Calculus is divided into two main parts - differential calculus and integral calculus. Differential calculus deals with differentiation of functions, while integral calculus is concerned with Integration. Integration can be understood as the reverse process of differentiation. It is also called anti-differentiation. Basically, integration is the summation of a infinitesimally small quantities to be integrated as a body. Integration is denoted by the symbol $\int $. The quantity obtained after integrating a function is known as its integral or anti-derivative. |

A definite integral is defined over an interval. It is the summation of a function between lower and upper limit of that interval. Let us consider a continuous function f(x) defined over a closed interval [a, b]. Definite integration of f(x) can be expressed as follows:

Few important integration rules are given below:

- $\int 1\ dx = x + c$
- $\int x^{n}dx$ = $\frac{x^{n + 1}}{n + 1}$$ + c$
- $\int e^{x}dx$ = $e^{x} + c$
- $\int a^{x}dx$ = $\frac{a^{x}}{\ln x}$$ +c$, for $a > 0$
- $\int $$\frac{1}{x}$$ dx = \ln x + c$
- $\int \sin x dx = -\cos x + c$
- $\int \cos x dx = \sin x + c$
- $\int \sec^{2}x dx = \tan x + c$
- $\int \csc^{2}x dx = -\cot x + c$
- $\int \sec x \tan x dx = \sec x + c$
- $\int \csc x \cot x dx = -\csc x + c$
- $\int \tan x dx = \ln \sec x + c = -\ln \cos x + c$
- $\int \cot x dx = \ln \sin x + c$
- $\int \sec x dx = \ln (\sec x + \tan x) + c$
- $\int \csc x dx = \ln (\csc x - \cot x) + c$
- $\int $$\frac{1}{\sqrt{1 - x^{2}}}$$ dx = \sin^{-1}x + c = -\cos^{-1}x + c$
- $\int $$\frac{1}{1 + x^{2}}$$ dx = \tan^{-1}x + c = -\cot^{-1}x + c$
- $\int $$\frac{1}{x\sqrt{x^{2} - 1}}$$ dx = \sec^{-1}x + c = -\csc^{-1}x + c$
- $\int \sinh x dx = \cosh x + c$
- $\int \cosh x dx = \sinh x + c$

**$\int a f(x)dx = a \int f(x) + c$**

Multiple of Constant Term:

Multiple of Constant Term:

**When we have to find the integration of product of two functions, we use integration by parts formula according to which, if we have two functions u and v, then according to the formula for integration by parts, we have:**

Integration by Parts:

Integration by Parts:

**Integration by Substitution:**At times, we come across unfamiliar integrands. In order to convert them to the standard familiar forms, we use substitution technique.

**Integration by Partial Fractions:**Sometimes, the integrand is in the form of algebraic fraction. In this case, it has to be decomposed into partial fractions which makes it similar to standard form.

Partial integration may be referred as integration by parts, which is discussed earlier. Let us understand this with the help of an example:

### Solved Example

**Question:**Integrate $x\ e^{x}$.

**Solution:**

^{x}

Then, by integration by parts formula:

$\int u v = u \int v - \int \left [ {u}'\ \int v \right ]$

Now, $\int x\ e^{x}\ dx$ = $x\int e^{x}-\int 1 \times e^{x}dx + c$, where c is a integration constant.

= $xe^{x} - e^{x} + c$

**Area between Two Curves:**It is useful in finding the area between two curves or area under a curve. We integrate the function for which, we need to find the area applying the appropriate limits.**Arc Length:**While finding arc length, we take a length element and integrate it along with its length.**Volumes of Revolution:**Integration is most commonly used in finding the volume of the shape formed by revolving a curve about a particular axis. In this case, we need to determine elementary volume and then, integrate it within suitable limits.**Moment of Inertia:**Integration is also used to compute moment of inertia of a body.**Average Value:**Integration is widely used in finding average value of a function, such as average temperature, average score etc.

Let us consider few examples based on integration.

### Solved Examples

**Question 1:**$\int \sin (3x-5)dx$

**Solution:**

By integrating with respect to x, we get

du = 3 dx

Now

$\int \sin (3x - 5)dx$ = $\int \sin u $$\frac{du}{3}$

= $\frac{1}{3}$$(-\cos u)+c$

Substituting back the value of u, we get

= $-\frac{1}{3}$$\cos (3x - 5) + c$

**Question 2:**Integrate $x \ln x$ with respect to x.

**Solution:**

Then, by integration by parts formula:

$\int u v = u \int v - \int \left [ {u}'\ \int v \right ]$

Now, $\int x \ln x dx$ = $\ln x\ \times $$\frac{x^{2}}{2}$$ - \int $$\left [ \frac{1}{x}\ \times \frac{x^{2}}{2}\right ]$$dx + c$, where, c is a constant

= $\frac{x^{2}}{2}$$\ln x\ - $$\frac{x^{2}}{4}$$ + c$