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Mathematics

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01_Straight_Line 02_Circle 03_Permutation_and_combination 04_Complex_numbers 05_Application_of_Derivative 06_Binomial_theorem 07_Progression__Series.pdf 08_Definite_Integral.pdf 09_Indefinite_Integral.pdf 10_Probability.pdf

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  1. DEFINITE INTEGRATION BASIC DEFINITION Let f (x) be a function defined on the closed interval [a, b] and F(x) be its antiderivative then b EVALUATION OF DEFINITE INTEGRAL BY SUBSTITUTION b Let us try out the procedure to evaluate the integral of the form : ff{g(x)}g'(x)dx Stepl : Put g(x) = t g'(x)dx = dt g(b) b Step 2: Change limits of integration in new system of variable i.e., ff{g(x)}g'(x)dx Step 3: Now evaluate the integral by conventional methods of integration and henceforth proceed forward. GEOMETRICAL INTERPRETATION OF THE DEFINITE INTEGRAL Let f(x) be a function defined on a closed interval b f (x)dx represents the algebraic [a, b]. Then sum of the areas of the regions bounded by the graph of the function y = f(x), the x axis and the straight lines x = a and x = b. The areas above the x axis enter into this sum with plus sign, while those below the x axis enter it with a minus sign. x Example 1 Solution: 6 : Evaluate: f(2x +3)dx . 2 Here, f(x) = 2x + 3 . f(6) = 15 and f(2) = 7 6 (6-2) f(2x +3)dx [15+7] 2 2 = 2 x 22 = 44. DEFINITE INTEGRAL AS THE LIMIT OF A SUM b b An alternative way of describing f f(x)dx is that the definite integral f f(x)dx is a limiting case of the b n—l = limhEf(a +rh) , where summation of an infinite series, provided f(x) is continuous on [a, b] i.e., b-1 h . The converse is also true i.e., if we have an infinite series of the above form, it can be expressed as n a definite integral. Method to express the infinite series as definite integral : Express the given series in the form E —f
  2. Definite Integration Then the limit is its sum when n +00, i.e. Replace — by x and — by dx and Limy n +00 2 1 Lim n +00 by the sign of The lower and the upper limit of integration are the limiting values of — for the first and the last n term of r respectively. Some particular cases of the above are: (a) (b) where = Lim— = 0 (as r = 1) and Lim—=p (as r = pn) n +00 n Note: The method to evaluate the integral, as limit of the sum of an infinite series is known as Integration by First Principle. Example 2 : Solution : Evaluate Lt Lt n 00 1 1 1 1 1 1 1 1 2n 1 2n n 00 1 1 = Lt n 00 r 1 1 n IMPROPER INTEGRALS 1 1 = loge2. Definite integral for which a or b or both are infinite are called improper integrals. These are evaluated in the following manner: b Lim , b b f f(x)dx Lim , f f(x)dx (iii) Another type of improper integral is that in which the integrand is not defined In both type of improper integral we take limiting values as shown in the following example. for a point Example 3 : Solution : dx Prove that f n ,(n>l). dx Let I Put x +
  3. 3 x dx 1 Also x dt = tdx Definite Integration dt 1 so that When when Hence 1 — Lim 2 + 00 1 0 2 dx = . 2t2 (n -l)tn I 1 and 1 1 1 1 1 dt 1 n PROPERTIES OF DEFINITE INTEGRATION Property 1 : Property 2 : Property 3 : The value of definite integral is independent of the change of variables i.e., b b If the limits of definite integral are interchanged then its value changes just sign only i.e., b ff(x) dx ff(x) dx b b b If a < c < b then ff(x) dx dx In general, we can restate it as : If a < Cl < c2 < ... < cn 1 < cn < b, then b b ff(x) dx —ff(x) f f(x) dx Example 4 : Solution: 4 Evaluate f min I tanx I , —x—2 dx. From the graph it is clear that the function 37t changes its definition at x = — and— 4 Hence, the given function can be written as f tanx dx + Tt/2 4 —x —2 dx It/4 cn-l 1 In I secx It 311/4 (3a/4, 1) (11/2) 311/4 371/4 4 f —x —2 dX+ f (—tanx)dx It/ 2 = In I secx + 2 371/4 Tt/2 It/ 4 9Tt 8 311/4 Tt/2 311 Tt 4-1 n u = — 4
  4. Definite Integration 4 Property 4 : Property 5 : b b ff(x) dx dx ff(x) dx [f(x) dx In particular we can say that : Iff(x) f(a—x) , then ff(x) dx 2 f f(x) dx f(a—x) , then ff(x) dx = 0 If f(x) Example 5 : Solution: Property 6 : Example 6 : Solution: x dx Evaluate 1 + cos2 x Let I = f 0 1+COS x Tt dx 1 -pcos2 x dx 1 + cos2 sec2 x dx 2+ tan2 x x (It —X) dx (by property 4 ) 01 -x) dx (by property 5) 1 + cos2 x Let tan x = t so that for x 0, t + 0 and for x Tt/2, t 00. dt Hence we can write, I = - —L Lt h tan 2 h Lt tan 0 tan ff(x) dx — dx In particular we can say that : 2ff(x) dx if f(x) is even as f(x) ff(x)dx 0 f(—x) if f(x) is odd as f(x) -k 1 -k 66 Evaluate dx . + 1 -k -k 66 Here I dx (x + +2 +20) (x +1 +22) dx -7/2 -7/2 dx (x 4-4) dx (x 4-4) dx — 7/21 (X + -k -k 4) | 7/21 (x +4) |
  5. 5 -3 Property 7 : Example 7 : Solution: Property 8 : Definite Integration -5/2 -7/2 -3 -7/2 1 -5/2 3 3 2 dx 3 3 dx -3 = 3 4-4) + 31 n 4-4) 4-3 In | 4-2 -7/2 Alternate: Put x = t —3 dx = dt 1/2 2t3 + dt which is clearly an odd function. Hence I = 0. - 4/21 t(t2 b 1 ff(x) dx — +a] dx 1 Evaluate fln(x 3)2dx +2fln(x +4)2dx . -5 1 In(x 3)2dx +2fln(x +4)2dx Let I = -5 1 1 = ( 5-3)2dx 1 1 = —2 fln4(x + 4)2dx + 2fln(x +4)2dx 1 1 1 = —2 f In 4dx — 2fln(x + 4)2dx + 2fln(x + 4)2dx If f (x) is periodic with the period T then f f(x) dx dxwhen n e I In particular we can say that : =-21n 4. Property 9 : Property 10 : Example 8 : Solution: Property 11 : If a = 0 ff(x) dx dx Ynel f f(x) dx —ff(x) dx If n = 1 f(x) dx (n dx Y m,nel b+nT b f f(x) dx ff(x) dx Evaluate f I cosx Idx . Note that Icos xl is a periodic function with period lt. Hence the given integral I =4fI cosx I dx (using property 8) 2 = 4 f cosx dx fcosx dx 2 Leibnitz's Rule [sin 2 [sin If f is continuous on [a, b] and gl(x) and g2(x) are differentiable functions whose values lie in interval [a, b] then we have :
  6. Definite Integration g2(x) d f f(t) dt g'2(x) dx 6 Property 12 : Example 9 : In particular we can say that : b If F(t) dx then as constant. 2 — ff(t) dt f(x) ög(x, t) d dx If f (x) = dt and dt = is derivative of g w.r.t. to t keeping x dx where 1, then find the value of f (2). Solution: Here f (x) = dt. Differentiating both the sides with respect to x we get, f' (x) so that f'(x) 1 1 f(x) Integrating both the sides from 0 to 2, with respect to x, we get 2 c Hence f (x) = 1 1 c c 2e 1 and f (2) = 2e 2 2 INEQUALITIES IN DEFINITE INTEGRATION Ineq—l : Ineq—2 : Example 10 : Solution : b b ff(x) dx dx Note that equality holds when f(x) is entirely of same sign on [a, b]. b b If f(x) > g(x) on [a, b] then ff(x) dx 2 fg(x) dx b In particular if f(x) > 0 then ff(x) dx > 0. 8 If f (x) is a continuous function such that f (x) > 0 V x e [2, 10] and = 0, then find f (6). 4 f(x) is above the x-axis or on the x-axis for all x e [2, 10]. If f (x) is greater than zero for any 8 8 sub interval of [4, 8], then must be greater than zero. But = 0 4 4
  7. Ineq—3 : Ineq—4 : Example 11 : Solution : Definite Integration For a given function f(x) continuous on [a, b] if we have continuous function fl(x) and f2(x) on [a, b] such that fl(x) 4— 4 > > 4—2x2 a) 4 sin dx Sin 4 (ii). Since the Hence, 4 — —x3 dx 4 — —x3 dx 4—x2 —x3 dx function f (x) dx 4 — 2x2 increases monotonically in the interval

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