Elementary Mathematics Past Paper PDF

1 ASPEE COLLEGE OF AGRICULTURE JUNAGADH AGRICULTURAL UNIVERSITY KHAPAT – PORBANDAR THEORY MATERIAL FOR FIRST SEMESTER B. Sc. (Hons.) Agri. COURSE No.: MATHS 1.1 (2+0) TITLE OF COURSE: ELEMENTARY MATHEMATICS PREPARED BY Ms. TR...

1 ASPEE COLLEGE OF AGRICULTURE JUNAGADH AGRICULTURAL UNIVERSITY KHAPAT – PORBANDAR THEORY MATERIAL FOR FIRST SEMESTER B. Sc. (Hons.) Agri. COURSE No.: MATHS 1.1 (2+0) TITLE OF COURSE: ELEMENTARY MATHEMATICS PREPARED BY Ms. TRUPTI J. LAKHLANI  2 CH-1. LIMIT, FUNCTION & DERIVATIVE  FUNCTION:- “ The relation between two variables or(two quantities) is known as a Function.”  How one quantity depends on other quantity, can be represented as Function.  WELL DEFINED FUNCTION:- “ If function is valid for all numbers then it is said to be Well defined function.”  The point where function become ‘ꚙ’ , is said to be Discontinuity Point. 2+𝑥 Ex. (i) f(x) = is not well defined at x=5. i.e. f(x) is discontinuous at x=5. 𝑥−5 𝑥 2 +2 (ii) f(x) = is not well defined at x =-9 & x =2. 𝑥 2 +7𝑥−18 (iii) f(x) = x2 – 2x + 17 is well defined function.  DOMAIN:- “For any function set of all permissible inputs is said to be Domain of that function.”  CO-DOMAIN:- “For any function set of all possible outputs is said to be Co-Domain of that function.”  RANGE:- “For any function set of all answers(outputs) is said to be Range of that function.” f(x) = x+1 Domain Image Co-Domain D = {1,2,3,4,5} Co-D.={1,2,3,4,5,6,7} Rg = {2,3,4,5,6} 1 1 2 2 3 3 F(1) =2  2 is image, 1 is pre-im. 4 4 F(2) =3 3 is image, 2 is pre-im. 5 5 ⁞ Pre- Image 6 F(5) =6  6 is image, 5 is pre-im. 8 6  IDENTITY FUNCTION:- “ If image of any number under the function ‘f’ is itself, then function is an Identity function.”  F(x) = x is Identity function. 6 Q-1 EVALUATE FOLLOWING FUNCTIONS. 1−𝑥 1 (1) If f(x) = x2+2x+1, then find f(x+1) – f(x-1). (3) If f(x) = , then find f(x) + f( ). 1+𝑥 𝑥 𝑥 1 (2) If f(x) = 3x2+9x+5, then find f(x+1) + f(2x). (4) If f(x) = , then find f(x) + f( ). 𝑥+1 𝑥 (5) For what values of x, f(x) = x2-3x+2 is zero?  3 L I M I T :- sin 𝑥 tan 𝑥 𝑒 𝑥 −1 lim =1 lim =1 lim =1 𝑥→0 𝑥 𝑥→0 𝑥 𝑥→0 𝑥 Q-2 FIND THE LIMIT OF FOLLOWINGS. 3𝑥 2 −4𝑥+1 𝑥 3 +1 (1) lim 𝑥 3 − 𝑥 2 + 1 (11) lim (21) lim 𝑥→1 𝑥→1 𝑥 2 −4x+3 𝑥→0 2𝑥 2 +5 2𝑥 3 −16 (2) lim 𝑥 (𝑥 + 1) (12) lim (22) lim 𝑥 ∙ 𝑠𝑒𝑐 𝑥 𝑥→3 𝑥→2 𝑥−2 𝑥→0 𝑥 2 +1 𝑥 2 −𝑥−6 2𝑥 2 +1 (3) lim (13) lim (23) lim 𝑥→1 𝑥+100 𝑥→3 𝑥−3 𝑥→∞ 6+𝑥−3𝑥 2 𝑥 2 −9 3𝑥 2 −4 (4) 𝒙→−𝟏 𝐥𝐢𝐦 (𝟏 + 𝒙 + 𝒙 𝟐 𝟏𝟎 + ⋯+𝒙 ) (14) lim (24) lim 𝑥→3 2𝑥 2 −5x−3 𝑥→∞ 2𝑥 2 −1 𝑥 2 −25 3𝑥 2 +1 (5) lim 𝑥 + 3 (15) lim (25) lim 𝑥→0 𝑥→5 x−5 𝑥→∞ 2𝑥 3 +6 𝑥 3 −8 2𝑥 3 −5𝑥 2 +𝑥−1 (6) lim 𝛱𝑥 2 (16) lim (26) lim 𝑥→1 𝑥→2 2𝑥 2 −8 𝑥→∞ 𝑥 3 −4𝑥 2 +3𝑥+1 𝑎𝑥 2 +𝑏𝑥+𝑐 2𝑥 2 +𝑥−3 𝑠𝑖𝑛 4𝑥 (7) lim (17) lim (27) lim 𝑥→1 𝑐𝑥 2 +𝑏𝑥+𝑎 𝑥→1 3𝑥 2 −4x+1 𝑥→0 𝑠𝑖𝑛 2𝑥 𝑥 2 −4 3𝑥 2 +𝑥−4 𝑠𝑖𝑛 𝑎𝑥 (8) lim (18) lim (28) lim 𝑥→2 𝑥 3 −4𝑥 2 +4x 𝑥→1 2𝑥 2 −3x+1 𝑥→0 𝑏𝑥 𝑥 3 −2𝑥 2 (𝑥+ℎ)2 −(𝑥)2 𝑠𝑖𝑛 (𝛱−𝑥) (9) lim (19) lim (29) lim 𝑥→2 𝑥 2 −5𝑥+6 ℎ→0 ℎ 𝑥→𝛱 𝛱(𝛱−𝑥) 3𝑥 2 −𝑥−10 𝑥 2 −5 𝑡𝑎𝑛 7𝑥 (10) lim (20) lim (30) lim 𝑥→2 𝑥 2 −4 𝑥→1 x+1 𝑥→0 20𝑥 √ 16𝑥 4 −8𝑥 3 +5𝑥 2 −2𝑥+1 (31) lim 𝑥→∞ −3𝑥 2 +2𝑥−4  4 C O N T I N U I T Y:- Condition of continuity: LHL = RHL = f(a) i.e. 𝐥𝐢𝐦− 𝐟(𝐱) = 𝐥𝐢𝐦+ 𝐟(𝐱) = f(a) 𝒙→𝒂 𝒙→𝒂 Q-3 FIND THE LIMIT OF FOLLOWINGS. 𝒙𝟐 −𝟗 2𝑥 + 3, 𝑥 ≤ 0 , 𝑥≠3 𝒙−𝟑 (1) 𝑓 (𝑥 ) = { (8) 𝑓 (𝑥 ) = { 3(𝑥 + 1), 𝑥 > 0 3 , 𝑥=3 𝒙𝟐 −𝟏𝟔 , 𝑥 ≠ −4 1 + 𝑥 2, 𝑥 ≤ 1 𝒙+𝟒 (2) 𝑓 (𝑥 ) = { (9) 𝑓 (𝑥 ) = { 2 − 𝑥, 𝑥 > 1 −8 , 𝑥 = −4 𝒙𝟐 −𝟐𝒙+𝟏 , 𝑥≠1 3𝑥 − 2, 𝑥 < 0 𝒙−𝟏 (3) 𝑓 (𝑥 ) = { (10) 𝑓 (𝑥 ) = { 𝑥 + 1, 𝑥 ≥ 0 2 , 𝑥=1 𝒙𝟐 −𝟏 , 𝑥≠1 2𝑥 − 1, 𝑥 < 0 𝒙+𝟏 (4) 𝑓 (𝑥 ) = { (11) 𝑓 (𝑥 ) = { 2𝑥 + 1, 𝑥 ≥ 0 2 , 𝑥=1 𝒙𝟐 −𝟒 , 𝑥≠2 𝒙−𝟐 𝑥 + 3, 𝑥 < 2 (5) 𝑓 (𝑥 ) = { (12) 𝑓 (𝑥 ) = { −𝑥 + 3, 𝑥 ≥ 2 4 , 𝑥=2 𝒙𝟐 −𝟏 , 𝑥≠1 𝒙−𝟏 𝑥 2 − 1, 𝑥 ≤ 1 (6) 𝑓 (𝑥 ) = { (13) 𝑓 (𝑥 ) = { −𝑥 2 − 1, 𝑥 > 1 2 , 𝑥=1 𝒙𝟒 −𝟏𝟔 , 𝑥≠2 𝒙, 𝒙 < 𝟏/𝟐 𝒙−𝟐 (7) 𝑓 (𝑥 ) = { (14) 𝑓 (𝑥 ) = { 𝟏/𝟐, 𝒙 = 𝟏/𝟐 32 , 𝑥 = 2 𝟏 − 𝒙, 𝒙 > 𝟏/𝟐  5 Q-4 FIND THE UNKNOWN NUMBER IF FOLLOWINGFUNCTIONS ARE CONTINUOUS. 𝑎𝑥 + 5, 𝑥 ≤ 2 𝑘𝑥 + 1, 𝑥 ≤ 5 (1) 𝑓 (𝑥 ) = { (4) 𝑓 (𝑥 ) = { 𝑥 − 2, 𝑥 > 2 3𝑥 − 5, 𝑥 > 5 𝑥2 + 2 , 𝑥 < 0 𝑥2 + 2 , 𝑥 < 0 (2) 𝑓 (𝑥 ) = { 𝑎𝑥 + 𝑏 , 0 ≤ 𝑥 < 1 (5) 𝑓 (𝑥 ) = { 𝑎𝑥 + 𝑏 , 0 ≤ 𝑥 < 1 3 + 2𝑥 − 𝑥 2 , 𝑥 ≥ 1 3 + 2𝑥 − 𝑥 2 , 𝑥 ≥ 1 𝑎 + 𝑏𝑥, 𝑥 < 1 (3) 𝑓 (𝑥 ) = { 4 , 𝑥 = 1 𝑏 − 𝑎𝑥, 𝑥 > 1 EXTRA: 2𝑥 2 − 3, 𝑥 > 2.5 Given 𝑓(𝑥 ) = { then find the values of f(2.5), f(-0.5), f(5), 4 + 𝑥 2 , 𝑥 ≤ 2.5 f(√3), f(-4), f(0), f(√10).  6 D I F F E N T I A T I O N:- 𝒅 𝒅 𝒅 Formulae: (𝒄) = 𝟎 (𝑒 𝑥 ) = 𝑒 𝑥 (sin 𝑥 ) = cos 𝑥 𝒅𝒙 𝒅𝒙 𝒅𝒙 𝒅 𝒅 𝒅 (𝑥 ) = 𝟏 (𝑎 𝑥 ) = 𝑎 𝑥 ∙ log 𝑎 (cos 𝑥 ) = −sin 𝑥 𝒅𝒙 𝒅𝒙 𝒅𝒙 𝒅 𝒅 1 1 𝒅 (𝑥 𝑛 ) = 𝑛 ∙ 𝑥 𝑛−1 ( )=− (tan 𝑥 ) = 𝑠𝑒𝑐 2 𝑥 𝒅𝒙 𝒅𝒙 𝑥 𝑥2 𝒅𝒙 𝒅 1 𝒅 1 𝒅 (log 𝑥 ) = (√𝑥) = (sec 𝑥 ) = sec 𝑥 ∙ tan 𝑥 𝒅𝒙 𝑥 𝒅𝒙 2√𝑥 𝒅𝒙 𝒅 (cot 𝑥 ) = −cosec 2𝑥 𝒅𝒙 𝒅 (cosec 𝑥 ) = −cosec 𝑥 ∙ cot 𝑥 𝒅𝒙 𝒅 𝒅 𝒅 Rules: (𝑓 ± 𝑔) = (𝑓 ) ± (𝑔) 𝒅𝒙 𝒅𝒙 𝒅𝒙 𝒅 (𝑓 ∙ 𝑔) = 𝑓 ′ 𝑔 + 𝑔′ 𝑓 (Product Rule) ( x) 𝒅𝒙 𝒅 𝑓 𝑔 𝑓′ −𝑓 𝑔′ ( )= (Quotient Rule) (÷) 𝒅𝒙 𝑔 𝑔2 𝑑𝑦 Q-5 FIND. (Simple Derivative) :- 𝑑𝑥 4 2 𝑥 2 x 𝑥 2 −1 (1) y = 2x – 23x + sin x + 𝑒 (7) y = x ∙ e (13) y = 𝑥 2 +1 𝑒 𝑥 𝑙𝑜𝑔𝑥 (2) y = x ∙ logx – 3 ∙ ex ∙ cos x (8) y = x3 – 27 (14) y = a−cos 𝑥 2𝑥+3 𝑥−3 (3) y = 5 sec x + 4 cos x (9) y = (15) y = 𝑥−2 𝑥 3 +5 𝑥+1 3−sin 𝑥 (4) y = 4√x - 2∙ logx + 3x (10) y = (16) y = 𝑥−1 3+cos 𝑥 𝑥 (5) y = sin x ∙ cos x (11) y = 1+tan 𝑥 1 1 2−3 tan 𝑥 (6) y = (x+ ) (√x + ) (12) y = x √x 3+2cot 𝑥 𝑑𝑦 Q-6 FIND. (Implicit function) :- 𝑑𝑥 (1) xy = c2 (4) 2x + 3y = sin x (7) 4x + 3y = logx – 3y (2) x2 +y2=100 (5) x5 +y5 = 5xy (8) y3 – 3xy2 = x3 + 3x2y (3) 2x + 3y = sin y (6) xy + y2 = tan x + y  7 𝑑𝑦 Q-7 FIND. (Chain Rule) :- 𝑑𝑥 (1) y = sin(3x+5) (5) y = sin2(2x-1) (9) y = etan√x (2) y = sin(logx) (6) y = [log (sin x)]2 (10) y = x10+ 2x+ logx – cos x2+ e2x (3) y = tan2x (7) y = e3x ∙cos 2x (4) y = log(2x-3) (8) y = log (cos x2) 𝑑𝑦 Q-8 FIND. (Logarithmic function) :- 𝑑𝑥 Hint: Formule: log xn = n logx If y = u + v 𝑑𝑦 𝑑𝑢 𝑑𝑣 log (x∙y) = logx + logy = + 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥 log ( ) = logx – logy 𝑦 (1) y = cos x ∙ cos 2x ∙ cos 3x (5) y = (sin x) x (9) y = x y (𝒙−𝟏)(𝒙−𝟐) (𝒙+𝟐)(𝒙−𝟓)𝟐 (2) y = √ (6) y x = x y (10) y = (𝒙−𝟑)(𝒙−𝟒)(𝒙−𝟓) (𝒙+𝟕)𝟓 (3) y = (x+3)2 (x+4)3 (x+5)4 (7) xy = e x - y (4) y = (logx)cos x (8) y = x x – 2sin x 𝑑2𝑦 Q-9 FIND. (2nd Order Derivative) :- 𝑑𝑥 2 (1) y = 12x3 – 28x2 +15x +19 (4) y = sin x – logx + ex (2) y = tan x + e2x + x16 –logx (5) y = x-3 (3) y = log (2x-3)6  8 M A X I M A & M I N I M A :- (1) f(x) is given. (3) f’’(x) < 0 f’’(x) > 0 f’’(x) = 0 (2) Take f’(x) = 0. Maximum. Minimum. Test fails. Q-10 FIND MAXIMUM & MINIMUM VALUES :- (1) f(x) = 3x4 – 10x3 +6x2 + 5 (7) f(x) = x3 – 3x2 + 3x – 25 (2) f(x) = x3 – 6x2 + 9x + 2 (8) f(x) = 4x2 – 4x + 4 (3) f(x) = 2x3 – 15x2 + 36x + 10 (9) f(x) = x3 – 6x2 + 9x + 15 (4) f(x) = 2x3 – 15x2 + 36x + 1 (10) f(x) = 2x3 – 24x + 107 in [1,3] (5) f(x) = 2x3 – 15x2 + 36x + 17 (11) h = 1/3 (1 + t∙ e -2t + 2 ) (6) f(x) = 4x3 + 9x2 – 12x + 5 Q-11 WORD PROBLEMS ON MAXIMUM & MINIMUM VALUES :- (1) Find two numbers whose sum is 16 and whose product is the largest possible. (2) Find two positive real numbers whose sum is 60 and product of one of them with the cube of the other is maximum. (3) A farmer wishes to enclose a rectangular paddock using only 100 meter of fencing. What is the largest area can be enclosed? (4) A farmer wishes to enclose a rectangular paddock using only 200 meter of fencing. What is the largest area can be enclosed? (5) A farmer wishes to enclose a rectangular paddock using only 150 meter of fencing. What is the largest area can be enclosed? (6) A farmer wishes to enclose a rectangular paddock using only 600 meter of fencing. What is the largest area can be enclosed? (7) Organic waste, when deposited in a lake, decreases the oxygen content of water. If t = time In days, after the waste deposited and oxygen content given by y = t3 – 30t2 + 6000 for 0 ≤ t ≤ 20. Find maximum and minimum values of y. (8) Prove that f(x) = sin x (1+ cos x) is maximum at x = Π/3.  9 GROWTH & D E C A Y :- 𝒇(𝒕𝟐 )−𝒇(𝒕𝟏 ) ∆t = t 2 − t1 Change = +ve Change = Growth 𝒕𝟐 − 𝒕𝟏 -ve Change = Decay Q-12 WORD PROBLEMS ON GROWTH & DECAY :- 6000 (1) The size of an bacteria population at time t(in days) is given by p(t) = 1000 +. 1+𝑡 2 Determine the change when t=2 and ∆t = 3. (2) p(t) = 100 (t+3)2. Find the average rate of growth between t=2 and t=5. (3) p(t) = 5000 + 3000t – 2000t2. Find (a) Initial population and (b) population after one hour. 4000 (4) p(t) = 6000 -. Find the average rate of growth when t=3 and ∆t = 2. 2+3𝑡−𝑡 2 600 (5) Find the increments of function f(t) = 1000 + for the intervals t = 2 & ∆t = 1. 1+𝑡 2 (6) The size of a bacteria population at time ‘t’ is given by p(t) = 10,000 + 1000t – 120t2. Determine the average rate of growth between t = 3hr and t = 5hr. (7) The yield of fruit from each tree of ‘Chiku’ orchard decreases as the density, which the Trees are planted increases, when there are ‘n’ trees per acre, the average number of Chikus per tree is known to be equal to (900 – 10n) for a particular variety os Chiku. What value of ‘n’ gives the maximum total yield? 3 (8) The average growth rate of certain population of bacteria is given at time ‘t’ by 2𝑡+5 mg/h. If the size of the population is 7mg at t = 0, calculate the weight at general time t. 3 (9) The average growth rate of certain population of bacteria is given at time ‘t’ by 𝑡−3 mg/h. If the size of the population is 7mg at t = 0, calculate the weight at general time t. 2 (10) The average growth rate of certain population of bacteria is given at time ‘t’ by 𝑡+2 mg/h. If the size of the population is 5mg at t = 0, calculate the weight at general time t. EXTRA: (1) Find velocity and acceleration at t = 0. If distance is given by S = 9t + 16t2 – 3t3.  10 CH - 2. INTEGRATION Formulae: ∫ 0 𝑑𝑥 = C ∫ sin 𝑥 𝑑𝑥 = – cos x + c ∫ 1 𝑑𝑥 = x + c ∫ cos 𝑥 𝑑𝑥 = sin x + c ∫ 𝑘 𝑑𝑥 = kx + c ∫ 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = tan x + c ∫ 𝑥 𝑛 𝑑𝑥 = 𝑛 ∙ 𝑥 𝑛−1 + c ∫ 𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = – cot x + c 1 ∫ 𝑑𝑥 = log x + c ∫ sec 𝑥 ∙ 𝑡𝑎𝑛𝑥 𝑑𝑥 = secx + c 𝑥 ∫ 𝑒 𝑑𝑥 = 𝑒 𝑥 + c 𝑥 ∫ 𝑐𝑜𝑠𝑒𝑐 ∙ 𝑥 cot 𝑥 𝑑𝑥 = – cosec x + c ∫ 𝑎 𝑥 𝑑𝑥 = 𝑎 𝑥 ∙ log a + c 𝐹(𝑎𝑥+𝑏) Rules: ∫ 𝑓 ± 𝑔 𝑑𝑥 = ∫ 𝑓 𝑑𝑥 ± ∫ 𝑔 𝑑𝑥 ∫ 𝑓(𝑎𝑥 + 𝑏) 𝑑𝑥 = 𝑎 +c ∫ 𝑘 ∙ 𝑓 𝑑𝑥 = 𝑘 ∙ ∫ 𝑓 𝑑𝑥 Q-1 FIND THE INTEGRATION (SIMPLE FUNCTION). 1 (𝑥+1)2 (1) ∫ 𝑥 2 + 5𝑥 − 7 𝑑𝑥 (9) ∫ √𝑥(𝑥 3 − ) 𝑑𝑥 (17) ∫ 𝑑𝑥 𝑥 𝑥 1/2 2 1 (2𝑥+1)2 (2) ∫ 3𝑥 2 + + + 3𝑥 𝑑𝑥 (10) ∫ (2𝑥 + 1)3 𝑑𝑥 (18) ∫ 𝑑𝑥 𝑥 𝑥2 𝑥 1 𝑥 3 −8 (3) ∫ 4𝑥 3 − + sin 𝑥 − 𝑒 𝑥 𝑑𝑥 (11) ∫ (𝑥 2 − 4)2 𝑑𝑥 (19) ∫ 𝑑𝑥 𝑥 𝑥 2 −2𝑥 5 1 𝑥 2 +𝑥−2 (4) ∫ 2𝑥 + − 𝑑𝑥 (12) ∫ (𝑥 + 5)2 𝑑𝑥 (20) ∫ 𝑑𝑥 𝑥 𝑥 1/3 𝑥−1 3𝑥 2 +5𝑥+6 𝑥 2 +5𝑥+6 (5) ∫ 𝑥 5/4 − 2𝑥 1/9 𝑑𝑥 (13) ∫ 𝑑𝑥 (21) ∫ 𝑑𝑥 𝑥 𝑥 2 +2𝑥 1 𝑥 5 +𝑥 −2 +2 (6) ∫ √𝑥 + + 3 𝑑𝑥 (14) ∫ 𝑑𝑥 √𝑥 𝑥2 𝑥 −7𝑥 2 +3𝑥−1 3 (7) ∫ 𝑥 4 + 12𝑥 + 83 𝑑𝑥 (15) ∫ 𝑑𝑥 √𝑥 1 2 (1−3𝑥)2 (8) ∫ (√𝑥 + ) 𝑑𝑥 (16) ∫ 𝑑𝑥 √𝑥 𝑥3 Q-2 FIND THE INTEGRATION (TRIGO. FUNCTION). 2+3 sin 𝑥 1 (1) ∫ 7 sin 𝑥 − 5 cos 𝑥 + 𝑠𝑒𝑐 2𝑥 𝑑𝑥 (7) ∫ 𝑑𝑥 (11) ∫ 𝑑𝑥 𝑐𝑜𝑠 2 𝑥 1+sin 𝑥 3 sin 𝑥 cos 𝑥 1 (2) ∫ sec 𝑥(sec 𝑥 + tan 𝑥) 𝑑𝑥 (8) ∫ + 𝑑𝑥 (12) ∫ 𝑑𝑥 2𝑐𝑜𝑠 2 𝑥 𝑠𝑖𝑛2 𝑥 1−sin 𝑥 𝑠𝑖𝑛3 𝑥+𝑐𝑜𝑠 3 𝑥 1 (3) ∫ 𝑡𝑎𝑛2 𝑥 𝑑𝑥 (9) ∫ 𝑑𝑥 (13) ∫ 𝑑𝑥 𝑠𝑖𝑛2 𝑥 ∙ 𝑐𝑜𝑠 2 𝑥 1+cos 𝑥 𝑠𝑒𝑐 2 𝑥 1 (4) ∫(tan 𝑥 − cot 𝑥 )2 𝑑𝑥 (10) ∫ 𝑑𝑥 (14) ∫ 𝑑𝑥 𝑐𝑜𝑠𝑒𝑐 𝑥 1−cos 𝑥 4+3 cos 𝑥 3𝑠𝑖𝑛2 𝑥−2 (5) ∫ 𝑑𝑥 (6) ∫ 𝑑𝑥 𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥  11 Q-3 FIND THE INTEGRATION (BY PARTS). FORMULA: ∫ 𝑢 ∙ 𝑣 𝑑𝑥 = 𝑢 ∙ ∫ 𝑣 − ∫[𝑢′ ∫ 𝑣 ] I Inverse L Log (1) ∫ log 𝑥 𝑑𝑥 (7) ∫ 𝑥 ∙ sin 𝑥 𝑑𝑥 A Arithmetic (2) ∫ 𝑥 ∙ log 𝑥 𝑑𝑥 (8) ∫ 𝑥 ∙ tan2 x 𝑑𝑥 T Trigo. (3) ∫ 𝑥 2 ∙ log 𝑥 𝑑𝑥 (9) ∫ 𝑥 ∙ 𝑒 𝑥 𝑑𝑥 E Exponential (4) ∫ √𝑥 ∙ log 𝑥 𝑑𝑥 (10) ∫ 𝑥 2 ∙ 𝑒 𝑥 𝑑𝑥 (5) ∫ 𝑥 2 ∙ cos 𝑥 𝑑𝑥 (11) ∫ 𝑥 ∙ 𝑒 −2𝑥 𝑑𝑥 (6) ∫ 𝑥 ∙ cos 𝑥 𝑑𝑥 (12) ∫ 𝑥 2 ∙ 2𝑥 𝑑𝑥 Q-4 FIND THE INTEGRATION (DEFINITE INTEGRALS). 5 −3 𝛱/3 (1) ∫2 𝑥 2 𝑑𝑥 (8) ∫−4 1⁄𝑥 4 𝑑𝑥 (15) ∫0 𝑡𝑎𝑛2 𝑥 𝑑𝑥 4 3 𝛱/2 (2) ∫0 𝑥 3 𝑑𝑥 (9) ∫−3 𝑥 5 𝑑𝑥 (16) ∫0 cos 𝑥 𝑑𝑥 4 4 𝛱/6 (3) ∫2 √𝑥 𝑑𝑥 (10) ∫1 1⁄ 𝑑𝑥 (17) ∫0 sin 2𝑥 𝑑𝑥 √𝑥 4 7 𝛱/2 (4) ∫2 𝑒 𝑥 𝑑𝑥 (11) ∫0 𝑥 2 + 6 𝑑𝑥 (18) ∫0 (3 cos 𝑥 − 4 sin 𝑥) 𝑑𝑥 3 3 𝛱/3 (5) ∫1 1⁄𝑥 𝑑𝑥 (12) ∫1 𝑥 2 + 2𝑥 + 5 𝑑𝑥 (19) ∫0 3 cos 3𝑥 − 4 sin 3𝑥 𝑑𝑥 3 𝛱/4 (6) ∫2 𝑥 2 + 𝑥 + 1 𝑑𝑥 (13) ∫0 𝑥 ∙ cos 𝑥 𝑑𝑥 2 𝛱/2 (7) ∫1 9𝑥 2 + 4𝑥 − 2 𝑑𝑥 (14) ∫0 𝑥 ∙ sin 𝑥 𝑑𝑥 Q-5 FIND THE AREA UNDER THE CURVE. (1) y = x between x = 0 and x = 4. (11) y = 2x + 4 between y = −5 and y = 1. (2) y = x−4 between x = −2 and x = 2. (12) y = x − 4 between y = 1 and y = 5. (3) y = 2x2 between x = 3 and x = 5. (13) x = y2 − 4 between y = 1 and y = 5. (4) y = 3x2 between x = 2 and x = 3. (14) y = 4x2 between x = 2 and x-axis. (5) y = x2 +1 between x = 0 and x = 4. (15) y = x2 −7x+10 and x- axis. (6) y = x2 +x+1 between x = 1 and x = 2. (16) y = 4x − x2 −3 and x- axis. (7) y = x2 −4x+1 between x = 1 and x = 2. (17) y = x2 −4x+5 betn x = 3, x = 6 & x-axis. (8) y = 3x2 betn x = −2, x = 2 and x-axis. (18) y = x2 – x betn x = –2, x = 2 & x-axis. (9) y = x2 +2x+1 betn x = 1, x = 4 & x-axis. (10) y = 2x2 betn x = 1, x = 2 and x-axis.