)
  • The point to the right of the point, that is, ‘a’, which generally shows the value of the function is the right-hand limit of the function at that point. It can be denoted as:
\(\lim_{x \to a_+} f(x) = A\)

Note: The value of a limit of a function f(x) at a point, that is, f(a) may vary from the value of f(x) at the point ‘a’.

Given below is the list of formulae for calculating limits:

  1. \(\lim\limits_{x \to 0}\)\(\frac{sinx}{x}\) = \(\lim\limits_{x \to 0}\)\(\frac{tanx}{x}\) = 1
  2. \(\lim\limits_{x \to 0}\)\(\frac{sin^{-1}x}{x}\)\(\lim\limits_{x \to 0}\)\(\frac{tan^{-1}x}{x}\)= 1
  3. \(\lim\limits_{x \to 0}\)\(\frac{1n(1+x)}{x}\)= 1
  4. \(\lim\limits_{x \to 0}\) \(\frac{a^x -1}{x}\)= 1na
  5. \(\lim\limits_{x \to 0}\)\(\frac{e^x -1}{x}\)= 1
  6. \(\lim\limits_{x \to 0}\)\(\frac{x^n - a^n}{(x-a)} = n.a^{n-1}\)
  7. \(\lim\limits_{x \to 0}\)(1 + x)\(^{\frac{1}{x}}\)\(\lim\limits_{x \to 0}\)\((1 + \frac{1}{x})^x\)= c
  8. \(\lim\limits_{x \to 0}\)\(\frac{(1+x)^m -1}{x} = m\)

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Properties of Limits

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The following figure summarises the properties of limits used for calculations:

If \(\lim\limits_{x \to a}\) f(x) = m and \(\lim\limits_{x \to a}\) g(x) = n

where ‘m’ and ’n’ are real and finite the

(i) \(\lim\limits_{x \to a}\) [f(x) = g(x)] = \(\lim\limits_{x \to a}\) f(x) = \(\lim\limits_{x \to a}\) g(x) = m ± n

(ii) \(\lim\limits_{x \to a}\) [cf(x)] = c . \(\lim\limits_{x \to a}\) f(x) = c.m

(iii) \(\lim\limits_{x \to a}\) (f(x) . g(x)) = \(\lim\limits_{x \to a}\) f(x) . \(\lim\limits_{x \to a}\) g(x) = m . n

(iv) \(\lim\limits_{x \to a}\)\(\frac{f(x)}{g(x)}\) = \(\frac{ \lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} g(x)}\) = \(\frac{m}{n}\)provided n ≠ 0

In the figure, f(x), and g(x) are functions, and ‘m’ and ‘n’ is real and finite constants. Properties of multiplication, division, addition or subtraction, and multiplication by a constant are provided above. 

Read More: Increasing & Decreasing Functions


Differential Calculus Formula

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Differential calculus can be defined as the study of the definition, properties, and applications of the derivative of a function. The process of finding the derivative is called differentiation. This involves the derivative equation that describes the rate at which a change occurs in a given function. Minimum or maximum values of given functions for optimization problems are one the prominent applications of differential calculus.

Given below is the list of differential calculus formulae that can be used to find the derivation of a function:

  • \(\frac{d}{dx}\)xn = nxn-1
  • \(\frac{d}{dx}\) (fg) = fg’ + gf’
  • \(\frac{d}{dx}\)\((\frac{f}{g}) = \frac{gf'-fg'}{g^2}\)
  • \(\frac{d}{dx}\)f(g(x)) = f’ (g(x))g’(x)
  • \(\frac{d}{dx}\)(sin x) = cos x
  • \(\frac{d}{dx}\)(cos x) = – sin x
  • \(\frac{d}{dx}\)(tan x) = sec2
  • \(\frac{d}{dx}\) (cot x) = – csc2
  • \(\frac{d}{dx}\) (sec x) = sec x tan x
  • \(\frac{d}{dx}\) (csc x) = – csc x cot x
  • \(\frac{d}{dx}\) (ex) = ex
  • \(\frac{d}{dx}\)(ax) = ax 1n a
  • \(\frac{d}{dx}\) 1n x = \(\frac{1}{x}\)
  • \(\frac{d}{dx}\) (arc sin x) = \(\frac{1}{\sqrt{1 - x^2}}\)
  • \(\frac{d}{dx}\) (arc tan x) =\(\frac{1}{1 + x^2}\)

Read More: Multivariable Calculus


Integral Calculus Formula

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Integral Calculus can be defined as the branch of calculus that is concerned with Integrals, the accumulation of quantities, and the areas under and between curves and their properties and their properties.

Given below is a list of Integral Calculus Formulae that can be used to find the integral of a function:

Function Formula
∫ xn dx (xn+1/n+1) + C, where n ≠ -1
∫ sin x dx - cos x + C
∫ cos x dx sin x + C
∫ sec2 x dx tan x + C
∫ cosec2 x dx -cot x + C
∫ sec x tan x dx sec x + C
∫ cosec x cot x dx -cosec x +C
∫ ex dx ex + C
∫ 1/x dx ln x+ C
∫11+x211+x2 dx tan x +C
∫ ax dx ln(ax) a + C

Read More: Properties of Definite Integral


Differential Calculus vs Integral Calculus

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Differential calculus utilizes differentiation to find the derivative of the function whereas integral calculus employs integration to find the integral of the function. Integration is the reverse of differentiation. It is also known as anti-derivative. Some of the differences between differential calculus and integral calculus are tabulated below: 

Differential Calculus Integral Calculus 
The instantaneous rate of change of a function is determined in differential calculus using derivatives. The area under a curve is determined using integral calculus using integrals.
Differentiation is the method of dividing complex functions into smaller pieces to monitor changes. The integration adds up infinitesimal components to determine the total area under a curve.
It helps determine whether the function is increasing or decreasing.  It helps determine volume, area, and central point. 
It is the opposite of integration It is the opposite of differentiation.  

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Things to Remember

  • Calculus formulae describe the rate of change of a function for the given input value using the derivative of a function or differentiation formula.
  • The process of finding the derivative of any given function is known as differentiation.
  • Rules and differentiation formulas help to calculate the derivative of a function and integration.
  • Calculus is a branch of mathematics that focuses on limits, functions, derivatives, integrals, and mainly infinite series.
  • Integration is also known as anti-derivative.
  • Integration determines volume, area, and central point. 

Previous Year Questions

  1. The value of sin⁡51+ sin⁡2 39∘ is… [KCET – 2020]
  2. If cos x = |sin x| then, the general solution is… [KCET – 2019]
  3. If 0≤ x< π​/2, then the number of values of x… [JEE Main – 2019]
  4. A vertical lamp-post at the midpoint D… [JEE Main – 2019]
  5. If tanA+cotA=2, then the value of…  [KCET – 2020]
  6. The value of cos245−sin215 is… [KCET – 2017]
  7. A, B and C are the angles opposite to the corresponding sides of lengths… [JKCET – 2017]
  8. The value of tan 8/π​ is equal to… [KCET – 2016]
  9. The value of tan⁡10∘ tan⁡20∘ tan⁡30∘ tan⁡40∘ tan⁡50∘ tan⁡60 [COMEDK UGET – 2012]
  10. A value of θ satisfying sin⁡5θ−sin⁡3θ+sinθ…  [KCET – 2011]

Sample Questions

Ques. Solve the Following Definite Integral ∫42 x2 dx : (2 Marks)

Ans. 4 2 x2 dx

42 x2 dx = [x3/3]42

= 64/3 – 8/3

= 56/3

= 18.66

Ques. What is the instantaneous velocity at time t = π/2 of a particle whose positional equation is represented by s(t) = 12tan(t/2 + π)? (5 Marks)

Ans. The instantaneous velocity is represented by the first derivative of the positional equation.

v(t) = s'(t) = 12 * (1/2) sec2(t/2 + π)

= 6sec2(t/2 + π) = 6/cos2(t/2 + π)

= 6/((–1)2 cos2(t/2)) = 6/cos2(t/2)

Based on the nature of the cosine, know that

6/cos2(t/2 + π) = 6/((–1)2 cos2(t/2)) = 6/cos2(t/2)

v(π/2) = 6/cos2(t/2)

= 6/((1/√2)2)

= 6/(1/2)

= 12

Ques. What is the instantaneous velocity at time t=π/2 of a particle whose positional equation is represented by s(t) = 12cos2(t/2 + π)? (5 Marks)

Ans. The instantaneous velocity is represented by the first derivative of the positional equation. This is found by using the chain rule both on the square of the cosine function and the function itself.

v(t) = s'(t) = 12 * 2 cos(t/2 + π) * (–sin(t/2 + π)) * (1/2)

= –12cos(t/2 + π)sin(t/2 + π)

Given, 

cos(t/2 + π) = –cos(t/2) and sin(t/2 + π) = –sin(t/2)

Therefore, v(t) = –12(–sin(t/2))(–cos(t/2)) = –12sin(t/2)cos(t/2)

v(π/2) = –12sin(π/4)cos(π/4)

= –12(1/√2)(1/√2)

= –12 x (1/2)

= –6

Ques. The position s of a particle at time t is given by s(t)=3t2−2t. What is the particle's velocity at time t=3. (1 mark)

Ans. The velocity function is given by the derivative of the position function. So here v=6t−2. Plugging 3 in for t gives 16.

Ques. Find the velocity function if the position function is given as: s(t)=3t2+3t+1. (3 Marks)

Ans. There are three terms in this problem that has to be derived. The derivative of the position function, or the velocity function, represents the slope of the position function.

The derivative of 3x2 can be solved by using the power rule, which is:

2⋅3x(2−1)

Therefore. the derivative of 3x2 is 6x.

The derivative of 3x is 3 by using the constant multiple rules.

The derivative of 1 is 0 since the derivatives of constants are equal to zero.

v(t)=6t+3

Ques.  Find the velocity of a function if the acceleration is: a(t)=4t+6. (2 Marks)

Ans. To find the velocity given the acceleration function, we will need to integrate the acceleration function.

∫a(t)dt=∫(4t+6)dt=2t2+6t

Ques. Find the velocity at t=2 if the acceleration function is: a(t)=1. (2 Marks)

Ans. The velocity function can be obtained by integrating the acceleration function.

v(t)=∫a(t)dt=∫1dt=t

Since we are finding the velocity at t=2, substitute this into the velocity function.

v(2)=2

Ques. Find the derivative of x(sin x) using differential calculus. (2 Marks)

Ans. y = x(sin x)
 dy / dx = f'(x)g(x) + f(x) g'(x)
f(x) = x, g(x) = sin x
dy / dx = sin x + xcos x

Derivative of x(sin x) is sin x + x(cos x)

Ques. Mention the basic calculus. (1 Mark)

Ans. There are two main branches of calculus. They are Differential Calculus and Integral Calculus.

Ques. What are the applications of differential calculus? (2 Marks)

Ans. Differential calculus has many diverse applications. Differential calculus is used in the majority of quantitative disciplines, including differential geometry, functional analysis, abstract algebra, and complex analysis. Differential calculus is used in biology to analyze the temporal evolution of the population of predators and prey. 

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CBSE CLASS XII Related Questions

  • 1.

    At a birthday party, children are being served orange juice in conical cups, as shown in the figure. 


    Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
    On the basis of the above information, answer the following questions :


      • 2.

        A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. 


          • 3.
            Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).


              • 4.
                If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


                  • 5.
                    Find:

                    The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]


                      • 6.
                        Find:

                        The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

                          • \(-\frac{\pi}{2}\)
                          • \(-\frac{\pi}{4}\)
                          • \(\frac{\pi}{4}\)
                          • \(\frac{\pi}{2}\)
                        CBSE CLASS XII Previous Year Papers

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