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SUBJECT : MATHEMATICS
CLASS : F.Sc
CHAPTER : ANTI-DERIVATIVES
LESSON No : 04 OF 07
TOPIC : APPLICATIONS OF
INTEGRATION
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?
Q. How can we find the area of a circle ?
Q. How can we find the area of a square ?
Area of square = Length x Width
Q. What do you observe in this movie ?
We find the area of curve by integration process.
APPLICATION OF INTEGRATION
 To find Velocity
 To find Distance
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
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
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
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?
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
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
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.
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
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.
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
APPLICATIONS OF INTEGRATION
 To find Velocity
 To find Distance
Questions /
Answers
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?
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
1
2
3
4
5
t2
x2
x1
t1
Assignment
Q. Find the velocity and distance of the car if its
acceleration is a= X3 + X2 + X + 2
when t=0 and s= 0m ?
OBJECTIVE QUESTIONS
Thank You
APPLICATION OF INTEGRATION

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APPLICATION OF INTEGRATION

  • 1. 1
  • 2.
  • 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?
  • 5. Q. How can we find the area of a circle ?
  • 6. Q. How can we find the area of a square ? Area of square = Length x Width
  • 7. Q. What do you observe in this movie ? We find the area of curve by integration process.
  • 8. APPLICATION OF INTEGRATION  To find Velocity  To find Distance
  • 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
  • 19. APPLICATIONS OF INTEGRATION  To find Velocity  To find Distance
  • 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
  • 23. 1 2 3 4 5 t2 x2 x1 t1 Assignment Q. Find the velocity and distance of the car if its acceleration is a= X3 + X2 + X + 2 when t=0 and s= 0m ?