Continuous Integration: definition, types, and solved questions

Jasmine Grover logo

Jasmine Grover

Education Journalist | Study Abroad Lead

Continuous integration is a method of adding small parts of a function to find the whole. Integration and differentiation are the two major operations in the field of calculus. Integration is a way to take the sum of very small sections to find the whole section. Integration is the reverse process of differentiation. It has a lot of applications from finding the area of asymmetric shapes and regions bounded by different mathematical functions. Continuous integration is assigning numerical values to some mathematical functions, which has some minimal value for it. We can find numerous parameters such as displacement, volume, area, energy, etc. 

Read: Differentiation and Integration Formula

Key takeaways: Continuous integration, calculus, differentiation, Basics of continuous integration, types of integration, definite integral, indefinite integral


What is Continuous Integration?

[Click Here for Sample Questions]

By assigning numerical values to the function, it helps us to represent various parameters such as displacement, area, volume, etc. Integrals and differentials are the backbones for many calculus applications. Continuous integrals are one of them. Continuous integrals are helpful in so many ways. Earlier, areas of only defined shapes could be found using algebra but after the development of calculus and continuous integration, we can find the area of any kind of curve given to us the function of that curve. We can find areas, volumes of many asymmetric objects just by knowing the function of that curve using continuous integration.

Also read: Integration by Partial Fractions

Integrals Detailed Video Explanation


Basics of Continuous Integration

[Click Here for Sample Questions]

When we integrate tiny pieces of data and assign it to a function, it represents something meaningful like displacement, area, volume. Integration and differentiation are two important aspects of calculus and we also know that integration and differentiation are two opposite things operations on a function. 

 Let us understand it using an example. Consider a function f(x) which is continuous in the interval a<= x <= b and let h(x) be a function such that d h(x)/dx = f(x) for every x in (a,b).

Let the interval [a,b] be I.

Now h(x) can be defined by 

h(x) = ab f(x)dx  

In general, 

A function F(x) can be defined by 

F(x) = axf(t)dt

 Let c be in I

Let x be very close to c in the interval I,

acf(t)dt = axf(t)dt + xcf(t)dt

acf(t)dt - axf(t)dt = xcf(t)dt

F(c) - F(x) = xcf(t)dt

So in continuous integration, let f(y) = y , u(k) = a, v(k) = k

In the case here, the function is independent of k and it is supposed to be substituted with f,u,v

In the case here, the function is independent of k and it is supposed to be substituted with f,u,v

 Also read Integration by Parts


Types of Integration

[Click Here for Sample Questions]

In general, there are only two types of integrals definite and indefinite integral. The fundamental theory of integral is finding large asymmetrical areas and finding areas under the curve for different functions.

Also Read Application of Integrals

Definite integrals: We can define definite integral to be exactly the value of area under the curve for the function within the given limits. All the integrals with limits given fall under definite integrals. Two values are required to know the limits of integration. One is the lower limit and the other is the upper limit. Let a be the lower limit and b be the upper limit. It is not necessary that b should be greater than a or a should be less than b. Both a and b combined can be called the limits of integration. 

The definite integral as a limit of sum:

The definite integral abf(x)dx is the area bounded by the curve y= f(x) , the ordinates x = a and x=b and the x axis is given by

abf(x)dx = (b-a) n∞1/n[ (f(a) + f(a+h) + ….. f (a +(n-1)h) ]

abf(x)dx = h0h[ (f(a) + f(a+h) + ….. f (a +(n-1)h) ]

Where h = (b-a) / n as n -> ∞ 

Indefinite Integrals: It is an integral of a function whose limits are not provided. This is the reverse method for differentiation in calculus. Its value is not always exact, that's why it has an arbitrary constant. It signifies a sense of ambiguity.

Some properties of Indefinite Integral:

  1. The process of differentiation and integration are reverse of each other,

 d/dxf(x)dx = f(x) and f’(x)dx = f(x) + c

Where c is any arbitrary constant

  1. Two indefinite integrals with same derivative lead to the same family of curves and so are equivalent . So if f and g are two functions such that 

 d/dxf(x)dx = d/dxg(x)dx then f(x)dx and g(x)dx are equivalent.

  1. The integral of the sum of two function equals to the sum of integrals of function i.e,

(f(x)+g(x))dx = f(x)dx + g(x)dx

  1. A constant factor may be written either before or after the integral sign i.e,

af(x)dx = a f(x)dx , where a is any constant

Also read: Methods of Integration


Things to Remember

[Click Here for Sample Questions]

  1. The fundamental theory of integral is finding large asymmetrical areas and finding areas under the curve for different functions.
  2. There are only two types of integrals definite and indefinite integral.
  3. All the integrals with limits given fall under definite integrals.
  4. The process of differentiation and integration are reverse of each other, d/dxf(x)dx = f(x) and f’(x)dx = f(x) + c
  5.  It is not necessary that upper limit should be greater than lower limit or vice versa
  6. Indefinite integral is an integral of a function whose limits are not provided.

Also Read:


Sample Questions

Ques. Integrate ( 2a/x - b/x2+ 3c3x2 ) w.r.t x (3 marks)

Ans. (2a/x - b/x2+ 3c3x2) dx

 = 2a(x)-1/2dx - bx – 2dx + 3cx2/3dx

= 4ax+b/x+9cx5/3/5+ c (ans)

Ques. evaluate 3ax/ (b2 +c2x2) dx (3 marks)

Ans. let v = b2 +c2x2 

Then dv = 2c2xdx

Therefore 

3ax/ (b2 +c2x2) dx = 3a/2c2dv/v

= 3a/2c2log|b2+c2x2| + c (ans)

Ques. verify the following using the concept of integration as an antiderivative.
x3dx /(x+1) =x - x2/2+x3/3 - log|x+1|+c (3 marks)

Ans. d/dx(x - x2/2+x3/3 - log|x+1|+c)

= 1 -2x/x + 3x2/3 - 1/(1+x)

=1 - x + x2 -1/|x+1| = x3/(1+x) (ans)

Ques. Evaluate dx/(x-)(-x) (3 marks)

Ans. put x- = t2

then -x = - (t2+) = - t2 - = -t2 -+ and dx =2tdt 

Now let I = 2tdt / t2(--t2) = 2dt/--t2 dt

 = 2dt/k2 - t2 , where k2 = -

=2sin-1t/k +c = 2sin-1(x-)/( - ) + c (ans)

Ques. evaluate tan8x sec4x dx (3 marks)

Ans. let I = tan8x sec4x dx

=tan8x(sec2x)sec2xdx

=tan8x(tan2x +1)sec2xdx

=tan10xsec2xdx + tan8xsec2xdx

= tan11x/11 + tan9x/9 + c (ans)

Ques. x3dx / (x4+3x2+2) dx (5 marks)

Ans. put x2 = t then 2xdx = dt

Now I = x3dx / (x4+3x2+2) dx

= 1/2tdt/(t2+3t+2)

Consider t/(t2+3t+2) = A / (t+1) + B/(t+2)

solving we get A = -1 , B = 2

Now let I = 1/2[2dt/(t+2) -dt/(t+1)]

 = 1/2[2log|t+2| - log|t+1|]

= log|(x2+2)|/|x2+1| + c

Ques. Evaluate (1+x)/(1-x)dx (5 marks)

Ans. let I = (1+x)/(1-x)dx 

Multiplying (1+x) on both numerator and denominator we get,

I = (1+x)/(1-x2) dx = 1/(1-x2)dx + x/1-x2 dx

I = sin-1x + I1 , where I1 = x/1-x2 dx

Let (1-x2) = t2 => -2xdx = 2tdt

Therefore we get, 

I1 = -dt = t+c =-1-x2 +C

I = sin-1x -1-x2 +C

Ques. Find dx /( 2sin2x+5cos2x) (5 marks)

Ans. Dividing numerator and denominator by cos2x we get

I = sec2xdx/(2tan2x + 5)

put tanx = t such that sec2xdx = dt

I = dt/(2t2+5) = 1/2dt/(t2+5/22)

 = 1/2*2/5tan-1(2t/5) + C

= 1/10tan-1( 2tanx/5)+C

Ques. evaluate 0/2tan7x /(cot7x + tan7x) dx (5 marks)

Ans. I = 0/2tan7x /(cot7x + tan7x) dx —--------------1

= 0/2tan7(/2-x) /(cot7(/2-x) + tan7(/2-x) dx

 =0/2cot7x /(tan7x + cot7x) dx —---2

Adding 1 and 2 we get,

2I = 0/2(tan7x +cot7x)/(cot7x + tan7x) dx

2I = 0/2 dx =/2

I = /4 (ans)

Ques. Find 2810 -x/(x+10-x) dx (5 marks)

Ans. we have 

I = 2810 -x/(x+10-x) dx —-------1

 = 2810 -(10-x)/((10-x+10-(10-x) )dx

 = 28x/(10-x+x) dx —----------------2

Adding 1 and 2 we get

2I = 281.dx = 8-2 = 6

I = 3 (ans)

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Check-Out: 

CBSE CLASS XII Related Questions

  • 1.
    If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).


      • 2.
        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 \( (1, 2, 3) \).


          • 3.

            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. 


              • 4.

                Find:
                Let \(A=[a_{ij}]\) be a \(2\times2\) matrix whose elements are given by \[ a_{ij}=\frac{(2i-j)^2}{3} \] Find the transpose matrix \(A'\).

                  • \(\begin{bmatrix} \frac{1}{3} & 3 \\ 0 & \frac{4}{3} \end{bmatrix}\)
                  • \(\begin{bmatrix} \frac{1}{3} & 0 \\ 3 & \frac{4}{3} \end{bmatrix}\)
                  • \(\begin{bmatrix} \frac{4}{3} & 3 \\ 1 & 0 \end{bmatrix}\)
                  • \(\begin{bmatrix} \frac{4}{3} & 0 1 & \frac{3}{3} \end{bmatrix}\)

                • 5.
                  Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).


                    • 6.
                      Find:

                      If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

                        • \(0\)
                        • \(-2\)
                        • \(-1\)
                        • \(2\)
                      CBSE CLASS XII Previous Year Papers

                      Comments


                      No Comments To Show