This document provides an overview of integration and its applications. It defines integration as the reverse process of differentiation and gives examples of integrals of common functions. Integration can be used to find velocity when given acceleration and distance when given velocity. Examples are provided to demonstrate calculating velocity from acceleration using integration and distance from velocity using integration. Common applications of integration include finding velocity, distance, area, and volume. Practice problems are included at the end to reinforce these concepts.
3. SUBJECT : MATHEMATICS
CLASS : F.Sc
CHAPTER : ANTI-DERIVATIVES
LESSON No : 04 OF 07
TOPIC : APPLICATIONS OF
INTEGRATION
4. Q. What is the definition of integration?
A. The reverse process of differentiation is called
“integration”.
Q. What is the integral of Sin(x) ?
A. - Cos X + C
Q. What is the integral of Cos(x) ?
A. Sin x + C
Q. What is the definition of acceleration ?
A. The rate of change in velocity w.r.t. to time?
9. To find Velocity
If acceleration of a moving body is given, we can find its
velocity by using integration. When a body is moving with
acceleration ‘a’ then
We can find its velocity as;
Acceleration = a =
Velocity = integral of acceleration = ∫ a dt
APPLICATION OF INTEGRATION
10. APPLICATION OF INTEGRATION
Q. The acceleration of a particle moving in a
straight line is given by a = 6t+ 4.
Find the formula for the velocity, given that
when t = 0, v = 6
11. Solution
Acceleration = a = 6t + 4
Since acceleration =
Therefore, v= 3t2 + 4t + c (integration w.r.t ‘t’)
When t = 0, v = 6
then c = 6 (by putting the values of ‘t’ and ‘v’)
Hence, velocity = v = 3t2 + 4t +6
dt
dv
APPLICATION OF INTEGRATION
12. A. V = 38 m/sec.
Q. What is the velocity when t= 3sec
and acceleration of the car is a = 3 t2 + 2t
when t = 0, v = 0?
13. Solution
Acceleration = = 3t2+ 2t (A)
As Velocity = integral of acceleration = ∫ a dt
Therefore integrating (A) w.r.t ‘t’ we have ,
v = t3 + t2 + c (B)
When t = 0 , s = 0
Therefore c = 0 so eq. (B) becomes
v = t3+ t2
V = 34 m/s
APPLICATION OF INTEGRATION
14. To Find Distance
If velocity of a moving body is given, we can find its distance by
using integration. When a body is moving with velocity ‘v’ then
we can find its distance as;
velocity = v =
Distance = integral of velocity = ∫ v dt
APPLICATION OF INTEGRATION
15. APPLICATION OF INTEGRATION
Q. If a particle is moving in a straight line, the equation of
motion is as v = 3 t2+ 4t + 6
Find the distance covered by particle when t = 0,s =0m.
16. Solution
Velocity = v = 3t2+ 4t + 6 (A)
As Distance = integral of velocity = ∫ v dt
Therefore integrating (A) w.r.t ‘t’ we have ,
s = t3 + 2t2 + 6t + c (B)
When t = 0 , s = 0
Therefore c = 0 so eq. (B) becomes
s = t3+ 2t2+6t
APPLICATION OF INTEGRATION
17. Q. What is the distance covered by a car when t= 2sec and velocity is
v = 2t + 5 when t = 0sec, S = 0m ?
A. d = 14 m.
18. Solution
Velocity = v = 2t+5 (A)
As Distance = integral of velocity = ∫ v dt
Therefore integrating (A) w.r.t ‘t’ we have ,
s = t2 + 5t + c (B)
When t = 0 , s = 0
Therefore c = 0 so eq. (B) becomes
s = t2+5t As t = 2sec
Therefore s = (2)2+5(2) = 4+10 = 14m
APPLICATION OF INTEGRATION
21. Q. What will be the distance covered by a moving car its
velocity is v = 3t + 5 ?
A. S = t2 + 5t + c
Q. How can we find out the velocity from given
acceleration?
A. By integration
Q. What will be the velocity of a moving car if its
acceleration is a = 3 t2 + 2t ?
A. V = t3 + t2 + c
Q. What is the definition of acceleration?
A. The rate of change of velocity w.r.t. to time?
22. Q. If a satellite is moving in its orbit
around the earth with the velocity
V = 3t2 + 2t.
Find the distance traveled by the satellite in
the fourth second ?
A. 80m