HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Properties of Definite Integrals

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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

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Properties of The Definite Integral [00:07:37]
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