What is Integration?

Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.

What is the use of Integration?

If you are standing at the bottom of a mountain and your friend is standing on the top of that mountain, what would be the shortest time between you and your friend? You might think that the shortest distance would be the straight line connecting you and your friend, but it would take longer time because the velocity of motion in this straight line will be very slow. This is because of the steepness of the mountain. That is why the roads are bent and curved for traveling to the mountain top. Even though the path is longer, the time it takes to reach the mountain top is shorter.

This type of problem is solved by integral calculus. Here, the lines joining you and your friend are summed together.

It is easy to evaluate the area, perimeter, surface area or volume of shapes of any two-dimensional or three-dimensional objects like square, cone, sphere, or circle. But how to calculate the area of any shape which has cylinder, cone and sphere attached together? This might appear very complicated. But it can be easily solved by the concept of integral calculus.


Imagine that you have the function’s derivative. How would you find the function itself? This is done by integrating the derivative of the function. Let us learn some formulae while finding the function.


Let F '(x) denote the derivate of the function F(x). Presume F '(x) = 3x.

The function F(x) is found by integrating F '(x) = 3x using the formula F' x dx=F x +c .

Here, F ' (x)is the integrand. The variable x is the integrator. The integral sign denotes the sum.

Integrating F' x gives F x dx= 3xdx

In the right side, 3 is the constant. It can be written outside the integral.

This gives 3 xdx .

Now, find the integration of x.

This is given by the formula x n dx= x n+1 n+1 +c where n1. Here, c is any constant value.

In the integral 3 xdx , the degree of the variable x is 1. Apply n=1in the above formula.

x 1 dx= x 1+1 1+1 +c xdx = x 2 2 +c

Apply xdx = x 2 2 +c in 3 xdx .

This gives 3 x 2 2 +c. Thus, the function F x is F x =3 x 2 2 +c.

Now, the constant c is found by using the initial condition. Let the initial condition be F 0 =1 when x=0.

This gives,

F 0 =3 0 2 2 +c 1=3 0 +c 1=0+c 1=c .

Thus, the value c of is c=1.

  • e x dx= e x +c
  • 1 x dx =log x +c
  • sinxdx=cosx+c
  • cosxdx=sinx+c
  • sec 2 xdx=tanx+c

Practice Problem

Find the integration of x 10 .

Some integration Techniques

When the function is in a complicated form, it is hard to integrate directly. But after changing the function into any easy form, it can be done easily. The important methods are given below.

  1. Integration by decomposing the function into sum or difference.
  2. Integration by replacing the variables.
  3. Integration by parts.
  4. Integration by reducing successively.

 Integration by decomposing into sum or difference.

Presume to integrate 1+ x 3 2 . There is no formula for this. But after decomposing into a sum of functions, it is can be done easily. The function 1+ x 3 2 can be expanded as 1+2 x 3 + x 6 . When this is integrated, this can be written as the sum of integrals.

1+2 x 3 + x 6 dx= 1dx+ 2 x 3 dx+ x 6 dx

Now this is easy to solve.

Integrating a constant is given as follows.

cdx =cx+k where k is a constant.

Apply this in the above integral.

1dx+ 2 x 3 dx+ x 6 dx = dx+ 2 x 3 dx + x 6 dx =x+2 x 3+1 3+1 + x 6+1 6+1 c =x+2 x 4 4 + x 7 7 +c

Practice Problem

Find the integration of x 2 x x .

Hint: Decompose into a difference of functions and then integrate.

Integration by replacing the variables.

This method is also called as substitution method.

Let us consider the function 2x 1+ x 2 . It can be easily integrated by replacing 1+ x 2 by u. That is, 1+ x 2 =u.

Differentiating this on both the side of the equation gives 2xdx=du.

Replace 2x 1+ x 2 by u and du.

2x 1+ x 2 dx= u du = u 1 2 du = u 1 2 +1 1 2 +1 +c = u 3 2 3 2 +c

Substitute 1+ x 2 =u.

This gives 2x 1+ x 2 dx= 1+ x 2 3 2 3 2 +c.

Practice Problem

Find the integration of x x+1 .

Hint: Replace 1+x=u and differentiate on both the sides to find dx.

Integration by Parts

Presume that the integrand is a product of two different types of functions. It cannot be integrated directly. This can be found using the formula

udv=uv vdu where u and v are two differentiable functions.

Let us find x e x dx .

The integrand if the product of function x and e x .

Take x=u. Differentiating on both sides gives dx=du

Take e x dx=dv. Integrating on both the sides gives e x =v

Apply this in the formula udv=uv vdu

This gives x e x dx=x e x e x dx

Now, e x dx= e x +c . Apply this in the above equation. This gives, x e x dx=x e x e x +c

Presume that the integrand is even more complex. Bernoulli’s formula is used for such cases.

It is given as follows.

udv=uv u v 1 + u v 2 ...

Here, u , u , u ,... are derivatives of u and v 1 , v 2 , v 3 ,... are integrals of v.

Let us find x 2 e 5x dx . Here, u= x 2 and dv= e 5x dx

Here, u =2x is the derivative of u= x 2 and u =2 is the derivative of u =2x


dv= e 5x dx gives v= e 5x 5

v dv= e 5x 5 dx gives v 1 = e 5x 5 2

Substitute these values in the formula.

This gives x 2 e 5x dx = x 2 e 5x 5 2x e 5x 5 2 +...

Thus, it is made simple by doing this way.

Practice Problem

Find 9x e 3x dx

Find x 4 e x dx

Integration by reducing successively

This method is used for integrating of functions raised to a power. These cannot be done directly.

Some real-life applications:Imagine you are traveling in a train from point A to point B. The journey starts at 1pm. At this time you are at point A. Since this is the initial time, it is takes as t=0 where t denotes the time in hours. The train is moving at the velocity v t =20t+10 kilometre per hour. How far the train would have travelled at 3 pm?

Solution: Velocity is given by ds dt where s denotes the position of the train.


ds dt =20t+10 ds= 20t+10 dt ds= 20t+10 dt s= 20 t 2 2 +10t+c

Substitute t=0 and s=0 in the above equation. This gives the value of constant as c=0

To find the distance after 2 hours, apply t=0 in s= 20 t 2 2 +10t

This gives

s= 20 2 2 2 +10 2 s=60

Thus, the train would have travelled 60 kilo meters at the end of 2 hours.

  1. The satellites at specific speed with an upward thrust so that it does not return back to earth. Integral calculus helps in determining the speed required for this.
  2. Suppose that a strain of virus growing in a laboratory doubles in one hour. The rate at which it increases after 2 hours is determined by integral calculus.

Common Mistakes:

  • Solve the given problem step by step instead of providing the answers directly, which may cause incorrect answer.
  • Do not cross out the entire solution of the problem if incorrect answer derived because the derivation may be partially correct, consider the mentor to correct it.
  • Problems with average velocity does not mean arithmetic mean value, so do not add, and divide the given problem. Instead, average velocity means change in position against change in time.
  • Add the constant of integration to all required solution ‘+C’ at the end.
  • Simplify the problem but do not use distributive laws unnecessarily.
  • Pay close attention to parentheses while solving problems.

Context and Applications:

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

  • B.Sc. in Mathematics
  • B.Sc. in Physics
  • M.Sc. in Mathematics

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