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QUESTION BANK mathematics(INTEGRATION)
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THIS PDF CONTAINS QUESTION BANK OF CLASS 12TH CHAPTER INTEGRATION ALONG WITH THE ANSWER KEY AT THE END
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IntegralsQUICK RECAP
INDEFINITE INTEGRAL
i.e., d F (x ) = f (x ) ⇒ f (x ) dx = F (x ) + C ,
8 Integration is the inverse process of dx ∫
differentiation. where C is the constant of integration.
Integrals are also known as antiderivatives.
,8 Some Standard Integrals (iii) ∫[ f (x) + g (x)]dx = ∫ f (x) dx + ∫ g (x) dx
X ∫ dx = x + C, where ‘C’ is the constant of
integration (iv)
∫ k ⋅ f (x) dx = k ∫ f (x) dx, k being any
n+1
n x real number.
X ∫x dx =
n +1
+ C , where n ≠ –1
x x METHODS OF INTEGRATION
X ∫ e dx = e + C 8 Integration by Substitution
x ax The given integral ∫ f (x ) dx can be
X ∫ a dx =
log e a
+ C , where a > 0
transformed into another form by changing
1
X ∫ x dx = loge | x | + C, where x ≠ 0 the independent variable x to t by substituting
x = g(t).
X ∫ sinx dx = − cos x + C
Integrals Substitution
X ∫ cosx dx = sin x + C
2
∫ f (ax + b)dx ax + b = t
X ∫ sec x dx = tan x + C
∫ f ( g (x)) g ′(x)dx g(x) = t
∫ cosec x dx = −cot x + C
2
X
f ′( x )
X ∫ sec x tan x dx = sec x + C ∫ f (x )
dx f(x) = t
X
∫ cosec x cot x dx = −cosec x + C n
∫ ( f (x)) f ′(x)dx f(x) = t
dx px + q = A(cx + d) + B.
X ∫ 2
= sin −1 x + C = – cos–1x + C,
∫ ( px + q) cx + d dx Find A and B by
1− x
where |x| < 1 px + q equating coefficients
dx or ∫ cx + d
dx
X ∫ 1 + x 2 = tan–1x + C = – cot–1 x + C of like powers of x on
both sides.
1 1
X ∫ dx = sec–1 x + C = – cosec–1x + C,
2 ∫ ( px + q) dx or cx + d = t2
x x −1 where |x| > 1 cx + d
1
X
∫ tan x dx = log|sec x| + C = –log |cos x| + C ∫ ( px 2 + qx + r ) dx
cx + d
X ∫ cot x dx = log|sin x| + C 1 1
X ∫ sec x dx = log|sec x + tan x| + C ∫ 2
dx px + q =
t
( px + q) cx + dx + e
x π
= log tan + + C
2 4 1 1
∫ dx x= and then
X ∫ cosec x dx= log|cosec x – cot x| + C ( px 2 + q) cx 2 + d t
x c + dt2 = u2
= log tan + C
2 px + q (px + q)
8 Properties of Indefinite Integral ∫ ax 2 + bx + c dx or
(i) ∫ f ′(x )dx = f (x ) + C d
px + q =A (ax 2 + bx + c) + B
∫ dx or dx
(ii) ∫ f (x )dx = ∫ g (x )dx + C , f and g are ax 2 + bx + c
indefinite integrals with the same
derivative. ∫ ( px + q) ax 2 + bx + c dx
,X Integration using Trigonometric Identities 1 1 x −a
When the integrand consists of trigonometric (v) ∫ x 2 − a2 dx =
2a
log
x +a
+C
functions, we use known identities to convert 1 1 x
it into a form which can be easily integrated. (vi) ∫ dx = tan −1 + C
Some of the identities useful for this purpose x 2 + a2 a a
are given below : 8 Integration by Partial Fractions
X If f(x) and g(x) are two polynomials such
x
(i) 2 sin2 = (1 − cos x ) that deg f(x) ≥ deg g(x), then we divide f(x)
2 by g(x).
x \ f (x ) = Quotient + Remainder
(ii) 2 cos2 = (1 + cos x ) g (x ) g (x )
2
(iii) 2 sin x cos y = sin (x + y) + sin (x – y) X If f(x) and g(x) are two polynomials such
that the degree of f(x) is less than the degree
(iv) 2 cos x sin y = sin (x + y) – sin (x – y)
f (x )
(v) 2 cos x cos y = cos (x + y) + cos (x – y) of g(x), then we can evaluate ∫ dx by
(vi) 2 sin x sin y = cos (x – y) – cos (x + y) f (x ) g (x )
decomposing into partial fraction.
X Some Special Substitutions g (x )
Expression Substitution
Form of the Form of the Partial
x = a sinq or a cosq Rational Function Fraction
a2 − x 2
x = a tanq or a cotq px + q A B
a2 + x 2 or (a2 + x 2 ) ,a ≠ b +
(x − a)(x − b) x −a x −b
x = a secq or a cosecq
x 2 − a2 px + q A B
2
+
a−x a+x x = a cos2q ( x − a) x − a (x − a)2
or
a+x a−x px + q A B C
+ +
x = a sin q or a cos q
2 2
x − a (x − a) (x − a)3
2
x a−x (x − a)3
or
a−x x
px 2 + qx + r A B C
x = a tan q or a cot q
2 2 + +
x a+x (x − a)(x − b)(x − c) x −a x −b x −c
or
a+x x
px 2 + qx + r A B C
x = a cos q + b sin q
2 2 + +
a−x x −b 2
x − a (x − a) (x − b)
or (x − a)2 (x − b)
x −b a−x
or (a − x )(x − b) px 2 + qx + r A Bx + C
+
2
x − a x + bx + c
X Integrals of Some Particular Functions (x − a)(x 2 + bx + c)
1 1 a+x where x2 + bx + c
(i) ∫ dx = log +C can not be factorised
a2 − x 2 2a a−x further
1 x
(ii) ∫ dx = sin −1 + C 8 Integration by Parts
a2 − x 2 a If u and v are two differentiable functions of
1 x, then
(iii) ∫ dx = log x + x 2 − a2 + C du
2
x −a 2
∫ (uv ) dx = u ⋅ ∫ vdx − ∫ dx ⋅ ∫ vdx dx .
1
(iv) ∫ 2 2
dx = log x + x 2 + a2 + C In order to choose 1st function, we take the
x +a letter which comes first in the word ILATE.
, I – Inverse Trigonometric Function FUNDAMENTAL THEOREM OF CALCULUS
L – Logarithmic Function
8 First Fundamental Theorem : Let f(x) be a
A – Algebraic Function
continuous function in the closed interval
T – Trigonometric Function [a, b] and let A(x) be the area function. Then
E – Exponential Function A′(x) = f(x), for all x ∈ [a, b].
X Integral of the type
8 Second Fundamental Theorem : Let f(x) be
x x
∫ e ( f (x) + f ′(x))dx = e f (x) + C a continuous function in the closed interval
[a, b] and F(x) be an integral of f(x), then
INTEGRALS OF SOME MORE TYPES b
a2 − x 2 dx =
x 2
a − x2
∫ f (x)dx = F (b) − F (a)
(i) ∫ 2 a
a2 x EVALUATION OF DEFINITE INTEGRAL BY
+ sin −1 + C
2 a
SUBSTITUTION
x 2 2 8 When definite integral is to be found by
(ii) ∫ x 2 − a2 dx = x −a
2 substitution, change the lower and upper
a2 2 2 limits of integration. If substitution is
− log x + x − a + C t = f(x) and lower limit of integration is a and
2
upper limit is b, then new lower and upper
(iii) x 2 limits will be f(a) and f(b) respectively.
∫ x 2 + a2 dx = x + a2
2
a2 SOME PROPERTIES OF DEFINITE
+ log x + x 2 + a2 + C INTEGRALS
2
b b
DEFINITE INTEGRAL (i) ∫a f (x)dx = ∫a f (t )dt
8 Let F(x) be integral of f(x), then for b a
any two values of the independent (ii) ∫a f (x)dx = − ∫b f (x)dx
variable x, say a and b, the difference a
F(b) – F(a) is called the definite integral of In Particular ∫a f (x)dx = 0
b b c b
f(x) from a to b and is denoted by ∫ f (x)dx . (iii) ∫a f (x)dx = ∫a f (x)dx + ∫c f (x )dx , where
a a<c<b
Here, x = a is the lower limit and x = b is the b b
upper limit of the integral. (iv) ∫a f (x)dx = ∫a f (a + b − x)dx
8 Definite Integral as a Limit of Sum a a
Let f(x) be a continuous real valued
(v) ∫0 f (x)dx = ∫0 f (a − x)dx
function defined on the closed interval 0 , if f (− x ) = − f (x )
(vi)a
[a, b]. Then ∫ f (x)dx = a
2 ∫0 f (x )dx , if f (− x ) = f (x )
−a
b
∫ f (x)dx = hlim
2a a a
h[ f (a) + f (a + h) +
→0
(vii) ∫0 f (x )dx = ∫ f (x )dx + ∫ f (2a − x )dx
0 0
a
... + f (a + (n − 1)h)], a
2a
(viii) f (x )dx = 2 ∫ f (x )dx , if f (2a − x ) = f (x )
b−a ∫ 0
where h = → 0 as n → ∞ 0
0, if f (2a − x ) = − f (x )
n
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