CHAPTER NAME- INTEGRATION
NCERT class 12 Maths
chapter 1 DEFINITE INTEGRATION GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL If f(x) > 0 for all x Î [a, b]; then
ò
b
a
f ( x ) is numerically
x =a
equal to the area bounded by the curve y = f(x), then x-axis
ò
and the straight lines x = a and x = b i.e.
In general
ò
b
a
b
a
x =b S C
f(x) L
f ( x)
+
+
f ( x) dx represents to algebraic sum of the
A
O
-
Q
figures bounded by the curve
M
-
B
D
R
y = f(x), the x-axis and the straight line x = a and x = b. The areas above x-axis are taken place plus sign and the areas below x-axis are taken with minus sign i.e, i.e.
ò
b
a
Note:
f ( x ) dx = area OLA - area AQM – area MRB + area BSCD
ò
b
a
f ( x ) dx = , represents algebraic sum of areas means, that if area of function y = f(x)
is asked between a to b. ÞArea bounded =
ò
b
a
b
f(x) dx andnot beenrepresented by ò f(x) dx a
e.g., If some one asks the area of y = x3 between -1 to 1. Then y = x3 could be plotted as; \ Area =
ò
0
-1
1
-x 3 dx + ò x 3 dx = 0
or, using above definition Area =
1
-1
1 2
O -1
ò
1
-1
1
x dx = 2ò x 3 dx 3
0
1
é x4 ù 1 =2ê ú = ë 4 û0 2 But if, we integrate x3 between -1 to 1.
y = x3
1
Þ
ò
1
-1
x 3 dx = 0 which does not represent area.
Thus, students are adviced to make difference between area and definite Integral.
2.1.1
FUNDAMENTAL THEOREM OF CALCULUS (NEWTON-LEIBNITZ FORMULA)
This theorem state that If f(x) is a continuous function on [a, b] and F(x) is any anti derivative of f(x) b
on [a, b] i.e. F' (x) = f (x) " x Î (a, b), then
Úf(x)dx
= F(b) - F(a)
a
The function F(x) is the integral of f(x) and a and b are the lower and the upper limits of integration.
Illustration 1: Evaluate
dx directly as well as by the substitution x = 1 /t. Examine -2 4 + x 2
ò
2
as to why the answer do not tally? Solution:
dx -2 4 + x 2
I= ò
2
2
é1 1 æ x öù = ê tan-1 ç ÷ ú = éë tan-1 (1) - tan-1 (-1) ùû è 2 ø û -2 2 ë2 1 é p æ p öù p p = ê - ç - ÷ú = Þ I = 2 ë 4 è 4 øû 4 4 On the other hand; if x = 1/t then, 1/2 1/2 dx dt dt =ò 2 = -ò 2 2 -2 4 + x -1/2 t (4 +1/t ) -1/2 4t 2 +1
I= ò
2
1/2
é1 ù =- ê tan-1(2t)ú ë2 û -1/2
1 p p p æ 1 ö = - tan -1 (1) - ç - tan -1 (-1) ÷ = - - = 2 8 8 4 è 2 ø 2
\I=
p
when x =
4
1 t
In above two results l = - p/4 is wrong. Since the integrand
1 > 0 and 4 + x2
therefore the definite integral of this function cannot be negative. Since x = 1/t is discontinuous at t = 0, the substitution is not valid
(\ I
= p/4). Note: It is important the substitution must be continuous in the interval of integration.
¥
¥
Illustration 2:
dx x 2 dx Let a = ò 4 then show that a = b and b = ò 4 x +7 x2 +1 x +7 x2 +1 0 0
Solution:
a=ò
¥
0
dx x + 7x 2 + 1 4
put x = 1/t Þ dx = –1/t2 then
1 dt ¥ ¥ 2 dt t 2 t 2dt t a=ò =ò 4 = 2 ò0 t 4 + 7t 2 + 1 = b 1 7 t + 7t + 1 ¥ 0 + +1 t4 t2 0
-
2.1.2
PROPERTIES OF DEFINITE INTEGRATION
1.
Change of variable of integration is immaterial so long as limits of integration remain the same b
i.e.
ò
b
f(x)dx =
a
b
2.
ò a
ò
f(t)dt
a
a
f(x)dx = -
ò
f(x)dx
b
3
b
3.
ò a
c
f(x)dx =
ò
b
f(x)dx +
a
ò
f(x)dx .
c
Generally we break the limit first at the points where f(x) is discontinuous and second at the points where definition of f(x) changes.
5p 12
ò [tan x ] dx , where [.] is the greatest integer function.
Illustration 3: Evaluate
0
5p 12
Solution:
ò [tan x] dx
Let I =
0
Value of tan x at x =
5p is 2 + 12
3
Value of tan x at x = 0 is 0 Integers between 0 and 2 +
3 are 1, 2, 3
\ tan x = 1, tan x = 2, tan x = 3 Þ x = tan-1 1, x = tan-1 2, x = tan-1 3 tan -1 1
tan -1 2
tan -1 3
5p 12
0
tan - 1 1
tan - 1 2
tan - 1 3
ò [tan x] dx + ò [tan x] dx + ò [tan x] dx + ò [tan x] dx
\I=
=
tan -1 1
tan -1 2
tan -1 3
5p 12
0
tan - 1 1
tan - 1 2
tan - 1 3
ò 0 dx + ò 1 dx + ò 2 dx + ò 3 dx
(
) (
)
æ 5p ö - tan -1 3 ÷ è 12 ø
= 0 + tan -1 2 - tan -1 1 + 2 tan -1 3 - tan -1 2 + 3ç
=
5p p - - tan -1 3 - tan -1 2 4 4
4
é
ù æ3 + 2ö ÷ + pú = -tan-1 (-1) è 1- 6 ø û
= p - êtan -1ç
ë
p . 4
=
b
4.
ò
a
b
f(x)dx =
a
Illustration 4:
ò
f(a + b - x)dx . In particular
ò
a
f ( x )dx =
0
a
ò
f ( a - x ) dx .
0
If f, g, h be continuous function on [0, a] such that f(a - x) = f(x), g(a - x) = - g(x) and 3h(x) - 4h(a - x) = 5, then prove that
ò Solution:
a 0
I=
f ( x ) g ( x ) h ( x ) dx = 0 . a
a
0
0
ò f ( x) g( x) h ( x) dx = ò f (a - x) g(a - x) h (a - x) dx a
ò
= – f ( x ) g( x ) h (a - x ) dx 0
7I = 3I + 4I a
=
ò f ( x) g( x) {3h ( x) - 4h (a - x)} dx 0
a
ò
= 5 f ( x ) g( x ) dx = 0, since f (a – x) g (a – x) = –f (x) g (x) 0
ÞI=0
p
Illustration 5:
æp
ö
ò x sin 2 x sin çè 2 cos x ÷ø dx 0
5
p
Solution:
æp
ö
ò x sin 2x sinçè 2 cos x ÷ø dx
Let I =
……….(1)
0
p
=
æp
ö
ò (p - x )sin 2(p - x )sinçè 2 cos(p - x )÷ø dx = 0
p
æ p
ö
ò (p - x )(- sin 2x )sinçè - 2 cos x ÷ø dx 0
p
=
æp
ö
ò (p - x )sin 2x sinçè 2 cos x ÷ø dx
……….(2)
0
Adding (1) & (2), we get p
æp ö cos x ÷ dx è2 ø
ò
2I = p sin 2x sinç 0
p 2
æp ö cos x ÷ dx è2 ø
ò
Þ I = p 2 sin x cos x sinç 0
p p cos x = z Þ - sin x dx = dz 2 2
Put
0
0
8 2z æ 2 ö 8 = p 2. z sin z dz = . ç - ÷ sin z dz = p p è pø pp p
ò 0
ò
2
2
a 2
a
5.
ò
f(x)dx =
ò
[f(x) + f(a - x) ]dx
0
a 2
a
Special cases: If f (x) = f (a – x), then
ò 0
f ( x )dx = 2
ò
f ( x )dx .
0
a
If f (x) = - f (a – x), then
ò
f ( x )dx = 0 .
0
6
a
6.
ò
a
f(x)dx =
-a
ò [f(x)+ f(-x)]dx 0
a
Special case:
ò
-a
ì a ï f ( x )dx = í2 f ( x )dx, if f ( x ) is even . ï0 0, if f ( x ) is odd. î
ò
4
Illustration 6:
ò (x
Evaluate
-4
4
Solution:
Let I =
4
ò
- 4 (1 +
-4
x5
æ x2 ö ç f (x) = ÷ ç x 2 + 16 ÷ø è
dx
5
4
x5
)
dx +
)
f (x)
ò (1 + e
-4 4
f (x)
ò (1 + e
=
dx 5 + 16 )(1 + e x )
ex )
f (x)
-4 4
2
f (x)
ò (1 + e
2I =
x2
-4
)
f (-x)
ò (1 + e
dx +
x5
( - x )5
)
dx
dx
4
=
ò f ( x) dx
-4
4
I=
ò 0
x2 dx x 2 + 16
= 4 – tan-11
p
2
Illustration 7:
dx æ5 - x ö Find the value of ò + ò log ç ÷ dx is cos x 1+5 è5 + x ø 0 -2
Solution:
Let I = I1 + I2 p
Consider I1 =
dx
ò 1+ 5
cos x
…(1)
0
7
p
Now I1 =
p
p
dx dx 5 cos x dx = = …(2) cos( p - x ) -cos x cos x 1 + 5 1 + 5 5 + 1 0 0 0
ò
ò
ò
Adding (1) and (2) , we get
p
p
p
dx 5 cos x dx + = 1.dx = p 1 + 5 cos x 0 5 cos x + 1 0 0
ò
2I1 =
ò
ò
I1 = p/2
2
æ5-xö
ò logçè 5 + x ÷ødx
Consider I2 =
-2
æ5-xö ÷ è5+ xø
Let g(x) = logç
æ 5 - (-x) ö 5-x ÷÷ = - log = -g( x ) 5+x è 5 + (-x) ø
Now g(-x) = logçç
\ g(x) is an odd function 2
\
ò g( x)dx =0
Þ I2 = 0
-2
I = I1 + I2 = p/2 + 0 = p/2
7.
b
1
ò a f (x)dx =(b - a)ò 0 f ((b - a)x + a)dx
Illustration 8: Evaluate Solution:
-5
ò
-5
-4
e(x+5) dx + 3 ò 2
2/3
1/3
e
2ö æ 9ç x - ÷ 3ø è
2
dx
I1 = ò e(x+5) dx 2
-4
8
1
= ( 5 - 4 ) ò e(
(-5+4)x-4+5 )
2
0
1
dx
I1 = ò e(x-1) dx 2
…(i)
0
Again let I 2 =
ò
2/3
1/3
e
æ 2ö 9ç x - ÷ è 3ø
2
éæ 2 1 ö
dx 1 2ù
2
æ 2 1 ö 1 9 êç - ÷ x + - ú I 2 = ç - ÷ ò e ëè 3 2 ø 3 3 û dx è 3 3ø 0
=
1 1 (x-1)2 e dx 3 ò0
=
1 ( -l1 ) 3
...(ii )
where I = I1 + 3I2
æ l ö = I1 + 3 ç - 1 ÷ = I1 -I1 è 3ø I=0
ò
-5
-4
8.
e
(x+5)2
dx + 3ò
2/3
1/3
e
æ 2ö 9ç x - ÷ è 3ø
2
dx =0
If f (x) is a periodic function with period T, then a + nT
T
a
0
ò f(x)dx = n ò f(x)dx
where nÎI ,
In particular, (i) if a = 0,
nT
T
0
0
ò f ( x)dx = n ò f ( x)dx a+T
(ii) If n = 1,
ò a
where n Î I
T
ò
f ( x )dx = f ( x )dx 0
9
10 p
Illustration 9:
Evaluate
ò sin x dx . 0
10 p
Solution:
Let I =
ò sin x dx 0
We know that |sinx| is a periodic function with period p p
ò
Hence I = 10 sin x dx
[ applying prop. 8 ]
0
Illustration 10:
If f(x) is a function satisfying f(x + a) + f(x) = 0 for all x Î R and constant a such that
ò
c +b
b
f(x) dx is independent of b, then find the least positive
value of c. Solution:
We have f(x + a) + f(x)
for all x Î R
Þ f(x + a + a) + f(x + a) = 0 Þ f(x + 2a) + f(x+ a) = 0
……(i)
[Replacing x by x + a]
….(ii)
….(iii)
Subtracting (i) from (ii), we get f(x + 2a) - f(x) = 0 for all x Î R. Þ f(x + 2a) = f(x) for all x Î R So, f(x) is periodic with period 2a It is given that
ò
c +b
b
f(x) dx is independent of b.
\ The minimum value of ‘c’ is equal to the period of f(x) i.e., 2a.
10