Use of Calculus in Real Life

In this article we are going to discuss about the use of calculus in real life problems step by step concept. The process of contraction the calculus comparable to over screening the real life problems concept and problem in calculus are referred as review use of calculus in real life. This article helps to improve the knowledge the real life for calculus and below the problems are using toll for the help with test calculus. Put down behind the useful get review real life use of calculus to this article. Use of calculus in real life solutions also show below.

Level One Example Real Life Calculus Use Problems:-

Real life calculus use problem 1:-

Integrate the following equation

`int 5x^4+4x^5+5x^3 dx`

Solution:-

Step 1:-

`int 5x^4+4x^5+3x^3 dx`

Step 2:-

`= int5x^4dx + int4x^5dx + int 3 x^3 dx`

Integrating the above equation

We get

Step 3:-

`= (5x^5)/(5) + (4x^6)/ (6) + (3x^4)/ (4)`

Step 4:-

`= (5x^5)/(5) + (4x^6)/(6) +(3x^4)/ (4)`

Step 5:-

`= x^5+ (2x^6)/ (3) + (3x^4)/ (4)`


Real life calculus use problem 2:-

Solve by differentiating the following equation and get the first derivative second derivative and third derivative

`y = 12x^3 + 8x^2+ 4x^5 + x`

Solution:-

Given equation is `y = 12x^3+ 8x^2 + 4x^5 + x`

Step 1:-

To get the 1st derivative differentiate the above given equation

`(dy)/(dx) = 36x^2+16x+20x^4`

Step 2:-

To get the 2nd derivative differentiate the ist derivative

`(d^2y)/ (dx^2) = 72x+16+80x^3.`


Level Two Example Real Life Calculus Use Problems:-

Real life calculus use problem 1:-

Integrate the following equation

`int 8x^4+4x^5+2x^3 dx`

Solution:-

`int 8x^4+4x^5+2x^3 dx`

Step 1:-

`= int 8x^4 dx + int 4x^5 dx + int 2 x^3 dx`

Integrating the above equation

We get

Step 2:-

`= (8x^5)/ (5) + (4x^6)/ (6) + (2x^4)/ (4)`

Step 3:-

`= (8x^5)/(5) + (2x^6)/(3)+ (x^4)/(2)`

Real life calculus use problem 2:-

Differentiate the following equation and get the 1st, 2nd and 3rd derived

`y = 8x^3+4x^2+2x^1+ 12`

Solution:-

Step 1:-

By differentiate the given equation with respect to x to get the 1st derivative

`y= dy/dx =24x^2+8x^1+2.`

Step 2:-

To get the 2nd derived differentiate the 1st derivative of the given equation.

`y = (d^2x)/ (dy^2) = 48x + 8`

Step 3:-

To get the 3rd derived differentiate the 2nd derivative of the given equation.

`y= (d^3x)/ (dy^3) = 48.`