Title CBSE - MATHEMATICS (PART - II).pdf ✅

S.No. Content Page 01. Chapter-7 : Integrals 01 - 46 02. Chapter-8 : Application of Integrals 47 - 68 03. Chapter-9 : Differential Equations 69 – 96 04. Chapter-10 : Vector Algebra 97 – 130 05. Chapter-11 : Three Dimensional Geometry 131 – 156 06. Chapter-12 : Linear Programming 157 – 176 07. Chapter-13 : Probability 177 - 216 08. CBSE Question Paper & Solution – 2022-23 217 – 240 09. CBSE Sample Paper & Solution – 2023-24 241 - 266
E 1 CHAPTER-7 : INTEGRALS INDEFINITE INTEGRAL 1. DEFINITION : If f(x) is derivative of F(x), then F(x) is said to be an anti-derivative or integral or primitive of f(x). Therefore differentiation and integration are inverse processes of each other. If d dx [F(x)] = f(x) then f (x) dx F(x)   Here the sign  is used for integration, x in dx denotes variable of integration, f(x) is said to be integrand and F(x) is integral. 2. INTEGRATION CONSTANT AND INDEFINITE INTEGRAL : We know that the derivative of any constant is zero, therefore d dx (C) = 0, where C is constant. If d dx [F(x)] = f(x) then d dx [F(x) + C] = f(x) Integrating both sides w.r.t. x, we get   d F(x) C dx f (x)dx dx            F(x) + C = f (x)  dx  f (x)  dx = F(x) + C Here C is said to be integration constant and value of C is not definite, so F(x) + C is called an indefinite integral of f(x). Note : It may happen that, by different methods of integration we obtain different integrals of same function but they differ from each other by a constant. 3. STANDARD FORMULAE OF INTEGRATION : Derivatives Integrals (Antiderivatives) (i) d dx (C) = 0, C is constant 0 dx C, C is constant   (ii) d dx (x) = 1 1 dx x C    (iii) n 1 d x n x dx n 1         , n –1 n 1 n x x dx C, n 1 n 1        (iv) d dx (ex ) = ex x x e dx e C    (v) x x e d a a dx log a       , a > 0, a 1 x x e a a dx C log a    , a > 0, a 1 (vi) d dx (loge |x|) = 1 x , x 0 e 1 dx log x C x    , x 0 (vii) d dx (–cos x) = sin x sin xdx cosx C    

E 3 Illustration 1: Evaluate the following integrals : (i) 1 1 cos 2x tan dx 1 cos 2x           (ii) 1 1 x x dx x x                (iii) 2 2x dx (2x 1)   (iv) sin 5x sin 3x dx  (v) cos 2x cos 2 dx cos x cos      Solution: (i) Let I = 2 1 1 2 1 cos 2x 2sin x tan dx tan dx 1 cos 2x 2cos x                    =  tan–1 (tan x) dx (We are assuming that 0 < x < 2  ) = 1 2 x dx x C 2    . (ii) Let I = 1 1 x x 1 x x dx x x dx x x x x x x                           = 3/2 1/2 1/2 –3/2 x dx x dx x dx x dx         = 2 2 5/2 3/2 2 x x 2 x C 5 3 x     (iii) Let I =   2 2 2 2x (2x 1) 1 1 1 dx dx dx (2x 1) (2x 1) 2x 1 (2x 1)                   1 log 2x 1 (2x 1) C 2 ( 1) 2         1 1 log 2x 1 C 2 (2x 1)            (iv) Let I = sin 5x sin 3x dx   1 2 2sin 5x sin 3x  dx = 1 2 cos 2x – cos 8x    dx = 1 2 sin 2x sin8x C 2 8         = 1 16 (4sin 2x – sin 8x) + C. (v) Let I = 2 2 cos 2x cos 2 (2cos x 1) (2cos 1) dx cos x cos cos x cos              dx = 2  (cos x + cos ) dx = 2 sin x + 2x cos  + C Illustration 2: If f '(x) = x2 – 2 1 x and f(1) = 1 3 , find f(x) Solution: Given f '(x) = x2 – 2 1 x 2 2 1 f (x) x dx x          [Integration on both sides]  f(x) 2 –2  (x – x )dx  3 1 x x 1 1 3 C x C 3 1 3 x         . Also f(1) = 1 3  1 3 × 13 + 1 1 + C = 1 3  1 3 + 1 + C = 1 3 C = –1 f(x) = 1 3 x 3 + 1 x – 1