Properties of Definite Integrals: Definition and Proof

Definite Integration is one of the important parts of Calculus, which applies to measuring the change in the function at a certain point. Mathematically, it forms a powerful tool by which slopes of functions are determined, the maximum and minimum of functions found, and problems on motion, growth, and decay, to name a few. These concepts of integration have been broadly applied in branches of mathematics, physics, engineering, economics, and biology.

In this article, we will cover the concept of Definite Integration. This concept falls under the broader category of Calculus, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), seven questions have been asked on this concept, including one in 2022, and six in 2023.

Properties of Definite Integration

Definitte inegration calculates the area under a curve between two specific points on the x-axis.

Let f be a function of x defined on the closed interval [a, b] and F be another function such that ddx(F(x))=f(x) for all x in the domain of f, then

∫abf(x)dx=[F(x)+c]ab=F(b)−F(a)is called the definite integral of the function f(x) over the interval [a, b], where a is called the lower limit of the integral and b is called the upper limit of the integral.

Definite integrals have properties that relate to the limits of integration.

Property 1

∫aaf(x)dx=0

If the upper and lower limits of integration are the same, the integral is just a line and contains no area, hence the value is 0

Alternatively

If ddxF(x)=f(x), then ∫aaf(x)dx=[F(x)]aa=F(a)−F(a)=0

Property 2

The value of the definite integral of a function over any particular interval depends on the function and the interval but not on the variable of the integration.

∫abf(x)dx=∫abf(t)dt=∫abf(y)dy

For example,

∫02x2dx=[x33]02=233−033=233∫02t2dt=[t33]02=233−033=233∫02y2dy=[y33]02=233−033=233

Property 3

If the limits of definite integral are interchanged, then its value changes by a minus sign only.

If ddxF(x)=f(x), then

∫abf(x)dx=[F(x)]ab=F(b)−F(a)

and, ∫baf(x)dx=[F(x)]ba

=(F(a)−F(b))=−(F(b)−F(a))
Hence,

∫abf(x)dx=−∫baf(x)dx

Property 4 (King's Property)

This is one of the most important properties of definite integration.

∫abf(x)dx=∫abf(a+b−x)dx

Proof:
In R.H.S, Put t=a+b−x⇒dx=−dt
Also, when x=a, then t=b, and when x=b,t=a

∴ R.H.S =∫baf(t)(−dt)=−∫baf(t)dt=−(−∫abf(t)dt)=∫abf(t)dt=∫abf(x)dx= L.H.S

Corollary:

∫0af(x)dx=∫0af(a−x)dx

Recommended Video Based on Properties of Definite Integration

Solved Examples Based on Properties of Definite Integration:

Example 1: What is the value of integral ∫−aaf(x)dx equal to?

1) −∫aaf(x)dx
2) −∫−aaf(x)dx
3) ∫−aaf(−x)dx

4) none of these

Solution

As we learned,

Fundamental Properties of Definite Integration -

Interchanging the limit of the definite integral does not change the absolute value but change the sign of the integral.

∫abf(x)dx=−∫baf(x)dx

- wherein

$\int_{-a}^a f(x) d x=\int_{-a}^a f(-x) d x

Example 2: Which of the following is NOT true?

1) ∫abf(x)dx=∫abf(y)dy

2) ∫abf(x)dx=−∫baf(y)dy

3) ∫abf(x)dx=∫abf(y)dx

4) None of these

Solution

For option (C) to be correct we should have dy in the second integral instead of dx. So it is wrong

Hence, the answer is the option (3).

Example 3: Which of the following is equal to integral ∫710f(x)dx

1) 1) ∫107f(x)dx

2) ∫710f(t)dt

3) ∫−7−10f(x)dx

4) None of these

Solution

∫710f(x)dx=∫710f(t)dt

As the change in a variable in integral does not change its value

Hence, the answer is the option 2.

Example 4: Which of the following integrals is not equal to the other integrals?

1) ∫abf(x)dx
2) −∫baf(y)dy
3) ∫baf(t)dt
4) ∫ahf(z)dz

Solution

Option (C):

∫baf(t)dt=−∫abf(x)dx

So this integral is the odd one out.

Hence, the answer is the option 3.

Example 5: Let f:R→R be a continuous function satisfying f(x)+f(x+k)=n, for all x∈R where k>0 and r is a positive integer. If I1=∫04nkf(x)dx and I2=∫−k3kf(x)dx, then :

1) Λ1+2Λ2=4nk
2) l1+2l2=2nk
3) I1+nl2=4n2k
4) !+nI=−6n2k

Solution

f(x)+f(x+k)=n

Put x→x+k

f(x+k)+f(x+2k)=n

Subtract these equations

f(x)−f(x+2k)=0

⇒f(x+2k)=f(x)

f(x) is periodic with a period $\mathrm{2k }$

Now I1=∫04nkf(x)dx=∫02n(2k)f(x)dx=2n∫02kf(x)dx

and I2=∫−k3kf(x)dx=∫−k4k−kf(x)dx=∫04kf(x)dx

=∫02⋅(2k)f(x)dx=2∫02kf(x)dx

Now ∫02kf(x)dx=∫0kf(x)dx+∫k2kf(x)dx

Put x=t+k in second integral

=∫0kf(x)dx+∫0kf(t+k)dt=∫0k(f(x)+f(x+k))dx=∫0kndx=nk∴I1=2n2k,I2=2nk
Checking options,

I1+nI2=2n2k+2n2k=4n2k

Hence, the answer is the option 3.

Summary

Definite integration is a powerful tool in calculus that allows us to calculate the area under a curve between two specific points. It provides a deeper understanding of mathematical ideas paramount for later developments in many scientific and engineering disciplines.