INTEGRAL CALCULUS

  INTEGRAL CALCULUS

Integral Calculus is the branch of calculus where we study integrals and their properties. Integration is an essential concept which is the inverse process of differentiation. Both integral and differential calculus are related to each other by the fundamental theorem of calculus. 

If we know the f’ of a function that is differentiable in its domain, we can then calculate f. In differential calculus, we used to call f’, the derivative of the function f. In integral calculus, we call f the anti-derivative or primitive of the function f’. And the process of finding the anti-derivatives is known as anti-differentiation or integration.

Integration can be classified into two different categories, namely,

• Definite Integral
• Indefinite Integral

e name suggests, it is the inverse of finding differentiation.

Definite Integral

An integral that contains the upper and lower limits (i.e.) start and end value is known as a definite integral. The value of x is restricted to lie on a real line, and a definite Integral is also called a Riemann Integral when it is bound to lie on the real line.

A definite Integral is represented as:

$\begin{array}{l}{\int }_a^{b}f(x)dx\end{array}$

Learn more about definite integrals here.

Indefinite Integral

Indefinite integrals are not defined using the upper and lower limits. The indefinite integrals represent the family of the given function whose derivatives are f, and it returns a function of the independent variable.

The integration of a function f(x) is given by F(x) and it is represented by:

∫f(x) dx = F(x) + C 

where R.H.S. of the equation means integral off(x) with respect to x

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is called the constant of integration.

x is the variable of integration.

Click here to understand more about indefinite integral.

It may seem strange that there exist an infinite number of anti-derivatives for a function f. Taking an example will clarify it. Let us take f’ (x) = 3x2. By hit and trial, we can find out that its anti-derivative is F(x) = x3. This is because if you differentiate F with respect to x, you will get 3x2. There is only one function that we got as the anti-derivative of f. Let us now differentiate G(x)= x3 + 9 with respect to x. Again we would get the same derivative, i.e. f. This gives us an important insight. Since the differentiation of all the constants is zero, we can write any constant with x3, and the derivative would still be equal to f. So, there are infinite constants that can be substituted for c in the equation F(x) = x3 + C. Hence, there are infinite functions whose derivative is equal to f. And hence, there are infinite functions whose derivative is equal to 3x2. C is called an arbitrary constant. It is sometimes also referred to as the constant of integration.

☛ Also Check:

• Integration of uv formula
• Definite integral formula

 

THIS IS A BRAVE ATTEMPT TO HELP ALL STUDENTS GET QUALITY STUDY MATERIAL FOR FREE.HOPE YOU APPRECIATE   

Integral Calculus Examples

1. Example 1. Find the integral of e^3x

   Solution:

   ∫ d/dx(f(x)) = ∫ d/dx( e^3x)

   We know this is of the form of integral, ∫ d/dx( e^ax) = 1/a e^ax + C

   ∫ d/dx( e^3x) = 1/3 e^3x + C

   Answer: The integral of e^3x = 1/3 e^3x + C

2. Example 2. Find the integral of cos 3x.

   Solution:

   ∫ d/dx(f(x)) =∫ cos 3x

   Let 3x = t

   thus x = t/3

   dx = dt/3

   The given integral becomes ∫1/3(cos t) dt

   = 1/3(sin t) + C

   = 1/3 sin (3x) + C

   Answer: The integral of cos 3x = 1/3 sin (3x) + C

3. Example 3. Evaluate the integral i = $\underset{2}{\overset{3}{\int }}$ (x+1) dx

   Solution:

   By the 2nd theorem of fundamentals of integrals we know that $\underset{a}{\overset{b}{\int }}F\left(x\right)dx=f\left(b\right)-f\left(a\right)$

   $\underset{2}{\overset{3}{\int }}$ (x+1) dx = f(3) -f(2)

   f(x) = x²/2 + x + C

   f(3) = 3²/2 +3 = 9/2 + 3 = 15/2

   f(2)= 2²/2 + 2 = 4/2 + 2 = 4

   f(3) -f(2) = 15/2 - 4

   = 7/2

   Answer: The value of the given integral I = 7/2

A Practical Example: Tap and Tank

1. 

   Let us use a tap to fill a tank.

   The input (before integration) is the flow rate from the tap.

   We can integrate that flow (add up all the little bits of water) to give us the volume of water in the tank.

   Imagine a Constant Flow Rate of 1:

   

   With a flow rate of 1, the tank volume increases by x. That is Integration!

   An integral of 1 is x

   With a flow rate of 1 litre per second, the volume increases by 1 litre every second, so would increase by 10 litres after 10 seconds, 60 litres after 60 seconds, etc.

   The flow rate stays at 1, and the volume increases by x

   And it works the other way too:

   If the tank volume increases by x, then the flow rate must be 1.
   

   The derivative of x is 1
   

   This shows that integrals and derivatives are opposites!

   

    

Now For An Increasing Flow Rate

1. Imagine the flow starts at 0 and gradually increases (maybe a motor is slowly opening the tap):

   

   As the flow rate increases, the tank fills up faster and faster:

   • Integration: With a flow rate of 2x, the tank volume increases by x²
   • Derivative: If the tank volume increases by x², then the flow rate must be 2x

   

   We can write it down this way:

   The integral of the flow rate 2x tells us the volume of water:		∫2x dx = x2 + C
   The derivative of the volume x2+C gives us back the flow rate:		ddx(x2 + C) = 2x

    

   

   And hey, we even get a nice explanation of that "C" value ... maybe the tank already has water in it!

   • The flow still increases the volume by the same amount
   • And the increase in volume can give us back the flow rate.

   Which teaches us to always remember "+C".

   
   

Other functions

1. How do we integrate other functions?

   If we are lucky enough to find the function on the result side of a derivative, then (knowing that derivatives and integrals are opposites) we have an answer. But remember to add C.

Example: what is ∫cos(x) dx ?

1. 

   From the Rules of Derivatives table we see the derivative of sin(x) is cos(x) so:

   ∫cos(x) dx = sin(x) + C

   But a lot of this "reversing" has already been done (see Rules of Integration).

Example: What is ∫x³ dx ?

1. On Rules of Integration there is a "Power Rule" that says:

   ∫xⁿ dx = xⁿ⁺¹n+1 + C

   We can use that rule with n=3:

   ∫x³ dx = x⁴4 + C

   Knowing how to use those rules is the key to being good at Integration.

   So learn the rules and get lots of practice.

   Learn the Rules of Integration and Practice! Practice! Practice!
   (there are some questions below to get you started)

Definite vs Indefinite Integrals

1. We have been doing Indefinite Integrals so far.

   A Definite Integral has actual values to calculate between (they are put at the bottom and top of the "S"):

   Indefinite Integral		Definite Integral

What is Integral Calculus Used for?

1. Calculus is a branch of Mathematics that determines how matter, particles, and celestial bodies move. The rate of change in real-time is calculated using calculus. Isaac Newton applied calculus directly to measuring physical structures.

   Mathematicians could use these methods to measure slopes, curvatures, rate of change, and motion. It allowed them to comprehend motion and dynamics in an evolving universe, as well as the elements in the earth and galactic planet orbits.

   Integral calculus is primarily used for two purposes. One is to find f$f$ from f′$f^{\prime }$ and the other is to find out the area under the curve.

   A general function f(x)$f(x)$ is sometimes positive and sometimes negative, so the integral calculates the signed area, that is, the total area above the x$x$-axis minus the total area below the x$x$-axis. The integral count areas above/below the x$x$-axis as positive/negative.

   

   Integral Calculus – Function

   

   Integral Calculus – Function

Integral Calculus Formulas

1. We have got some Integral formula which is generally used while calculating integral. Some of the important integral calculus formulas are given below:

   (i) Constant Rule: ∫kdx=kx+C$\int kdx=kx+C$

   (ii) Constant Multiple Rule: ∫kf(x)dx=k∫f(x)dx,$\int kf(x)dx=k\int f(x)dx,$ where k$k$ is constant.

   (iii) Sum/Difference Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx$\int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx$

   (iv) Power Rule: ∫xndx=xn+1n+1+C,n≠−1$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1$

   (v) Log Rule: ∫1xdx=ln|x|+C,x≠0$\int \frac{1}{x}dx=\ln |x|+C,x\ne 0$

   (vi) Exponent Rule: ∫akxdx=akxklna+C,x≠0$\int a^{kx}dx=\frac{a^{kx}}{k\ln a}+C,x\ne 0$

   (vii) Trigonometric Rule:
   ∫sinxdx=−cosx+C$\int \sin xdx=-\cos x+C$
   ∫cosxdx=sinx+C$\int \cos xdx=\sin x+C$
   ∫tanxdx=ln|secx|+C$\int \tan xdx=\ln |\sec x|+C$ or −ln|cosx|+C$-ln|\cos x|+C$
   ∫cotxdx=ln|sinx|+C$\int \cot xdx=\ln \left|\sin x\right|+C$
   ∫secxdx=ln|secx+tanx|+C$\int \sec xdx=\ln |\sec x+\tan x|+C$
   ∫cosecxdx=ln|cosecx−cotx|+C$\int \mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}xdx=\ln |\mathrm{cosec}x-\cot x|+C$
   ∫sec2xdx=tanx+C$\int {\sec }^{2}xdx=\tan x+C$
   ∫secxtanxdx=secx+C$\int \sec x\tan xdx=\sec x+C$
   ∫cosec2xdx=−cotx+C$\int {\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}}^{2}xdx=-\cot x+C$
   ∫tan2xdx=tanx−x+C$\int {\tan }^{2}xdx=\tan x-x+C$

   For Definite Integral
   ∫baf(x)dx=F(b)−F(a)${\int }_a^{b}f(x)dx=F(b)-F(a)$
   
   Where,
   F(x)$F(x)$ is Antiderivative or Primitive
   F(b)$F(b)$ is value of Integral at Upper limit
   F(a)$F(a)$ is value of Integral at Lower Limit
   f(x)$f(x)$ is the Integrand
   dx$dx$ is the Integrating agent 
   x$x$ is the variable of Integration

What are the Applications of Integral Calculus?

1. Calculus is used in actuarial science, computer science, statistics, engineering, economics, business, medicine, and demography, among other fields.

   Calculus is used in many disciplines, including physics, chemistry, medicine, economics, biology, engineering, space exploration, statistics, and pharmacology. Architects and engineers can not construct stable structures without calculus. One can find the application of integral calculus in the area under curves, arc length, surface area, volume, probability, etc.

Solved Examples

1. Q.1. If  f′(x)=2x$f^{\prime }(x)=2x$ and f(0)=2,$f(0)=2,$ then find f(x).$f(x).$
   Ans: Since ∫2xdx=x2+C$\int 2xdx=x^2+C$
   And f(0)=2$f(0)=2$
   So, C=2$C=2$
   Hence, f(x)=x2+2$f(x)=x^2+2$

   Q.2. Evaluate ∫8x−−√dx$\int 8\sqrt{x}dx$
   Ans: Since antiderivative of f(x)=x−−√$f(x)=\sqrt{x}$ is F(x)=23x32$F(x)=\frac{2}{3}{x}^{\frac{3}{2}}$
   So, ∫8x−−√dx=8∫x−−√dx=8(23x32+C)=(163x32+k)$\int 8\sqrt{x}dx=8\int \sqrt{x}dx=8(\frac{2}{3}{x}^{\frac{3}{2}}+C)=(\frac{16}{3}{x}^{\frac{3}{2}}+k)$
   Where k$k$ is some other constant

   Q.3. Evaluate ∫213dx${\int }_1^{2}3dx$
   Ans: ∫213dx=3x|21=3(2)−3(1)=3${\int }_1^{2}3dx={3x|}_1^2=3(2)-3(1)=3$

   Q.4. Evaluate ∫e11xdx${\int }_1^e\frac{1}{x}dx$
   Ans: ∫e11xdx=lnx|e1=ln(e)−ln(1)=1−0=1${\int }_1^e\frac{1}{x}dx={\ln x|}_1^e=\ln (e)-\ln (1)=1-0=1$

   Q.5. Evaluate ∫π20sinxdx${\int }_0^{\frac{\pi }{2}}\sin xdx$
   Ans: ∫π20sinxdx=−cosx|π20=−(cos(π2)−cos(0))=1${\int }_0^{\frac{\pi }{2}}\sin xdx=-\text{cosx}{|}_0^{\frac{\pi }{2}}=-(\cos \left(\frac{\pi }{2}\right)-\cos (0))=1$

Summary

1. Integral calculus is the branch of calculus where we deal with the theory, properties and applications of Integral. Definite Integral and indefinite integral are two useful categories that are used to solve many real-life problems and the fundamental theorem of integral calculus connects the derivative and the integral.

   We learnt about how Integral calculus is used in every area of our lives to assist us in solving problems, forecasting the future, exploring the unknown, and comprehending our world and universe.

Frequently Asked Questions

1. Q.1. Who is the father of integral calculus?
   Ans: Isaac Newton (and also Gottfried Wilhelm Leibniz) is considered to be the father of integral calculus.

   Q.2. Why is calculus so hard?
   Ans: Calculus isn’t always intuitive. It requires thinking about mathematics in terms of movement in multiple directions. It also requires thinking in terms of infinite and infinitesimals, most people think more concretely than abstractly, which makes Calculus hard.

   Q.3. How do you solve an integral in calculus?
   Ans:  We solve an integral by applying the methods of integration. Basic and derived formulae are used to solve an integral.

   Q.4. What is integral calculus and what is its process?
   Ans: Integral calculus is the branch of calculus where we learn about the theory, properties, and applications of Integral. It is closely related to differential Calculus and together leads to the foundation of mathematical analysis.

   Q.5. Is integral calculus hard?
   Ans: Integral calculus is not so difficult. To have command of this, you need to devote sufficient time and practice to comprehend basic concepts. It involves critical thinking sometimes. You can have command on this if you have cleared your basic concepts.

   Q.6. What are the different parts of calculus?
   Ans: Differential and integral calculus are two parts of calculus.

   Q.7. Who is the father of integral calculus?
   Ans: Isaac Newton (and also Gottfried Wilhelm Leibniz) is considered to be the father of integral calculus.

.

integral Calculus Formulas

Similar to differentiation formulas, we have integral formulas as well. Let us go ahead and look at some of the integral calculus formulas.

Methods of Finding Integrals of Functions

We have different methods to find the integral of a given function in integral calculus. The most commonly used methods of integration are:

• Integration by Parts
• Integration using Substitution

It is also possible to integrate the given function using the partial fractions technique.

Uses of Integral Calculus

Integral Calculus is mainly used for the following two purposes:

1. To calculate f from f’. If a function f is differentiable in the interval of consideration, then f’ is defined. In differential calculus, we have already seen how to calculate derivatives of a function, and we can “undo” that with the help of integral calculus.
2. To calculate the area under a curve.

Application of Integral Calculus

1. The important applications of integral calculus are as follows. Integration is applied to find:

   • The area between two curves
   • Centre of mass
   • Kinetic energy
   • Surface area
   • Work
   • Distance, velocity and acceleration
   • The average value of a function
   • Volume

Find the integral of cos2n with respect to n.

Solution:

Let f(n) = cos2n 

we know that 2 cos2A = cos 2A + 1

So, f(n) = (1/2)(cos 2n + 1)

Let us find the integral of f(n).

∫f(n) dn = ∫(1/2)(cos 2n + 1) dn

= (1/2) ∫(cos 2n + 1) dn

= (1/2) ∫cos 2n dn + (1/2)∫1 dn

= (1/2) (sin 2n/2) + (1/2) n + C

= (sin 2n/4) + (n/4) + C

(1/4)[sin 2n + n] + C

Fundamental Theorems of Integral Calculus

We define integrals as the function of the area bounded by the curve y = f(x), a ≤ x ≤ b, the x-axis, and the ordinates x = a and x =b, where b>a. Let x be a given point in [a,b]. Then $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ represents the area function. This concept of area function leads to the fundamental theorems of integral calculus.

• First Fundamental Theorem of Integral Calculus
• Second Fundamental Theorem of Integral Calculus

First Fundamental Theorem of Integrals

A(x) = $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ for all x ≥ a, where the function is continuous on [a,b]. Then A'(x) = f(x) for all x ϵ [a,b]

Second Fundamental Theorem of Integrals

If f is continuous function of x defined on the closed interval [a,b] and F be another function such that d/dx F(x) = f(x) for all x in the domain of f, then $\underset{a}{\overset{b}{\int }}f\left(x\right)dx$ = f(b) -f(a). This is known as the definite integral of f over the range [a,b], a being the lower limit and b the upper limit.

Types of Integrals

Integral calculus is used for solving the problems of the following types.

a) the problem of finding a function if its derivative is given.

b) the problem of finding the area bounded by the graph of a function under given conditions. Thus the Integral calculus is divided into two types.

• Definite Integrals (the value of the integrals are definite)
• Indefinite Integrals (the value of the integral is indefinite with an arbitrary constant, C)

Indefinite Integrals

These are the integrals that do not have a pre-existing value of limits; thus making the final value of integral indefinite. ∫g'(x)dx = g(x) + c. Indefinite integrals belong to the family of parallel curves.

Definite Integrals

The definite integrals have a pre-existing value of limits, thus making the final value of an integral, definite. if f(x) is a function of the curve, then $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=f\left(b\right)-f\left(a\right)$

properties of Integral Calculus

Let us study the properties of indefinite integrals to work on them. 

• The derivative of an integral is the integrand itself. ∫ f(x) dx = f(x) +C
• Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. ∫ [ f(x) dx -g(x) dx] =0
• The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the individual functions. ∫ [ f(x) dx+g(x) dx] = ∫ f(x) dx + ∫ g(x) dx
• The constant is taken outside the integral sign. ∫ k f(x) dx = k ∫ f(x) dx, where k ∈ R.
• The previous two properties are combined to get the form: ∫ [k${}_1$f${}_1$(x) + k${}_2$f${}_2$(x) +... k${}_n$f${}_n$(x)] dx = k${}_1$∫ f${}_1$(x)dx + k${}_2$∫ f${}_2$(x)dx+ ... k${}_n$ ∫ f${}_n$(x)dx

Integrals Formulas

We can remember the formulas of derivatives of some important functions. Here are the corresponding integrals of these functions that are remembered as standard formulas of integrals.

• ∫ xⁿ dx=xⁿ⁺¹ /n+1+C, where n ≠ -1
• ∫ dx =x+C
• ∫ cosxdx = sinx+C
• ∫ sinx dx = -cosx+C
• ∫ sec²x dx = tanx+C
• ∫ cosec²x dx = -cotx+C
• ∫ sec²x dx = tanx+C
• ∫ secx tanxdx = secx+C
• ∫ cscx cotx dx = -cscx+C
• ∫1/(√(1-x²))= sin^-1 x + C
• ∫-1/(√(1-x²))= cos^-1 x + C
• ∫1/(1+x²)= tan^-1 x + C
• ∫-1/(1+x²)= cot^-1 x + C
• ∫1/(x√(x² -1))= sec^-1 x + C
• ∫-1/(x√(x² -1))= cosec^-1 x + C
• ∫ e^xdx=e^x + C
• ∫dx/x=ln|x| + C
• ∫ a^x dx=a^x/ln a + C

Methods to Find Integrals

There are several methods adopted for finding the indefinite integrals. The prominent methods are:

• Finding integrals by integration by substitution method
• Finding integrals by integration by parts
• Finding integrals by integration by partial fractions.

Finding Integrals by Substitution Method

A few integrals are found by the substitution method. If u is a function of x, then u' = du/dx.

∫ f(u)u' dx = ∫ f(u)du, where u = g(x).

Finding Integrals by Integration by Parts

If two functions are of the product form,  integrals are found by the method of integration by parts.

∫f(x)g(x) dx = f(x)∫ g(x) dx - ∫ (f'(x) ∫g(x) dx) dx. 

Finding Integrals by Integration by Partial Fractions

Integration of rational algebraic functions whose numerator and denominator contain positive integral powers of x with constant coefficients is done by resolving them into partial fractions.

To find ∫ f(x)/g(x) dx, decompose this improper rational function to a proper rational function and then integrate.

∫f(x)/g(x) dx = ∫ p(x)/q(x) + ∫ r(x)/s(x), where g(x) = a(x) . s(x)

Applications of Integral Calculus

Using integration, we can find the distance given the velocity. Definite integrals form the powerful tool to find the area under simple curves, the area bounded by a curve and a line, the area between two curves, the volume of the solids. The displacement and motion problems also find their applications of integrals. The area of the region enclosed between two curves y = f(x) and y = g(x) and the lines x =a, x =b is given by

Area = $\underset{a}{\overset{b}{\int }}\left(f\right(x)-g(x\left)\right)dx$

Let us find the area bounded by the curve y = x and y = x² that intersect at (0,0)and (1,1).

The given curves are that of a line and a parabola. The area bounded by the curves = $\underset{0}{\overset{1}{\int }}(y_2-y_1)dx$

Area = $\underset{0}{\overset{1}{\int }}(x-x^2)dx$

= x² /2- x ³/3

= 1/2-1/3

= 1/6 sq units.

Important Notes

• The primitive value of the function found by the process of integration is called an integral.
• An integral is a mathematical object that can be interpreted as an area or a generalization of area.
• When a polynomial function is integrated the degree of the integral increases by 1.

Properties of Indefinite Integral
(i) ∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx
(ii) For any real number k, ∫k f(x) dx = k∫f(x)dx.
(iii) In general, if f1, f2,………, fn are functions and k1, k2,…, kn are real numbers, then
∫[k1f1(x) + k2 f2(x)+…+ knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫ f2(x) dx+…+ kn ∫fn(x) dx

Basic Formulae

Integration using Trigonometric Identities
When the integrand involves some trigonometric functions, we use the following identities to find the integral:

• 2 sin A . cos B = sin( A + B) + sin( A – B)
• 2 cos A . sin B = sin( A + B) – sin( A – B)
• 2 cos A . cos B = cos (A + B) + cos(A – B)
• 2 sin A . sin B = cos(A – B) – cos (A + B)
• 2 sin A cos A = sin 2A
• cos2 A – sin2 A = cos 2A
• sin2 A = (1−cos2A2)
• sin2 A + cos2 A = 1
• sin3A=3sinA−sin3A4
• cos3A=3cosA+cos3A4

Integration by Substitutions
Substitution method is used, when a suitable substitution of variable leads to simplification of integral.
If I = ∫f(x)dx, then by putting x = g(z), we get
I = ∫ f[g(z)] g'(z) dz
Note: Try to substitute the variable whose derivative is present in the original integral and final integral must be written in terms of the original variable of integration.

Integration by Parts
For a given functions f(x) and q(x), we have
∫[f(x) q(x)] dx = f(x)∫g(x)dx – ∫{f'(x) ∫g(x)dx} dx
Here, we can choose the first function according to its position in ILATE, where
I = Inverse trigonometric function
L = Logarithmic function
A = Algebraic function
T = Trigonometric function
E = Exponential function
[the function which comes first in ILATE should taken as first junction and other as second function]

Note
(i) Keep in mind, ILATE is not a rule as all questions of integration by parts cannot be done by above method.
(ii) It is worth mentioning that integration by parts is not applicable to product of functions in all cases. For instance, the method does not work for ∫√x sinx dx. The reason is that there does not exist any function whose derivative is √x sinx.
(iii) Observe that while finding the integral of the second function, we did not add any constant of integration.

Integration by Partial Fractions
A rational function is ratio of two polynomials of the form p(x)q(x), where p(x) and q(x) are polynomials in x and q(x) ≠ 0. If degree of p(x) > degree of q(x), then we may divide p(x) by q(x) so that

p(x)q(x)=t(x)+p1(x)q(x), where t(x) is a polynomial in x which can be integrated easily and degree of p1(x) is less than the degree of q(x) . p1(x)q(x) can be integrated by expressing p1(x)q(x) as the sum of partial fractions of the following type:


where x2 + bx + c cannot be factorised further.