APPLICATION PROBLEMS IN INTEGRAL CALCULUS

About "Application Problems in Integral Calculus"

Application Problems in Integral Calculus :

Here we are going to see some application problems in integration.

Here we discuss how integration is used to find the position and velocity of an object, given its acceleration and similar types of problems. Mathematically, this means that, starting with the derivative of a function, we must find the original function.

Many common word which indicate derivative such as rate, growth, decay, marginal, change, varies, increase, decrease etc.

Let us look in to some example problems to understand the concept.

Application Problems in Integral Calculus - Examples

Question 1 :

A ball is thrown vertically upward from the ground with an initial velocity of 39.2 m/sec. If the only force considered is that attributed to the acceleration due to gravity, find

(i) how long will it take for the ball to strike the ground?

(ii) the speed with which will it strike the ground? and

(iii) how high the ball will rise?

Solution :

Given that :

Initial velocity = 39.2 m/sec

By differentiating "distance" we get "velocity", by differentiating "velocity" we get "acceleration".

By doing the process vice versa, we get "velocity" by integrating "acceleration", integrating "velocity" we get "distance".

When a ball is thrown vertically upward, it reaches the maximum height then it will strike the ground.

Acceleration  =  9.8 m/sec²

When we throw a ball in upward direction, it moves against the gravity, so we have to consider the acceleration as -9.8.

Velocity  = ∫a dt

a = -9.8       

velocity  =  ∫(-9.8) dt

 v(t) =  -9.8 t + c₁ ------(1)

When an object reaches its maximum height, its velocity will become zero at the point.

Initial velocity = 39.2 at t = 0

 39.2 =  -9.8 (0) + c₁

c₁  =  39.2

 v(t) =  -9.8 t + 39.2

By applying v(t)  =  0, we may find the time taken by the ball to reach its maximum height.

0  =  -9.8 t + 39.2

-9.8 t  =  -39.2  ==> t  =  39.2/9.8  =  4

Hence it takes 4 seconds to reach maximum height.

Distance  =  ∫ v(t) =  ∫(-9.8 t + 39.2) dt

Distance  =  -9.8t²/2 + 39.2 t + c₂

D(t)  =  -4.9 t² + 39.2 t + c₂

When t = 0, D(t)  =  0, then c₂  =  0

Hence D(t)  =  -4.9 t² + 39.2 t 

(i) how long will it take for the ball to strike the ground?

Time taken by the ball to strike the ground  =  8 seconds 

[It takes 4 seconds to reach its maximum height and 4 seconds to strike the ground.]

(ii) the speed with which will it strike the ground? 

 v(t) =  -9.8 t + 39.2

when t = 8

 v(8) =  -9.8 (8) + 39.2

  =  39.2 m /sec

(iii) how high the ball will rise?

When t = 4

D(4)  =  -4.9 (4)² + 39.2 (4) 

  =  -4.9(16) + 156.8

  =  -78.4 + 156.8

  =  78.4 m

Hence the maximum height reached by the ball is 78.4 m.

Question 2 :

A wound is healing in such a way that t days since Sunday the area of the wound has been decreasing at a rate of -3/(t+2)² cm² per day. If on Monday the area of the wound was 2cm²

(i) What was the area of the wound on Sunday?

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?

Solution :

Let "A" be the are of wound and "t" be the days

By integrating the wound on Sunday, we get the area of wound on Monday.

Rate of change (dA/dt)  =  -3/(t+2)² cm²/ per day

∫dA  =  -3 ∫(t+2)^-2 dt

A  =  3 (t+2)^-1 + c

A  =  3/(t+2) + c    -----(1)

When A = 2, t = 1

2  =  3/(1+2) + c 

2  =  1  + c

c = 1

Applying c = 1 in (1)

A  =  3/(t+2) + 1

(i) What was the area of the wound on Sunday?

Since it is starting day t = 0

A  =  (3/(0+2)) + 1

A  =  (3/2) + 1

A = 2.5 cm²

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?

When Sunday = 0, then Thursday  =  4

A  =  (3/(4+2)) + 1

A  =  (3/6) + 1

A = 1.5 cm²

After having gone through the stuff given above, we hope that the students would have understood, "Application Problems in Integral Calculus"

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