Challenging Definite Integral

Problem1: Evaluate the definite integral:

Solution:

The integral contains the greatest integer.

For limit 0 to 1 greatest integer function [x] = 0

For limit 1 to 2 greatest integer function [x] = 1

So, we split the above integral into two parts using the following definite integral property.

=

= [8/3] – [1/3]

= 7/3

Problem 2: Evaluate:

Solution:

The integral contains a mod function.

For limit 1 to 2 let x=1, the function f(x) = x2 – 2x = 1-2 = -1 is negative.

For limit 2 to 3 let x=3, the function f(x) = x2 – 2x = 9-6 = 3 is positive.

So, we split the above limit of integral into two parts using the following definite integral property.

= 2/3 + 4/3

= 2

Problem 3: Evaluate:

Solution:

Let √x = t

1/ (2√x) dx = dt

dx/√x = 2dt

Limits: If x=0, t=0 and x=π²/4, t=π/2

= 2(1 – 0)

= 2

Problem 4: Evaluate:

Solution:

I =

I =

I =———-(1)

Using property

I =

I =———-(2)

Adding (1) and (2)

2I =

2I =

2I =

2I =

2I = (π/3) – (π/6)

2I = π/6

I = π/12

Problem 5: Evaluate:

Solution:

The integral contains mod function.

sin x – cos x = 0

sin x = cos x

tan x = 1

x = π/4

Between limit 0 to π/4, |sinx – cosx| is negative and between π/4 to π/2, |sinx – cosx| is positive

So, we divide the above limits of integral using the following formula

= 2√2-2

= 2(√2-1)

Problem 6: Prove that:

Solution:

Let I =———(i)

⇒ I =

Using property:

⇒ I =———(ii)

Adding (i) and (ii)

2I =

2I =

2I ==0

I = 0

Problem 7: Evaluate:

Solution:

I =

|cos x | is an even function

Using property:

I = 2

I = 2 {}

cos x is negative if π/2< x ≤ π

I = 2{}

I = 2{}

I = 2 + 2 = 4

Problem 8: Evaluate:

Solution:

I =———-(i)

Using property

I =

I =———(ii)

Adding (i) and (ii)

2I =

2I =

2I = 2 – 1

2I = 1

I = 1/2

Problem 9: Show that:

Solution:

I =

Using property

I =

I =

I =

Adding (i) and (ii)

2I =

Using property

2I =

2I =

2I =

Using property

2I =

2I =

2I =

I =

Problem 10: Evaluate:

Solution:

I =

Using property

I =

I =

Adding (i) and (ii)

2I =

Let t = sin²x ⇒ dt = 2sinx cosx dx

Limits: x= 0, t = 0 and x = π/2, t =1

2I =

2I =

2I =

I =

I =

Also, Check

• Definite Integral
• Area as Definite Integral

Definite Integrals are used to find areas of the complex curve, volumes of irregular shapes, and other things. Definite Integrals are defined by, let us take p(x) to be the antiderivative of a continuous function f(x) defined on [a, b] then, the definite integral of f(x) over [a, b] is denoted by and is equal to [p(b) – p(a)].

= p(b) – p(a)

The numbers a and b are called the limits of integration where a is called the lower limit and b is called the upper limit. The interval [a, b] is called the interval of the integration.

Note

• Constant of Integration is not included in the evaluation of the definite integral.
• is read as “integral of f(x) from a to b”