Integral calculus in physics

By Admin Koushi
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April 18, 2025

Integral calculus in physics

Integral Calculus:

Let us consider, a physical quantity y depends on another physical quantity x i.e., y = (x). The graph shown in figure represents the variation of y with respect to x and P and Q are two points on the curve. PS and QR are the perpendiculars on OX axis from points P and Q respectively. The value of x at point S is a and at point R is b respectively.

To calculate the area of PQRS we divide SR into N numbers of equal small length

Δx = . From the end of each small length Δx we draw straight lines parallel to y

axis which intersects the curve and from that point of intersection we draw perpendicular to the next parallel line. So, we get N numbers of rectangular bars.

The sum of N numbers of rectangular bars is

= ƒ(a) Δx + ƒ(a + Δx) Δx + ƒ(a + 2 Δx) Δx + ——– + ƒ(a + (N – 1) Δx) Δx

= f( ) Δx where is a, a +Δ x, a + 2Δ x, ——– (b -Δ x).

But is not the exact value of area of PQRS because in each rectangular bars we have not considered the small triangles which are just above the rectangular bars.

To neglect those triangles, we have to take N tends to infinity so that Δx tends to zero.

Then the area of PQRS is A =f( ) Δx = f(x) dx =ydx

Where a and b are the lower limit and upper limit of the integration respectively.

Some common rules of integrations are respectively:

If u and v are (x) then,

i) (u+v)dx =udx + vdx

(ii)(u-v) dx = udx – vdx

(iii)(ax+b) dx = =

Applications of integral calculus to solve physics problems:

1. Graphical problem using calculus

(i) The graph represents the curve y = . Calculate the area under the curve between the limit x = a to x = b.

We consider an elementary area dA = ydx

Or, dA

Or, =DA = dx

Or, = =

A =

(ii) The acceleration-displacement graph of a particle is given in figure. What is thevelocity when displacement is 2m?

From graph we can write the equation of straight line (y = mx + c) as a = – s + 4 (m = tan = – and is in clockwise direction so it is negative.)

Or, a = v = – s + 4

Or, 3vdv = (- 4s + 12)ds

Or, 3vdv = (- 4s+12)ds

Or, 3 = – 4 + 12

Or, = = -2[ ] + 12[2]

Or, = (24 – 8)

∴v = = = 4m

2. Problems on variable physical quantity

(i) The mass per unit area of a disc of radius is = kx where k is a constant and x is the distance from the centre of the disc. Calculate the mass of the disc.

We can consider an elementary ring of the disc of radius x thickness dx and the area of the elementary ring is dA = 2 xdx

The mass of that elementary ring is dm =dA = kx(2 xdx) = 2kdx

The mass of the disc is m = = 2k

m = dm =2k dx =

∴ m = 2k =

(ii) The acceleration of a particle is a = – k where k is a constant and v is the velocity of the particle which is initially moving with velocity . What is the displacement of the particle when it becomes rest?

a = – k

Or, v = – k

Or, dv = – kds

∴ s = .

(iii) A point mass m is placed at a distance x from a long rod of mass per unit length . The angle created by the ends of the rod with x are respectively and . Find the gravitational force acting on m due to rod.