INTEGRALS - NCERT
INTEGRALS
7 Chapter
7.1 Overview
7.1.1
Let
d dx
F (x) =
f
(x). Then, we write
f
( x) dx = F
(x) +
C. These integrals are
called indefinite integrals or general integrals, C is called a constant of integration. All
these integrals differ by a constant.
7.1.2 If two functions differ by a constant, they have the same derivative.
7.1.3 Geometrically, the statement f ( x) dx = F (x) + C = y (say) represents a
family of curves. The different values of C correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. Further, the tangents to the curves at the points of intersection of a line x = a with the curves are parallel.
7.1.4 Some properties of indefinite integrals
(i) The process of differentiation and integration are inverse of each other,
i.e.,
d dx
f
(
x) dx =
f
(x)
and
f '( x) dx = f ( x) + C ,
where
C
is
any
arbitrary constant.
(ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. So if f and g are two functions such that
d dx
f
( x) dx
=
d dx
g(x) dx
, then
f
( x) dx
and
g ( x) dx are equivalent.
(iii) The integral of the sum of two functions equals the sum of the integrals of
the functions i.e., ( f ( x) + g ( x) ) dx = f ( x) dx + g ( x) dx .
144 MATHEMATICS
(iv) A constant factor may be written either before or after the integral sign, i.e.,
a f ( x) dx = a f ( x) dx , where `a' is a constant.
(v) Properties (iii) and (iv) can be generalised to a finite number of functions f1, f2, ..., fn and the real numbers, k1, k2, ..., kn giving
(k1 f1 ( x) + k2 f2 ( x) + ...+, kn fn ( x)) dx = k1 f1 ( x) dx + k2 f2 ( x) dx +... + kn fn ( x) dx
7.1.5 Methods of integration
There are some methods or techniques for finding the integral where we can not directly select the antiderivative of function f by reducing them into standard forms. Some of these methods are based on
1. Integration by substitution 2. Integration using partial fractions 3. Integration by parts.
7.1.6 Definite integral
b
The definite integral is denoted by f ( x) dx , where a is the lower limit of the integral a
and b is the upper limit of the integral. The definite integral is evaluated in the following two ways:
(i) The definite integral as the limit of the sum
(ii) 7.1.7
b
f ( x) dx = F(b) ? F(a), if F is an antiderivative of f (x).
a
The definite integral as the limit of the sum
b
The definite integral f ( x) dx is the area bounded by the curve y = f (x), the ordia
nates x = a, x = b and the x-axis and given by
b
a
f
( x) dx =
(b ? a)
lim
n
1 n
f
(a)
+
f
(a + h) +... f (a + (n ? 1) h)
INTEGRALS 145
or
b
f
( x) dx =
lim
h0
h
f
(a)
+
f
(a + h)
+... +
f
(a + (n ? 1) h)
,
a
7.1.8 (i)
(ii)
(iii)
where h = b ? a 0 as n . n
Fundamental Theorem of Calculus
Area function : The function A (x) denotes the area function and is given
x
by A (x) = f ( x) dx . a
First Fundamental Theorem of integral Calculus
Let f be a continuous function on the closed interval [a, b] and let A (x) be the area function . Then A (x) = f (x) for all x [a, b] .
Second Fundamental Theorem of Integral Calculus
Let f be continuous function defined on the closed interval [a, b] and F be an antiderivative of f.
7.1.9
b
f ( x) dx = [F( x)]ba = F(b) ? F(a).
a
Some properties of Definite Integrals
b
b
P0 : f ( x) dx = f (t ) dt
a
a
b
a
a
P1 : f ( x) dx = ? f ( x) dx , in particular, f ( x) dx = 0
a
b
a
b
c
b
P2 : f ( x) dx = f ( x) dx + f ( x) dx
a
a
c
146 MATHEMATICS
b
b
P3 : f ( x) dx = f (a + b ? x) dx
a
a
a
a
P4 : f ( x) dx = f (a ? x) dx
0
0
2a
a
a
P5 : f ( x) dx = f ( x) dx + f (2a ? x) dx
0
0
0
P6 :
2a
f
( x) dx
=
a 2
f
( x) dx,if
0
0
f (2a - x) = f (x) ,
0, if f (2a - x) = - f (x).
a
a
P7 : (i) f ( x) dx = 2 f ( x) dx , if f is an even function i.e., f (?x) = f (x)
?a
0
a
(ii) f ( x) dx = 0, if f is an odd function i.e., f (?x) = ?f (x) ?a
7.2 Solved Examples
Short Answer (S.A.)
Example
1 Integrate
2a x
?
b x2
+ 3c 3
x2
w.r.t.
x
Solution
2a x
b ?
x2
+ 3c 3
x2
dx
=
?1
2a ( x) 2 dx ?
bx?2 dx +
2
3c x 3 dx
5
= 4a x + b + 9cx3 + C . x5
INTEGRALS 147
3ax
Example 2 Evaluate b2 + c2 x2 dx Solution Let v = b2 + c2x2 , then dv = 2c2 xdx
Therefore,
b2
3ax + c2 x2
dx
=
3a 2c2
dv v
=
3a 2c2
log
b2
+
c2 x2
+C.
Example 3 Verify the following using the concept of integration as an antiderivative.
x3dx
=
x2 x?
+
x3 ? log x + 1 + C
x +1
23
Solution
d x2 dx x ? 2
+ x3 3
?
log
x
+
1
+
C
= 1 ? 2x + 3x2 ? 1 2 3 x+1
1
x3
= 1 ? x + x2 ?
=
x +1
x +1 .
Thus
x
?
x2 2
+
x3 3
? log
x
+1
+
C
=
x3 dx
x +1
Example 4
Evaluate
1+ 1?
x x
dx
,
x
1.
Solution Let I =
1+ x dx = 1? x
1 1 ? x2 dx +
x dx 1 ? x2 = sin?1 x + I1 ,
148 MATHEMATICS
x dx
where I =
.
1
1? x2
Put 1 ? x2 = t2 ?2x dx = 2t dt. Therefore
I1 = ? dt = ? t + C = ? 1 ? x2 + C
Hence
I = sin?1x ? 1 ? x2 + C .
Example 5 Evaluate
dx
, >
(x ?)( ? x)
( ) Solution Put x ? = t2. Then ? x = ? t2 + = ? t2 ? = ? t2 ? +
and dx = 2tdt. Now
I =
2t dt
( ) t2 ? ? t2
=
2 dt
( ? ?t2 )
dt = 2 k 2 ? t2 , where k 2 = ?
=
2sin?1 t k
+
C = 2sin?1
x? ?
+ C.
Example 6 Evaluate tan8 x sec4 x dx
Solution I = tan8 x sec4 x dx
( ) = tan8 x sec2 x sec2 x dx
( ) = tan8 x tan2 x + 1 sec2 x dx
INTEGRALS 149
= tan10 x sec2 x dx + tan8 x sec2 x dx
= tan11 x + tan9 x + C .
11
9
Example 7 Find
x4
+
x3 3x2
+
2
dx
Solution Put x2 = t. Then 2x dx = dt.
Now
I =
x4
x3dx + 3x2
+
2
=
1 2
t dt t2 + 3t + 2
Consider
t2
+
t 3t
+
2
=
t
A +1
+
t
B +2
Comparing coefficient, we get A = ?1, B = 2.
Then
I =
1 2
2
t
dt +2
?
t
dt +1
=
1 2
2
log
t
+
2
-
log
t +1
= log
x2 + 2 + C x2 +1
dx
Example 8 Find 2sin2 x + 5cos2 x Solution Dividing numerator and denominator by cos2x, we have
sec2 x dx
I = 2tan2 x + 5
150 MATHEMATICS
Put tanx = t so that sec2x dx = dt. Then
I =
dt 2t2 +
5
=
1 2
dt
t2
+
5 2
2
=
1 2
2 5
tan ?1
2t 5
+
C
=
1 10
tan ?1
2
tan 5
x
+
C
.
2
Example 9 Evaluate (7x ? 5) dx as a limit of sums. ?1
2 +1 Solution Here a = ?1 , b = 2, and h = n , i.e, nh = 3 and f (x) = 7x ? 5. Now, we have
2
(7x ? 5)dx
=
lim
h0
h
f
( ?1)
+
f
(?1
+ h)
+
f
( ?1
+ 2h)
+... +
f
(?1+ (n
? 1)h)
?1
Note that
f (?1) = ?7 ? 5 = ?12
f (?1 + h) = ?7 + 7h ? 5 = ?12 + 7h
f (?1 + (n ?1) h) = 7 (n ? 1) h ? 12.
Therefore,
2
(7
x
?
5)
dx
=
lim
h0
h
(
?12)
+
(7h
?
12)
+
(14h
?
12)
+
...
+
(7
(
n
?
1
)
h
?
12)
.
?1
=
lim
h0
h
7
h
1
+
2
+ ... + (n ? 1) ? 12n
................
................
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