Calculus has two parts. There are integrtaion and differentiation. The definition of differentiation is "rate of change of input". The Process of differentiation is called integration. For example, f(x) is a function, The integration of given function can be expressed as `int` f(x) dx .Here, f(x) is a continuous function. Integral calculus has two types. One is definite and indefinte integral. In this article, we shall discuss about application of standard integration.
Basic standard integration formulas:
1. `int` x n dx = `(x^n+1) / (n+1)`
2.`int`cos x .dx = sin x + c
3. `int`sin x.dx = - cos x + c
4.`int`sec2x dx = tan x + c
5. `int`cosec x . cot x .dx = - cosec x + c
6. `int` sec x . tan x. dx = sec x + c
7. `int`cosec2 x . dx = -cot x + c
8.`int`a f(x) dx = a `int`f(x) dx
9. `int`[f(x) ± F(x)] dx =`int` f(x) dx ± `int`F(x) dx
10. `int` `(1/x)` dx = ln x + c
These above formulas helps to solve the standard integration. Area and Volume of the sphere, cylinder, average value these are the main application of integrals.
2.`int`cos x .dx = sin x + c
3. `int`sin x.dx = - cos x + c
4.`int`sec2x dx = tan x + c
5. `int`cosec x . cot x .dx = - cosec x + c
6. `int` sec x . tan x. dx = sec x + c
7. `int`cosec2 x . dx = -cot x + c
8.`int`a f(x) dx = a `int`f(x) dx
9. `int`[f(x) ± F(x)] dx =`int` f(x) dx ± `int`F(x) dx
10. `int` `(1/x)` dx = ln x + c
These above formulas helps to solve the standard integration. Area and Volume of the sphere, cylinder, average value these are the main application of integrals.
Application integration standards - problems:
Application integration standards - problem 1:Determine the area of shaded portion of given graph.
Solution:
Given y = x2
The boundary limits are 2 and 5
So we can integrate the given equation is `int_a^b ` x2 dx.
we know the lower limit is 2 and upper limit is 5.
So, `int_2^5 ` x2 dx. = `[(x^3)/3]_2^5`
= `[(5^3)/3]`.- `[(2^3)/3]`.
= `125/3` - `8/3` .
= `(64 - 1)/3` .
= `117/3` .
= 39 .square units.
Answer: Area of shaded portion is 39 square units.
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Application integration standards - problem 2:
Determine the total area of graph x3 + x2 - x and the x axis. the points x = -1 and x = 2
Solution:
Given x3 + x2 - x
the limits are x = -1 and x = 2
In the above graph x axis -1< x< 0 and 0 < x < 2.
So, the total area = `int_(-1)^0` [x3 + x2 - x] dx + ` int_0^2` [x3 + x2 - x] dx.
`int_(-1)^0` [x3 + x2 - x] dx = ` [(x^4/4) + (x^3)/3 - ((x^2)/2)]_(-1)^0`
= `[0] - [(-1)^4/4 + (-1)^3/3 - ((-1)^2/2)] `
= ` - [1/4 - 1/3 - 1/2] `
= ` - [(3-4-6)/12] `
= ` - [(-7)/12] `
= ` [(7/12)] ` .
int_0^2` [x3 + x2 - x] dx. = `[(x^4/4) + (x^3)/3 - ((x^2)/2)]_0^2` ..
= `[(2^4/4) + (2^3/3) - (2^2/2)] - [0]` .
= `[(16/4) + (8/3) - (4/2)]` .
= `[(4) + (8/3) - (4/2)]` .
= `[(24 + 16 - 12)/6]` = `[(28/6)]` .
`int_(-1)^0` [x3 + x2 - x] dx + ` int_0^2` [x3 + x2 - x] dx. = ` [(7/12)+(28/6)] ` .
= `[(7+56)/12]` .
= `63/12` .
Answer: Total area is `63/12` .square units.
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