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Relations and Functions
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Algebra
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Calculus
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Determinants
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Linear Programming
Continuity and Differentiability
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Probability
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Integrals
- Introduction of Integrals
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Vectors
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Three - Dimensional Geometry
- Introduction of Three Dimensional Geometry
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- Equation of a Plane Passing Through Three Non Collinear Points
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- Overview of Three Dimensional Geometry
Linear Programming
Probability
Text
1) `P_0 : int_a^b f(x) dx = int_a^b f(t) dt`
Proof: It follows directly by making the substitution x = t.
2) `P_1: int_a^b f(x) dx =- int_b^a f(x) dx. "In particular" , int_a^a f(x) dx = 0`
Proof: Let F be anti derivative of f. Then, by the second fundamental theorem of
calculus, we have `int_a^b f(x) dx = F(b) - F(a) = [F(a)-F(b)] = -int_b^a f(x) dx`
Here, we observe that, if a = b, then `int_a^a f(x) dx = 0`
3) `P_2 : int_a^b f(x) dx = int_a^c f(x) dx + int_c^b f(x) dx`
Proof: Let F be anti derivative of f. Then
`int_a^b f(x) dx = F(b) - F(a)` ...(1)
`int_a^c f(x) dx = F(c) - F(a)` ...(2) and
`int_c^b f(x) dx = F(b) - F(c)` ...(3)
Adding (2) and (3), we get
`int_a^c f(x) dx + int_c^b f(x) dx = F(b) - F(a) = int_a^b f(x) dx `
This proves the property `P_2.`
4) `P_3 : int_a^b f(x) dx = int_a^b f(a + b - x) dx `
Proof: Let t = a + b – x. Then dt = – dx. When x = a, t = b and when x = b, t = a. Therefore
`int_a^b f(x) dx = - int _b^a f(a+b-t) dt`
= `int_a^b f(a+b-t)dt ("by" P_1)`
=`int_a^b f(a+b-x) dx ("by" P_0)`
5) `P_4: int_0^a f(x) dx = int_0^a f(a - x) dx `
(Note that `P_4` is a particular case of `P_3`)
Proof: Put t = a – x. Then dt = – dx. When x = 0, t = a and when x = a, t = 0. Now proceed as in `P_3`.
6) `P_5: int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx `
Proof: Using `P_2`, we have `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_a^(2a) f(x) dx`
Let t = 2a – x in the second integral on the right hand side.
Then dt = – dx. When x = a, t = a and when x = 2a, t = 0.
Also x = 2a – t. Therefore, the second integral becomes
`int _a^(2a) f(x) dx = -int_a^0 f(2a - t) dt = int_0^a f(2a - t) dt = int_0^a f(2a - x) dx `
Hence `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx `
7) `P_6: int_0^(2a) f(x) dx = 2 int_0^a f(x) dx , if f(2a - x) = f(x) and 0 if f(2a - x) = -f(x) `
Proof: Using `P_5` , we have `int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(2a - x) dx ` ...(1)
Now, if f(2a – x) = f(x), then (1) becomes
`int_0^(2a) f(x) dx = int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx,`
and if f(2a – x) = – f(x), then (1) becomes
`int_0^(2a) f(x) dx = int_0^a f(x) dx - int_0^a f(x) dx = 0 `
8) `P_7`:
i) `int_(-a)^a f(x) dx = 2 ``int_0^a f(x) dx` if f is an even function , i.e., if f(-x) = f(x)
ii) `int_(-a)^a f(x) dx = 0` , if f is an odd function , i.e., if f(-x) = -f(x).
Proof : Using `P_2` we hane
`int_(-a)^a f(x) dx` = `int_(-a)^0 f(x) dx` + `int_0^a f(x) dx` Then
t = – x in the first integral on the right hand side.
dt = – dx. When x = – a, t = a and when
x = 0, t = 0. Also x = – t.
Therefore `int_a^(-a) f(x) dx = - int_(a)^0 f(-t) dt + int_0^a f(x) dx `
`= int_(0)^a f(-x) dx + int_0^a f(x) dx ` `("by" P_0)` ...(1)
(i) Now, if f is an even function, then f(–x) = f(x) and so (1) becomes
` int_(-a)^a f(x) dx = int_0^a f(x) dx + int_0^a f(x) dx + int_0^a f(x) dx = 2 int_0^a f(x) dx `
(ii) If f is an odd function, then f(–x) = – f(x) and so (1) becomes
`int_(-a)^a f(x) dx = - int_0^a f(x) dx + int_0^a f(x) dx = 0 `
