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VOOZH | about |
A definite integral has fixed upper and lower limits and is used to find the area under a curve over a given interval. It also has several useful properties that make calculations easier and help in understanding functions.
f(x) = F(b) − F(a)
Various properties of the definite integrals are added below.
Proof:
f(x) dx.......(1)
Suppose x = y
dx = dy
Putting this in equation (1)
f(y) dy
Proof:
f(x) dx = F(b) - F(a)........(1)
f(x) dx = F(a) - F(b).......... (2)
From (1) and (2)
We can derive f(x) dx = - f(x) dx
Proof:
f(x) dx = F(b) - F(a) ...........(1)
f(x) dx = F(p) - F(a) ...........(2)
f(x) dx = F(b) - F(p) ...........(3)
From (2) and (3)
f(x) dx +f(x) dx = F(p) - F(a) + F(b) - F(p)
f(x) dx + f(x) dx = F(b) - F(a) = f(x) dx
Hence, it is Proved.
Proof:
Suppose
a + b - x = y............(1)
-dx = dy
From (1) you can see
when x = a
y = a + b - a
y = b
and when x = b
y = a + b - b
y = a
Replacing by these values he integration on right side becomes f(y)dy
From property 1 and property 2 you can say that
f(x) dx = f(a + b - x) dx
Property 4.2: If the value of a is given as 0 then property 4.1 can be written as
f(x) dx = f(b - x) dx
Proof:
We can write f(x) dx as
f(x) dx = f(x) dx + f(x) dx .............. (1)
I = I1 + I2
(from property 3)
Suppose 2a - x = y
-dx = dy
Also when x = 0
y = 2a, and when x = a
y = 2a - a = a
So, f(2a - x)dx can be written as
f(y) dy = I2
Replacing equation (1) with the value of I2 we get
f(x) dx = f(x) dx + f(2a - x) dx
Proof:
From property 5 we can write f(x) dx as
f(x) dx =f(x) dx + f(2a - x) dx .............(1)
Part 1: If f(2a - x) = f(x)
Then equation (1) can be written as
f(x) dx =f(x) dx + f(x) dx
This can be further written as
f(x) dx = 2 f(x) dx
Part 2: If f(2a - x) = -f(x)
Then equation (1) can be written as
f(x) dx=f(x) dx - f(x) dx
This can be further written as
f(x) dx= 0
Proof:
From property 3 we can write
f(x) dx as
f(x) dx = f(x) dx + f(x) dx .........(1)
Suppose
f(x) dx = I1 ......(2)
Now, assume x = -y
So, dx = -dy
And also when x = -a then
y= -(-a) = a
and when x = 0 then, y = 0
Putting these values in equation (2) we get
I1 = f(-y)dy
Using property 2, I1 can be written as
I1 = f(-y)dy
and using property 1 I1 can be written as
I1 = f(-x)dx
Putting value of I1 in equation (1), we get
f(x) dx = f(-x) dx +f(x) dx ..........(3)
Part 1: When f(-x) = f(x)
Then equation(3) becomes
f(x) dx = f(x) dx + f(x) dx
f(x) dx = 2f(x) dx
Part 2: When f(-x) = -f(x)
Then equation 3 becomes
f(x) dx = -f(x) dx +f(x) d
f(x)dx = 0
Example 1: I = x(1 - x)99 dx
Solution:
Using property 4.2 he given question can be written as
(1 - x) [1 - (1 - x)]99 dx
(1 - x) [1 - 1 + x]99 dx
(1 - x)x99 dx
= 1/100 - 1/101
= 1 / 10100
Example 2: I = cos(x) log
Solution:
f(x) = cos(x) log
f(-x) = cos(-x) log
f(-x) = -cos(x) log
f(-x) = -f(x)
Hence the function is odd. So, Using property
f(x)dx = 0; if a function is odd i.e. f(-x) = -f(x)
cos(x) log = 0
Example 3: I = [x] dx
Solution:
0 dx + 1 dx + 2 dx + 3 dx + 4 dx [using Property 3]
= 0 + [x]21 + 2[x]32 + 3[x]43 + 4[x]54
= 0 + (2 - 1) + 2(3 - 2) + 3(4 - 3) + 4(5 - 4)
= 0 + 1 + 2 + 3 + 4
= 10
Example 4: I = |x| dx
Solution:
(-x) dx + (x) dx [using Property 3]
= -[x2/2]0-1 + [x2/2]20
= -[0/2 - 1/2] + [4/2 - 0]
= 1/2 + 2
= 5/2
Example 5 : Evaluate ∫[-π to π] cos(x)dx and ∫[-π to π] sin(x)dx
Solution :
Solution:
cos(x) is even:
∫[-π to π] cos(x)dx = 2∫[0 to π] cos(x)dx = 2[sin(x)]|[0 to π] = 2(0 - 0) = 0
sin(x) is odd:
∫[-π to π] sin(x)dx = 0 (by property of odd functions)
1).Evaluate ∫[0 to 4] (2x + 3)dx using the properties of definite integrals.
2).If ∫[0 to 2] f(x)dx = 5 and ∫[2 to 4] f(x)dx = 7, find ∫[0 to 4] f(x)dx.
3).Calculate ∫[-1 to 1] |x|dx using the properties of even functions.
4).Given that ∫[0 to 1] f(x)dx = 2, evaluate ∫[0 to 1] [3f(x) - 2]dx.
5).Prove that ∫[0 to π/2] sin(x)dx = ∫[0 to π/2] cos(x)dx using properties of definite integrals.
6).If ∫[0 to 1] f(x)dx = 3 and ∫[0 to 1] g(x)dx = 2, calculate ∫[0 to 1] [2f(x) - g(x)]dx.
7).Show that ∫[-a to a] x³dx = 0 for any positive real number a.
8).Given ∫[0 to 1] f(x)dx = 2 and ∫[1 to 2] f(x)dx = 3, find ∫[2 to 0] f(x)dx.
9).Evaluate ∫[0 to π] sin²(x)dx using the identity sin²(x) = (1 - cos(2x))/2 and properties of definite integrals.
10).If ∫[0 to 1] f(x)dx = 4 and ∫[0 to 2] f(x)dx = 10, calculate ∫[1 to 2] f(x)dx.
