Tiêu đề 7. INTEGRALS.pdf ✅

TOPIC-1 Indefinite Integral Concepts Covered:  Meaning of Integral of function, Integration by Substitution, Partial fraction, Parts.  Formulae for Indefinite Integral, Second fundamental theorem, Properties of definite integral. tab Revision Notes z Meaning of Integral of Function If differentiation of a function F(x) is f(x) i.e., if d dx [F(x)] = f(x), then we say that one integral or primitive or anti-derivative of f(x) is F(x) and in symbols, we write, f x() () dx = + F x C. ∫ Therefore, we can say that integration is the inverse process of differentiation. z Methods of Integration (a)Integration by Substitution Method: In this method, we change the integral f x dx ( ) ∫ , where independent variable is x, to another integral in which independent variable is t (say) different from x such that x and t are related by x = g(t). [Board, 2022, 23, 24] i.e., f x dx ( ) ∫ = f g t g t dt [ ( )] '( ) ∫ , where x = g(t). (b)Integration by Partial Fractions: Consider f x g x ( ) ( ) defines a rational polynomial function. [Board, 2022, 23, 24] z In rational polynomial function if the degree (i.e., highest power of the variable) of numerator (Nr.) is greater than or equal to the degree of denominator (Dr.), then (without any doubt) always perform the division i.e., divide the Nr. by Dr. before doing anything and thereafter use the following: Numerator Denominator = Quotient + Remainder Denominator Table Demonstrating Partial Fractions or Various Forms Form of the Rational Function Form of the Partial Fraction px q x a x b a b + − − ≠ ( )( ) , A x a B − x b + − px q x a + ( ) − 2 A x a B − x a + ( ) − 2 px qx r x a x b x c 2 + + ( ) − − ( )( ) − A x a B x b C − x c + − + − px qx r x a x b 2 2 + + ( ) − − ( ) A x a B x a C − x b + − + ( ) − 2 px qx r x a x bx c 2 2 + + ( ) − + ( ) + where x2 + bx + c can't be factorized further. A x a Bx C − x bx c + + + + 2 INTEGRALS LEARNING OBJECTIVES After going through this Chapter, the student would be able to learn:  Meaning of Integral of function  Integration by Substitution, Partial fraction, Parts.  Formulae for Indefinite Integral  Second fundamental theorem  Properties of definite integral. 7 CHAPTER LIST OF TOPICS Topic-1: Indefinite Integral Topic-2: Definite Integral
MATHEMATICS, Class-XII (c)Integration by Parts: If U and V be two functions of x, then (I) (II) U V dx , ∫ = U Vdx dU dx Vdx dx ∫ ∫ ∫ −       KEY FORMULAE Formulae for Indefinite Integrals (a) x dx x n C n n n = + + ≠ − + ∫ 1 1 1, (b) 1 x dx = + x C ∫ log| | (c) a dx a a C x x = + ∫ 1 log (d) e dx a e C ax ax = + ∫ 1 (e) sin (ax) c dx os ( ) a = − ax + C ∫ 1 (f) cos (ax) s dx in ( ) a = + ax C ∫ 1 (g) tan l xdx x = + og|sec | C ∫ or – log|cos x| + C (h) cot l xdx x = + og|sin | C ∫ or – log |cosec x| + C (i) sec l xdx x = + og|sec tan |x C+ ∫ or log tan π 4 2 +       + x C (j) cosec c xdx x = − osec x C+ ∫ log| cot | or log tan x C 2 + (k) sec2 xdx x = + C ∫ tan (l) cosec2 xdx x = − + C ∫ cot (m) sec . x x tan dx = + x C ∫ sec (n) cosec . x x cot c dx = − osec x C+ ∫ (o) 1 1 2 1 x x dx x C − = + − ∫ sec (p) 1 1 2 2 1 a x dx a x a C + =       + − ∫ tan (q) 1 1 2 2 2 a x dx a a x a x C − = + − + ∫ log (r) 1 1 2 2 2 x a dx a x a x a C − = − + + ∫ log (s) 1 2 2 2 2 x a dx x x a C − = + − + ∫ log (t) 1 2 2 2 2 x a dx x x a C + = + + + ∫ log (u) 1 2 2 1 a x dx x a C − =       + − ∫ sin (v) 1 1 ax b dx a ax b C + = + + ∫ log| | (w) λ λ dx = + x C ∫ , where 'l' is a constant. (x) x a dx x x a a x x a C 22 22 2 2 2 2 2 − = − − + − + ∫ log (y) x a dx x x a a x x a C 22 22 2 2 2 2 2 + = + + + + + ∫ log (z) a x dx x a x a x a C 22 22 2 1 2 2 − = − +       + − ∫ sin
Integrals Example-1 Integrate the following functions w.r.t. x: tan sec 4 2 x x x Sol. Let I = tan sec 4 2 x x x dx        ∫  Put x = t ⇒ 1 2 x dx = dt So, I = −∫ 2 4 2 t t t t dt tan sec I =− ∫ 2 4 2 tan s t t ec dt Substitute tan t = u ⇒ sec2 tdt = du Thus, I = − ∫ 2 4 u du = 2 5 5 u + c = 2 5 5 tan t + c = 2 5 5 tan x + c Example-2 Find dx 3 1 x x3 10 2 + − ∫ Sol. Let I = dx 3 1 x x3 10 2 + − ∫ = dx 3 x x 13 3 10 3 2 + −       ∫ = 1 3 13 6 17 6 2 2 dx x +       −               ∫ Put x + 13 6 = t ⇒ dx = dt \ I = 1 3 17 6 2 2 dt ( )t −               ∫ = 1 3 2 17 6 17 6 17 6 1 × × − + log + t t c = 1 17 13 6 17 6 13 6 17 6 1 log x x c + − + + + = 1 17 6 4 6 30 1 log x x c − + + = 1 17 3 2 5 1 17 1 3 1 log log x x c − + + + = 1 17 3 2 5 log x x c − + + Where,c = 1 17 1 3 1 log + c SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) 1. f and g are functions such that f(x) = d dx (2020 – x ) and g(x) = d dx (x + 2020). Find [ (f x) ( + g x)]dx. ∫ R [CFPQ] Sol. [ (f x) ( + g x)]dx ∫ = d dx x d dx ( ) 2020 − + ( ) x d + 2020 x      ∫  = d dx x d dx ( ) 2020 − + ( ) x d + 2020 x ∫ ∫ 1⁄2 = 2020 2020 − + x x + = 4040 − + x x 1⁄2 2. Find 2 5 10 x x 1 1 x + − − ∫ dx. R [Outside Delhi Set-1, 2, 3, 2020-21] Sol. I = 2 5 1 5 ( ) ( ) 2 − − −      ∫  x x dx = − + + 2 5 5 1 5 2 2 x x C log ( )log 1 [Marking Scheme Outside Delhi Set-1, 2, 3, 2020-21] 3. Find: dx x x + ∫ R [Outside Delhi Set-1, 2020-21] 4. Find sin .cos 5 2 2 ∫             x x dx R&U [Outside Delhi Set-2, 2020-21] 5. Find e x x dx x ( c 1 ot ) 2 − + ∫ cosec R&U [SQP 2020-21] Sol. Let e x (1 – cot x + cosec2 x) f(x) = 1 – cot x This Question is for practice and its solution is given at the end of the chapter.