In a practical application, how would you use integrtion to find an area or volume?

I am going to explain the meaning of derivation and integration in this simple example:

Suppose Patrick is doing his homework on a piece of paper, and for some reason he gets mad and tears the price of paper to pieces. This act is called deriving in mathematics. After that Patrick realizes that he had written his homework on it. So he just gathers the pieces together and glues them. This act is called integration.

In mathematics, integration is just gathering pieces to create a function that covers a larger domain.

Now as to how do u use it in real life, there are many examples

Suppose u have an empty tank and ur filling it with water. The flow of water coming from the hose suppose it is tm^3/s => in 1 second, t m^3 flows out.

U can integrate here with respect to the Area*Height of the tank

and the limit of integration here is the time. Though u can choose whatever limit u want, like volume, quantity...

so here u can say :∫pi*R^2 and the limits are from 0 to the maximum height of the tank

or ∫t from 0 to the time u want to pump water

So it depends on the type of the problem u have, u can find many integrals in almost all cases

Hello Patrick, let us find the area of the circle using integration

Let us turn the radius vector through an angle d@. So the arc made by this rotation is r d@

The area of this triangle is 1/2 * r * r d@ = 1/2 * r^2 d@

Now to get the total area of the circle we have to sweep the radius through angle 2 pi

So integrating, 1/2 * r^2 (@) with limits 0 to 2pi

Plugging the limits, 1/2 *r^2 (2pi-0) = pi r^2

Same way we can find the volume of sphere

Let us take a slice of thickness dx at a distance x right from its centre

Let y be the height at the sliced part.

So volume of the slice will be pi y^2 dx

y is related to x by the expression, y^2 = r^2 - x^2

So volume V = pi (r^2 - x^2) dx

Now to get the volume of the sphere we have to integrate the above one within limits -r to +r

Or 2 times the integral within limits 0 to r

Hence 2 * pi * ( r^2 x ) - 2 * pi * x^3 / 3

Applying limits, 2 pi r^3 - 2/3 * pi * r^3

====> (2 - 2/3) pi r^3 = 4/3 pi r^3

A practical application would be a velocity function, v(t), that has been graphed. You can find the distance/displacement by using an integral function due to velocity being the first derivative of position.

Integration allows us to solve many real world problems through the use of differential equations, which is essentially mathematical modeling. It is very useful in the engineering field.

You should check out this site for more applications. http://www.intmath.com/applications-integration/ap...

Things like Voltage equations, Buoyancy force, Volumes of Revolution, Area between curves, finding Work done on an object, Harmonic motion, Spring constants, etc.

Real world example:

A car is driving at a velocity of 15 m/s. How far does the car go from t = 0 to 5.

Velocity is defined as the first derivative of a position function with respect to time: ds/dt.

So:

ds/dt = 15 m/s

Through separation of variables, we can find the area under the velocity curve by integration.

ds = ∫ 15 dt

ds = 15t from 0 to 5.

15(5) - 15(0) = 75 meters