Solving Integrals with the rules of Integration

Various formulas can be used to calculate the area under a straight line on a graph. However, have you ever wondered how to calculate the area under a curve line?

We already know that we can use differentiation to calculate the slope of a curve line. Let us now try to deduce the formula for the area under a curve line.

How do you calculate the area under the curve from 2 to 5 on and above the x-axis?

The sum of the areas of definite rectangles under the curve line, according to Riemann, gives the approximate area under that curve.

However, this is insufficient. Let’s reduce the width of rectangular slabs.

This area will be more precise than the previous one. But what if we reduce it even more? So much so that it’s almost nil!

To compute instantaneous slope, antiderivatives are used. We’ll use that concept to calculate the area under this curve. This is referred to as integrals. There are several integral calculators out there. You can try out integral calculator by allmath.com to save yourself some time if you are not interested in lengthy calculations.

Integrals – Rules and Laws

There are some rules that help to solve integrals in the same way we use rules of differentiation. Rules of integrals are quite related to the rules we use to solve derivatives.

1.    Power Rule

When a function is raised to some power then the rule used for integration is:

∫ fx.dx = (xⁿ⁺¹)/n+1

It is derived from the power rule of differentiation. Let’s first prove that this rule is the reverse of the power rule for differentiation.

The derivative of a function is 6x². Let’s revise the process of differentiation and do the opposite of it.

• One is subtracted from the power of a function, so to find its inverse we add one in the power i.e 6x²⁺¹ = 6x³.
• The coefficient is the same as the power so to reverse it we will divide by the same number i.e 6x³/3 = 2x³.

When we see a function that is in multiplication with some constant number, we apply a rule. According to this rule, during integration, the constant term is taken out of the integral notation.

∫k.fx dx =k ∫fx dx

A function is f(y) = 6y² + 12y. You can see the constant numbers 6 and 12. Let’s take 6 common.

= 6 (y² + 2y)

Apply integration:

∫ 6y² + 12y .dy = 6 ∫ y² + 2y .dy

Now on applying power rule:

= 6 ∫ y² + 2y .dy

= 6 {(y²⁺¹)/2+1 + 2y¹⁺¹/1+1}

= 6 {y³/3 + y²}

= 2y³ + 6y² + c

2.    Sum Rule

When two functions are in addition then by integration, they can be integrated separately and then added afterward.

∫ (fx + gx).dx = ∫ fx.dx + ∫ gx.dx

Find the integral of u⁵ + 2u.

Solution:

 ∫u⁵ + 2u.du = ∫u⁵.du + ∫2u.du

= (u⁵⁺¹/ 5+1) + 2∫u.du

= u⁶/6 + 2u¹⁺¹/2

= u⁶/6 + u² + c

3.    Difference Rule

Similar to the sum rule, when functions are in subtraction, they can be integrated separately.

∫ (fx – gx).dx = ∫ fx.dx – ∫ gx.dx

Find the integral of u⁵ – 2u.

Solution:

 ∫u⁵ + 2u.du = ∫u⁵.du – ∫2u.du

= (u⁵⁺¹/ 5+1) – 2∫1u.du

= u⁶/6 – 2u¹⁺¹/2

= u⁶/6 – u² + c

You can see from the example above, the only difference between the sum and difference rule is the sign symbol.

4.    Substitution Rule

This rule is also known as the reverse chain rule. This rule is applied in special situations. Sometimes the function can be set up in a way where the substitution rule can be applied.

The rule is:

∫ f((gx)).g’(x).dx = ∫f(u).du

To understand this formula look at the operations involved.

g’(x) = d/dx g(x)

For convenience, let

u = g(x)   then    du = g’(x)

It is to be noted that this formula is only applicable when there are both u and u’ in the function to be integrated.

Find the integral of (x + 1)².

Solution:

Understanding the function we can see that the derivative of x + 1 is 1. This means we can multiply 1 to use the substitution rule.

∫(x + 1)².dx = ∫(x + 1)².1dx

We now have u and u’ that are x + 1 and 1 respectively. Using the formula:

∫(x + 1)².1dx = ∫ u². du

= ∫ u²⁺¹ / 3

= ∫u³/3

Putting back the value of u:

= ∫(x+3)³ / 3

What is Integration by parts?

It is the inverse of the product rule of differentiation. Let’s derive its formula.

The product rule is:

(uv)’ = uv’ + u’v

Apply integration on both sides.

∫(uv)’.dx = ∫ uv’.dx + ∫ u’v.dx

uv = ∫ uv’.dx + ∫ u’v.dx

∫ uv’.dx = uv – ∫ u’v.dx

Evaluate  ∫ x.cosx .dx

Solution:

1. Identify u and v’.

u = x

v’ = cosx

1. Find the values of u’ and v.

For u’, find the derivative of u.

u = x

u’ = 1

For v, find the integral of v’.

v = cosx

∫v.dx =  ∫ cosx.dx = sinx

1. Solve.

∫xcosx.dx = x.sinx –  ∫1.sinx.dx             

= x.sinx – cosx

1. Add constant.

= x.sinx – cosx + c

Wrapping Up

Integrals are used to find the area under a curve. It is represented by the ∫ symbol. Integral rules help in the process of integration. Integrals can be definite or indefinite depending on the limit applied.