# Integration Rules Defined and Explained With Examples

Integration is a calculus technique to find the area under a curve line. It involves the application of limit function and is closely related to the concept of derivatives.

## Integral – Definition

Integration is performed to find masses, volumes e.t.c. It is the process of calculating integrals. An integral can be defined as:

“It is either a numerical value equal to the area under the graph of a function for some interval or a new function the derivative of which is the original function.”

For a better understanding, look at the graph below. If we want to calculate the area under it, what should we do?

One way is to use Riemann’s technique, dividing the area into small blocks and adding their areas.

This technique is not very efficient as it leaves some areas uncalculated. This is why it needs to be modified a little. We divide it into blocks of very small areas:

But you can still see some space uncovered. This is where we make use of limits. A limit function is applied to each block so that the difference between the two sides of a block is approximately zero.

By doing so, there will be an infinite number of blocks, which if you remember is the opposite of the derivative concept.

### Integral types

Integrals are classified into two types: definite and infinite. Definite integrals are used to find the area between two specific points of the curve. While the indefinite integrals are used to find the whole area under the curve.

Indefinite integrals are also known as antiderivatives. Indefinite integration of a previously derivate function gives the original function. To find step by step solution of an antiderivative, you can use an online antiderivative calculator.

### Integral notation

The symbol used for integration is “”. It is a fancy s. When applied to a function, the function looks like this:

∫ fx.dx + c

### Integration coefficient

The alphabet “c” is used at the place of any constant. In derivatives, the differentiation of a constant number is zero.

On computing the indefinite integral of a function, you cannot know if there was a constant number previously. For example:

The derivative of a function let’s say 6x + 4 is:

= 6 (using power rule)

The integration of 6 is:

= 6x (inversing derivative power rule)

You can see that this is not the original function. If we add a constant to it, we will get the original function.

But the problem here is you cannot possibly know “what was the constant number?”. The derivative of 6x + 1 or 6x + 50 is also 6.

So to remove any type of error, we simply use “c”. Such as:

= 6x + c

## Integration rules and laws

If you want to find an integral without any rule or help, you will have to understand the function very carefully and spend some time thinking about possible solutions.

For a simple function like 2x, you might make the assumption easily that the original function was x2 using derivative rules. But as the functions get difficult and difficult, the integration will become time taking and troublesome.

And your calculations can be wrong. This is why it is important to use rules of integration. Some of the most used rules are given below.

### Power Rule

Power rule is used when you see a coefficient or a variable or both as the case may be e.g 3, 8x, 4x3. In differentiation, we apply the power rule. So to find its inverse, integral in other words, we have to go backward.

In the power rule, we subtract ‘1’ from the variable’s power and multiply the coefficient by the initial power e.g. 4x3 12x2.

However, for integration, we add one to the power and then divide the variable by the resulting power.

∫ fx.dx = (xn+1)/n+1

Applying this on 12x2 we will get the original function 4x3.

Example:

Integrate x2.

Solution:

It involves power, so applying power rule:

∫ fx.dx = (xn+1)/n+1

∫ x2.dx = (x2+1)/2+1

= (x3)/3

= x3/3 + c

### Constant Rule

When there is a constant value in the function, then on integration, this constant is taken outside of the integral notation and is multiplied at the end.

∫k.fx dx =k ∫fx dx

### Example:

9x2

Solution:

∫k.fx.dx = k.∫ fx.dx

∫9.x2.dx = 9.∫ x2.dx

Applying power rule:

= 9.∫ (x2+1)/3.dx

= 9. (x3/3)

= 3x3 + c

### Sum and Difference Rule:

This rule is used when there are sum and difference operations involved between two functions.

In both of these rules, integration is applied separately on the functions and then they are subtracted or added accordingly.

∫ (fx +/- gx).dx = ∫ fx.dx +/- ∫ gx.dx

### Example:

y3 + 2

Solution:

Applying sum rule

∫ (fx + gx)dx = ∫ fx.dx + ∫ gx.dx

∫ (y2 + 2).dy = ∫ y2.dy + ∫ 2.dy

= y3/3 + 0                     (Applying power and constant rule)

= y3/3 + c

### Integration by parts:

Integration by part is a little complex rule. It can be applied when two functions are in multiplication. It is derived from the product rule of differentiation.

Let’s derive the equation for integration by parts. The product rule is:

(ab)’ = ab’ + a’b

On applying integration:

∫(ab)’dx = ∫ab’dx + ∫a’bdx

ab = ∫ab’dx + ∫a’bdx

∫ab’dx = ab –  ∫ab’dx

### Example:

1 – xsinx

Solution:

Applying difference rule:

= ∫ 1.dx – ∫ x.sinx.dx

= 0 – ∫ x.sinx.dx

Solving x.sinx.dx separately.

1. Identify a and b’:
2. Find a’ and b.

For a’, find the derivative of a.

a = x

a’= 1

For b, find the integral of b’.

b’ = sinx

∫b’.dx =  ∫ sinx.dx = – cosx

1. Solve.

∫x.sinx.dx = x.-cosx –  ∫1.-cosx.dx

= x.-cosx + sinx

= sinx – x.cosx

Putting in the original functions:

= 0 – sinx + x.cosx + c

You can also solve it using the integral calculator.

### Substitution Rule:

Lastly, we have the reverse chain rule. It is applied in specific situations and sometimes function is molded in such a way that this rule can be applied.

∫ f((bx)).b’(x).dx = ∫f(b).db

It is applied when the derivative and the original function are present in multiplication. For example here;

Derivative is = b’(x)

Original function = b(x)

In such cases, the function is integrated first generally and then the values are put.

Example:

(x2+ 2).2x

Solution:

Observing the above question we can see that 2x is the derivative of x2+2. This completes the condition for the substitution rule.

To use this rule, we will have b = x2 + 2, so:

∫(x2+2).2x.dx = ∫ b. db

= ∫ b1+1/2. db

= b2/2

Putting the value of b, we have;

= (x2+2)2/2 + c

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