Edexcel A Level Maths Integration Year 1/AS

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(x ²4+ 2x V)__ -1²2² +22²²² 1/2 3/2 =25x+4x¾/12 11: Find (8x²+6x) dx, giving each term in its simplest form. S (8₂²³46x - 3) dx = √ (8x² + 6x - 3z¹/²)dx 112 || (8x²³ +6x-3x²¹¹3) dx = 82² +6₂²-3x²¹² +c 4 2 1/2 2x+3x²-6x¹2 +c 2x4+3x²-6√√x tc✓. 12: a) Show that (2+5√5)² can be written as 4+K5x+25x, where it is a constant to be found. (2+55x)² = 4+20√√x +25x K=20 b) Hence find S (+55) ²dx $(4+20√x + 25x)= $(4+20x²+25a) dx 5 (4+20x²2² +25x) dx = 4x + 20x²³/²2² +25x²³² +c 3/2 13: Given that y = 3x²³ -41, x20, find & ydx in its simplest form. √ (3x²³-4) dx = √ (32²³-42= √2) dx .4x $(3x³- (3x³-4x ¹/²) dx = 3x6-4x²/2 1/2 6 14 1 62² + pq) dx = 2 + 10x + -2P = 2 x = 4x + •40 x * ² + 25x²³+ ✓ 3 -2p=2 P= -1 P9x = 10x (1gx=10x (-1) = 10 9 = -10 Find the value of of p and the value of q. $(² + pq) dx = √ (2₂_² +pq) da S (P₂²² + pq) dx = Px²² + pqx +c✓ -2 = 1x6-85x+c ✓ +C =-2е трух те x = P tpqx tc 2x = 2 +10x+c -P=2 2x x 1=2 and pq=10 -올 P = -4 9==25 9=10=(4) 05/01/21 Finding Functions (+C) integrate function substitute coordinates at a point on the curve solve equation to find c to find constant of integration, C: - Examples: 1) The curve * with equation y=f(od passes through the point (2,15). Given that f '(x) = 5xx ²3x, find the equation of X f(x)=51² +31² + ( 3 2 f(2)=15 15 = 5(2)² +3(2)² + 2 2) The curve X with equation y=f6d passes through the point (4,5) Given that f'(=x²-2, find the equation of X 15= 58 +( c = - 4 1/3 + +6x) = 522²³² + 322²-41/3 3 1'(x)-2² 1/2 512 f(x) = x³²² - 2x²¹2 112 312 y = 3x²² -27² + 1/2 3/2 y=652-4x²²2² +₁ 10= 65W-4 (4)2+( 10=-68+c (=48✓ 5 8(x) = = 1x²² S= 5/2 1/2 (512_45x+c S= 2 (4512-45(4) + ( 13C 4: The curve with equation y=f(x) passes through the point (-1,0). Given that f'(x)=9x²14x-3, find f(x). f(x) = 9x²³² +4x²-3x+( 24 +C 5 16x)= 3x²³²+2x²-3x+c O3(-1)³ +26-1) ²-3(-1) + 0 = 24c C = -2 f(x)= 3x²³+2x²-3x-2/ 4(4)=5 5: dy = 3x¹/2 - 2x5x₁x>0. Given that y=10 atx=4, find y in terms of x insimplest form. y = 65x-42²³²/²2 +118 C= = 1/5 so f (x) = 3/5 2²2/²2-42 ²/³+1/5 6: Given that 6x +5x³2 can be written in the form 6x² + 5x², Sx a) Write down the value of p and the value of 9. 6x15x²³¹2=6x¹/2 +Sx 5x 1/2 and P= 9=1 7: The displacement of a particle at time t is given by the function f(t), where f(C)=0. Given that the velocity of the particle is given by &"(t)= 10-st, a Find flt) f(0) = 10t-st² 2 b) Determine the displacement of the particle when t=3 f(3)=10(3)-S (3) ² 2 (3)-7.5 @ Find f(t) ✓ 8: The height, in metres, of an arrow fired herizontally from the top of a castle is mochelled by the function fit), where f(0) = 35. Given that f'(t) = -9.8t, f(t)=-49+²+35 f(0) = -9.8+² + +C f(t) = -4.9+² +C✓ b) Determine the height of the arrow when t= 1.5. f(1.5)=-4.9(1.5) ² +35 35=-4.9+( C = 35 height of the arrow = - 441 +35 40 -35=-4.94² +²=50 7 = 23.975 m I Write down the height of the castle according to this model. H(0)=35m height of castle= 35m d) Estimate the time it will take the arrow to hit the ground. O= -4.9+² +35 time. - arrow to hit the ground = 2.67s +2+ 2.67 e) state one assumption used in your calculation. We assume the ground is flat. = (-12 + 200²) - (- 12+26²) - 23-1 = 5.25 (2x + 3x) dx 13D 2: Evaluate the following definite integrals. a) S² ( 23 + 3x) dx = = S 5² (2x² ³ + 3x) dx = [2x²² + 32²³+ ( -2 = [+1 +2 + ²x² ] ₁ b) [² (2₂2² 4²x+5) dx = [24² - 42²° +52] -2 (2)² +5(1)-(260³4-2 (0)²4+5(0)) (2(24- 4 = 10-0 = 10 S (52-6) dx = [2²²-6x²27) = [² x * 2² + 4] ³ -1 24 (3 (39³/² + 6)-(3 (0)³ ² +6 ) = 56 - 44 6 =71/6/ 2 :d) 3: Evaluate the following definite integrals: @ 5²_²³²+²=² dx = [²2² +² ] ₁ (13)² + (30³²) - (43³² + (1²7) = 18 - 12/25 = 50/3 [ 2²2² +²²] = [ +*²*² + 2²] -Jo (4 (1)/2 +1²) - (20/² + (0)² - 11/14-0 - 11/14 54 = 6: Evaluate 4] 312 [42²¹12 - Ag" 1 d) 5² ( x ² ² + ²x -1 ) dx = [2 ²1² + 2x² - XI] 43 4. Given that A is a constant and f4 (652-A) doe= A², show that there are two possible velves for and fled these values. (65x-AE622 (4243³12 - A²) - (414312-43) =(3²-42) - (4 -44-) (232 (8) 4³5+ (85²-18) - (2 4²³ + ² - (0)=62-¾/2 -60.5 b) 5 (x-²)² dx = [2² - 6x + 0x²] ² (40³-216)-6(6)) 2 24 Sx 9.¹2 5²² ½ do = 5t² (2x²42) 3. 24 J te (20x²42) = ([282944 24 47712 = ((9²-2(9³)²-(².2(15³2) = 27 (1) = 28 [45] 4 = (4512)- (454) = -8 +8√3. a=-8 b= 8√ d 2+√x dx = S2+ St $" (2x²+₂x²³¹²)dx 1 5: Use calculus to find the value of $³(2x-3√52)de 5² (2x-3570) dx = [22² - 32²x127 3² 3/2 J1 =[₂²-2x²³²³1 (6)-7) -(3)³ - 2(3) -6 (3)-1)=-59-1 46.5 [2x² + x ना74 -4274 = 1/72] 1 = [ 72²2 - 2 x 47 4 1 ( 2-2(4 5¹1) - (-Z-2105²¹12) = -22-(-4) = $V dx, giving your answer in the form at b53, where a and bare integers. = 28-3A = 4² -34128=A² A= -7 A-+257 A-4 08 Jours • A = JPGc) dar Dreas Ender lowes Example: 1) Find the area of the finite region between the curve with equation. y=20-x-x² and the x-abis. y=(4-x)(8+x) A= 160-x-2x - [ox -22² ] = = (2014) - (W² - (U²³) - (20 (5) - (-52²-(3)³) A = 243 units² 2 1: The finite region R, is bound by the x-axis and the curve with equation y= 27-2x-95x-16 270 -16 The curve crosses the x-axis at the points (1,0) ud (4,0) X² Use integration to find the exact value for the area at R. (27-22-952-16) b= 5" (27-2x - 3x²² - 16x²) dx ds [27x-22²-92³1²-162 y = x(x+4)(x-2) x=0 → стіділ x=-4-A 3/21 312 -√2+x-x² - 6x³1 ² + 161² = " =48-36 = 12 units² = /27 (4)-(4)²-6(4) ³/² +16² A = +20 + 128 3 A = 148/3 units ² ✓ area of R 2. The curve ( with equation y = x(x+4)(x-2) crosses the x-axis at the origin points A and B. a) Write down the x-coordinates of the points A and B." 3₁ +46) =(27 (1)-(10³²-66 x=2-B The finite region is bounded by the curve ( and the x-axis. b) Use integration to find the total area of the finite region. 10 (x (x+4) 6x-2)) dx = 5² G²1x²-8x) dx =128 3 and at the coordinates of the points A and B are (0, -4) and (0.2) = S² (x² + 2x²³ -8x²) S (x² + 2x² - 42²) dx = 5₁4 (²²+3x²³-4₂²3) dx = 0 - ((4) 4 + 2 (W³²-4 (-4²) :-) - (1² + 12 ( 0 ² -4 (09) - (0)² + (0)²-467) =-20 → 20L since area cannot be - ● 13E 1: Find the area between the curve with equation y=f6x), the x-axis and the lines x=a and x=b in each of the following cases: (a) f(0 =-3x²+17x-10; a= 1 and ²3 6=3 5³ (-3x² +17x -10 <= [-32²2 +₂² - 10x] = [ 2²³ +172² 107] ₁ b) Had 1=2x²³ +7x²-4x; a=-3, b=1 [²x² (272²+72²-40) dx = [204 +3726²³ -42²] [²+²-²], @fG0 = 1; a=-4, b=-1 1-1 dx= = x= -2 x=2 5-2 (c) f(x) = -x²4+7x²³ - 11x² +5x; a=0,b=4 5² (² x ² 17x²2²-412²+5x) dx = [=-=-²³ +72 17³ +52] ९५- JO (4) = 6 units 2 y = x(x²_u) y = x (x + ²)(x-2) → x=6 = (-(3) ³+1767)² -1063)- (- (² + 176²- = 39-(-5) = 22 units² ^²(x²-4x) dx = [2² - 47²7² ₁² = -3 = (²+²x+³2(1) 1-26- 362 units? = (-( 4 ) ³ + 7 (40)² - 18 (4) ³ + 5 (w)²) = (10) ² + 7 (0)² - 14 (0)² + 5 (0) ²) S 3 18-0 =728/15 units ²1 = 0-(-4) = 4 units ² 2: The sketch shows part of the curve with equation y=x (x²-4). Find the area of the shaded region y=x ³-4x = = ((0²-4(0)") - (204-4 (2²) = 62/3 units ² 26/02/21 Areas Under the x-axis • when area bavided by a curve and the x-axis is below the x-axis, fydx <0 Examples: 1) Find the area of the finite region banded by the curvey==(x-3) and the x-axis. A= So x(x-3dx y= xGetz) 32 (x²+2x) dx do = 2) Find the area of the finite region bound the x-axis. A = = 4/3 units2 c) y=(x(+3)x (x-3) = do x(x-1)(3) dx -[***] 4 3 = (42²5 +2612³-X13) (0² +20²²-301²) = -7/12 (x³-9xx) dac 23-=-124-92²273 J³ (x²-3x/dox 13 F 1 Sketch the following and find the total area of and the x-axis." a) = ( 22 ²³ + (-2) ²)-(0²³+ (²) - (0²-3(27) - (01³-201²) = -4.5 so area = 4. 5 units 2 A = 1 x G-DG13) 0-3 = [2 +2₂² +22²] = ( (0)² + 2(01² - 3(0)") - (-2+² +2 (3) ²-3(-3) 4 45/4 - total area = 45/4 + 7/12 = 71/6 84/4 units2 1512²-9x)dx= [24-22²] = 81/4 totalarea= 40. Sunits 2 1 by the curus y=xGx-Wats) and at the b) the finite region or regions banded by the curves y=(x+1)(x-4) 14(x²-3x-4)dx = [(141² - 2040²_414) - (1² X-² uts) = -125/6 so area-125/6 units² y=x²(x-2) ५ = 23-252 =[#² -²²=²² -(10-260³) do - ((2) ²1-2(2)³) (4 3 =-43-0. area= 4/3 units² 1/e) y = x(x-2)(x-s) y=x²-7x²410x ²2(x²³-7x² Headdx = = [2² - +2 ² + 102²"] = [1/4 x² - 7/32²³ +52²7] 2 = (1/4(20²- 7/3 (0)³ + 3(4²) - (10" 7'3 (²³+5ca³²) = 36/3 15²6x²³-7x²+16x)dx = [12x²4-7/3x² + 5x²] { -(1/4 (5) 4 - 7/3 (5) ³ + 5(5/9) -(1/4(2) 4-7 / 3 (2)²³ +5 (2) ²) =(-125/12)-(16/3) =63/4 total area- 16/3 +63/4 = 253/12 units ² y = x(x+3)(2-x) x=0 x=-3 2: The graph shows a sketch of part of the wrve (with equation y = x (x+3)(2-x). The curve ( crosses the x-axis at the origin O and at points A and B. a) Write down the x-coordinates of A and B. A=(3,0) B = (20) 2=2 The finite region, shown shaded, is bounded by the curve ( and the x-axis. b) Use integration to find the total area of the finite shaded region. A= √²-x²³x²+6x)dx = [-2²-2²³ +6₂²] ² = (1/4 (234-1213 +3 (23²)-(2014-0³ + 3(0)²) = 16/3 A = √²/² (-2²³-2² +6x) dx =[2²+¹ = (1/24 (2)² - 122² + 3(2) ²) - ((-3/4 - (3² + 3(-3)) = total area 16/3+63/4 - 253/12 units² 3: f6d=-xзна2+1х-30 The graph shows a sketch of part of the curve with equation y = -x ²³+4x²³ +11x - 30. a) Use the factor theorem to show that bet3lis a factor of f(x) if (x+3) is a factor of f(x) then f(3) = 0 Let fod=y. -63)³ +4 (3)² +1463)-30 20 =634 herce (ct3) is a factor at f(x).✓ -x²+7x-10 (2x131-xx³+4x²+11x-30 -1-x²-Bx²) b) Write foc) in the form (x+3)(Ax³²+ Bx tc). f(x) = -x ²4₂ ²+11x-30 f(x)=(et 3) (-2²772-18) A=-1~ B = 7 ✓ |C=-10~ 7x² +11x -(7x²+2x) -10x-30 3:4) Hence, factorise fox) completely. f6₂)=(x+3)(-x²47x-10) f(x)=(x+3)(x-5) (x-2) d) Hence, determine the x-coordinates where the curve intersects the x-axis. roots of f(x) are (-3,0), (5,0) and (2,0) e) Hence, determine the total shaded area shown on the sketch. A = [² (-x²³ +42²³ +11x - 30) da [2² +₂²³ +442²-30x] 5 = (-(5)² +4 ()² + 1 (ST²-30(51)-(-(²³²+ + 4(2) ² +11 (0)² -30(2)) ૫(s)? (2)4 = (-25/12)-(-94(3) 117/4 units A = √²₂ (²x²³ +4₂²³² + 11x²-30) dx 2-3 = [²2² +42³ +147²-30x] ²₂ 2 = ( (224 + 4 (2)3³ + 11/621²-30621)- (-(021²4 + 4 (-3) ³ + 14 (-3)^² - 306-3)) 2 (943)-(333/4) = = -1373 12 total area = 147 +1,375 = 863 units? 12 01/03/201 1 Areas Between Lowes & Lines. Examples: 1) Determine the area between the lines with equations y = x(4-x) andy=x you can eather: * fydx - fline da ide y=x line (3,0) x=x(4-x) x=4x-x² x²-3x=0 x(x-3)=6 x=0 or x=3 Area = x²-11x +18-0 (x-9)(x-2)=0 x=9 andx=2 y=x(x^²y) 0=(0,0) A=(3,0) 2) The diagram shows a stretch of the curve with equation y=x(x-3) and the line with equation y=2x. Find the area of the shaded region OAC. y=2x 2x=x²-3x x25x-6 x(x-5)=0 B=(5,0) (= ($10) = ·1² (10x-2²-8/dx - point B= 10-9) = 1 Area = "ux-2²'da - (32²) 3/12 1. Figure 2 shows the line with equation y = 10-x and the curve with equation y = 10x=2²8 The line and the curve intersed at the points A and B, and O is the origin. a) Caliclate the coordinates of A and the coordinates AB. 10x-x²-8-10-x point A= 10-(2) = 8 (2,8) (P40-1) * * fydx -aread triangle O [22² =2²] = (Z(3)²-(3)³)-9/2 = 9/2 units32 area of GAC- 25- 2 = 10x² - 2²-8x1²₂ - [10x - 7²1²2² 0-x) dx. area of triangle = 10x5 2 = 25 units² 1 The shaded area Ris bounded by the line and the curve, as shown in Figure 2. b) (alculate the exact area of R. •5³ 2²-3x d dx 3 = 25 = 25-113 (5) ³-7/2₂ (519) - ( 1305³-3/2(3²) -25-26/3 = 49/3 units² O = [5x²-₂²-8x] - [10 x-V/x²] 22 2 (SC)2-1/3 (95³-869))-(2)²-1/7 (2³-8(2)]-[(10(9)-1/26919-(1062) - (1/2₂ (2)²7] = (30-4/3)-(99/2-18) 343/6 units2 y=10x-x28 By=10x 13 6 1. The diagram shows part of the curve with equation y=z²+2 and the line with equation y=6. the paints B. a) Find the coordinates of the points A and B 2²+2=6 2²-4=0 (x+(x-2)=0 x=-2 and x=2 b) Find the area at the finite region bounded by line AB and the curve. 2²42 point A = y-coordinate=6 (2,6) point B = (2,6) 2-2 1²6x - 5²₂ 2²³ +²x = [12- (-12) - [(+4) - ( 23 −4)] 1-2 2-2 3 32/3 units2 2: The diagram shows the finite region, R. bounded by the curve with equation. y=3. The line cuts the curve at the points A and B. as Find the coordinates of the points A and B. 3= A = (1,3) B-(3,3) 2²-4x+3=0 -3)(x-1)-0 x=3 x=1 b) Find the apa of A. 1³ (4x-x²) 1²³ = [4 ₂² -3²] ³²-³ -²₂²²-[x] ²₁ - = (213)²-(02²) - ((4)²-(4²) - (13(3) - 2 (41) = - (5-3) - 4/3 units 21 C 1y = 4x=x² and the lind 3: The diagram shows a sketch of part of the curve with equation y = 9-3x - 5² _x²³ and the line with equation y=4-4x. The line cuts the curve at the points AC-1, 8) and B(1,0). Find the area of the shaded region between AB and the curve. 1² (13-3x - 5₂2²-2²³) - 5²₂₁ • 4- 4x = [3x-30³² - 50²³ - +24] ² - [we- 4x It Гих J-1 12 20/3 units? 6.b) Find the area of R₂ DS² (²4) dx + 5² ( = (964) - 3/2 (4) ² - $ (10³ - 1 (2014) - 13(1) -3/2₂-13²-5-²² (4(1)-2(1)²) = ((4-1)-26-19 = √67 44 = (762) - (23²³) -(19€(3) -(-3³})- (U4)² + (23) - (<23)²¹+3) = (12)-(-34/2) - (14/3 - (- 12)) 275/2-50/3=125/6 units² 9. The diagram shows fait a verify that the line wnarx=4, y=2 b) Find the area of 13√x - √₂2³ + [2²/2-2/53 =(2640³¹/2-2/5( = 86-12 S 36/Sunit 7-x=x²41 x²+x-6=0 (x-2)(x)=0 x=2 (x-3) 6. The cliagram shows askelch of part of the core with equation y=x²+1 and the line with equation y = 7ox. The finite region, R₁ is bounded by the line and the core. the finite region, he is below the cure and the line and is bounded by the posite x a) Find the area of Rs. [(-x). [^² x ²41 = [x - 7²"] , [x² +*], The sketch show the curve and + 2-axis and A and a Find the as y=x²(x+ x=0 X=-4 1 y-axs/ Find the 4%ड फार्म 19 and the 6. b) Find the area of he [² (₂²³4) dx + √² (+-2) dx = [ 3² + ₂] + [7²-4 ] 9. The diagram shows part of the curve with equation y=35x -5x³+4 and the line with equation y=4-1/2x. a) Verify that the line and the curve cross at the point A (4,2). Whctx=4, :y=2 so 3.54-√4³ +4=2 4-1/24=2 hence point A is a point of intersection between the line and the curve. b) Find the area of the finite region banded by the curve and the line. 1355-√₂2³ +4-144-42x S 36/5 units² [2x2²³/2_2/52²³/2² +4₂] - [4x - 4/4x²] " = = (2640) ³/12-2/5(4) ¾/²2 +4(49)-((0) ³/2-2/5 (0)52 +4 (0)) - (4(4)-1/4 (4²) - (4(0)-1/4 (0) ²) = 96-12 a Find the area of Rj. y=x²(x+4) x=0 x=-4 = (121² + (2)-(0²16) + (7 (7) - (2) - (2764)-2²) 10. The sketch shows part of the curve with equation y=x² (x+4). The finite region Ry is bounded by the curve and the negative x-axis. The finite region R₂ is bounded by the crve, the positive 2-axis and AB, where A (2,24) and B(b, 0). The area of R₁ = the area of R₂. = 14 + 49 14. 2 - 103/6 units ² b) Find the value of b 5²4²³+4x²³ dx + ² AB = ܐ 44/3 + AB arra of R₁ = 1° (x² + 4x²) dx (6-2)x 24x6.5=29/3 126-24-20/3 126-92/3 b= 23/9 units 2 = (²+4 (0²) - (-²*+-²) 0-(-64/3) 64/3 units² = [x² +42] + AB = 64/3 2 = 64/3 6413 4. The line with equation y = 10-2 cits the curve with equation y=2x²-5x+4 at the points A and B. a) Find the coordinates of A and the coordinates of B. lesat410-x 2xx²-4x-6=0 (x-3)(xH) = 0 x=3 x=-1 y=7 y=11 A(-1,14) B (3,7) The shaded region R is bounded by the line and the curve as shown. b) Find the exact area of R. 1²40-x dx - [² (2₂²-5x+4) da = [0x-2²1²³₁-[2²2²³-5₂²³² +4x] = (1003)-(322)-10-0-0²) - (R(3)³ - 5 (3)² + 4(3)) - (2(-1)³ -54-102²³ +46-50) = (51/2-(-24/2)) - (15/2-(-43/6)) = 36-44/3 = 64/3 units ² 26/02/24 Areas Under Cowes 13 E 3: The diagram shows a sketch of the curve with equation y= 3x +_-6₂-5, x>0. The region R is bounded by the curve, the x-axis and the Find the area of R. 1₂² (3x + 6²-5Jdx= √²³ (3x + 6x^²-3)dx y= (3x)(1+x) = 2x+3-x² y=(x-3)(x+1) x=3,x=1 ở hay sử ys =0 x = 4 = (363)² +6 (0)² = 5(3)) - (23(0)² + 6 (15+ 5(1)) -1 -1 -- / - (2) = 6 units ² 4: Find the area of the finite region between the curve with equation y=(3-x)(1+x) and the x-axis. 2² lines x=1 and x= 3. [ (x²+²x+3) dx = [=32²³² +2₂² + 3x)²₂ = 32/3 units² 5: Find the area of the finite region between the curve with equation y=xGx-4)² and the x-axis. "gu (2²³-8x² + 16x) dx = [2²-82² HE]" x3-8x²+16x 3 (²³²-2x²+²x) dx = 8 = (-(02)² + (0)² + 3651) - (- (+1³² + (1) ² + 3(-1)) 6: Find the area of the finite region between the curve with equation y = 2x² y= 2x²-3x³ 194³ (2x² 3x²³) dx = [22²² = 34²9" ³ y = x²(2-3x) -3x³ x=0 x= 2/3 لا (K)²-(6) ²+2(K)) -(0)³-(0)² +260)) = 8 кз-кчек =8 K3-4²+2k-8-0 (2)²-(2)2 +212)-8=0) so = 2 = [+2²" - 3x²+x²7" = 4 4 3 1 ²) (4x400) = 64 +0 =64/3 onits²/ = (2(43)3 - 3 (2014)-(2(0)² - 3(014) = 4-0 = 4/81 units² 7. The shaded area under the graph of the function f(x)=3x²-2x+2, banded by the curve, the x-axis and the lines x=0 and x=k, is 8. Work at the value aftr. (3x²-2x+2) dx = 8 k>0 and the x-axis, 8. The finite region R is bounded by the x-axis and the curve with equation y=-x²+2xx +3₁ x 7,0. The curve meets the x-axis at points A and B. a Find the coordinates of point A and point B A = (1,0) B = (3,0) -x²+2x+3 y = y = (x-3)(x+1) x=3 and x=-1 b) Find the area of the region R 1²" (x²+²x+3) dx = [²2²³² + 2x² + 3x] ²³ [3x²+2²+37] = ( 12 (3) ³+ (5) ² + x(3)) - (-1 (²-1)³ + (-1)² + 3(-11) = 9- 1- (-¾/3) = = 32 units? 3 The 9. The graph shows part of the curve ( with equation y=x² (2x) 2 region 1R, shave shaded, is bounded by (and the x-axis. Use caldus to find the exact area of R. y=x²(2-x) y=-x²+2x² x=0 x=2 4 $ ² (-2²4+2x²) dx = [=z=²" + 27²³ ] [ O = (-402²4 +2(4³5) - (-101²4 +2(0)³) =4-0 3 = 4 units' <2 = ● Hixed Excercise 15: The diagram shows a sketch of the curve with equation y=12x¹²²_x²³² for @≤ x ≤ 12. a) Show that dy = 3x² ¹/² (4-x) da 2 12/03/21 y=122¹22 x ²/2 1/2 dy = 6x2-3x² dx 3/2x¹¹2(4-x) = b) At the point B on the curve the tangent to the curve is Afind the coordinates of the point B/2 (4-x)=0 at point B, dy=0. 3/27/1/2_3/2x4/2=0. Ex 615x y=12x²/2-x³/2 y = 12(4) ¹/2-(4) ³/2 y=16 6x = [12x²12-25/2752 82 L312 5/2 JO. - [8x3/2-2/575/273 = x(8-x=12 8x-x²=12 para 3/253 lel to the x-axis. 4x122x42 x = 4 J Find, to 3 significant figures, the area of the finite region bounded by the carve A= So •12 12x²=12-x²³/2 dx and the x-axis. Point Bis (4,16) A=(8 (10³2/5 (122)-(8 (0)³2_2/5(0)³7₂) = 133.021502-0 Area of finite region & 133 units 16: The diagram shows the curve ( with equation y=x (8-x) and the line with equation y=12 which meet at the points L and M. a Determine the coordinates of the point M. point Mis (6, 12). x²-8x+12=0 (x-6)(x-2)=0 x=6₁x=2 b) Given that N is the foot of the perpendicular from Mon to the x-axis, calculate. the area of the shaded region which is banded by NM, the curvel and the x-axis. N is at (6,0) Area of shaded region= x(8-x) dx =1²8x-x²4x = [4x² - 4/3x³] ² = (4(2) 4-4/3 (8)35-(4 (6) ²-4/3 (6³) 256/372 .40/3 units ²1 17. The diagram shows the line. 2 y=x-1 meeting the curve with equation y=(x-1) Ge-5) at A and (.. The curve meets the x-axis at A and B. @ Write down the coordinates of A and B and find the coordinates of C. A(1,0) and B(5,0) x-1=6x-4(x-S) x-1=x²-6x+5. x²-7x+6=0 4146)=0 y=x-1 point Cis (6,5) 2=1₁x=6 y=(6)-1 9 = 5 b) Find the area of the shaded region banded by the line, the curve and the x-axis. A= 1₁ x-1 dx = 1² 2²-6x + sdx A = ((6)² = (6))-(442² -(1)) -(5)² -3 (5)² +5(5) = -x]-[²-3x²+x]{ 31³-3(2+36-1)) 25/2-(-32/3) 139/6 units² 10%6 units? 18: The diagram shows part of the curve with equation y=p+10x-x², where p is a constant, and part of the line I with equation y=qpx +25, where a is a constant. The line I cuts the core at the paints A and B. The x-coordinates of A and B are le and 8 respectively. The line through A parallel to the x-cusis intersects the core again at the point C a show that p=-7 and calculate the value of q. ( 25-p=32 P=-7 -10+9=12 9=-2 p+ 10x-x²=qx+25 x²-10x tqx +25+p=25₁ x² - 10+q x + 25-p=(x-4) (x-8) x²-1019x +25-p=x²-12x+32 6) Calculate the coordinates of C y=qx+25 y=-2(4)+25 y=17 x²-10x+ 24-0 c) the shaded region in the diagram is bundled by the curve and the line segment AC Using integration and showing all your working, calculate the area of the shaded. region. A= 56-7+10x-x² dx - √²17 dx 24 106/3-34 = 4/3 units ² = · [7X+5x² - 4/3x²] 6 - [17x] 4₂ = ( 7 (6) +566) ² - 1/3 (61)-(-7 (4) +5 (2) ² - 1/3(4) ³) - (17(0)-(17(4)) t y=-7+10x²x² 17=-7+10x-x² 19. Given that f(x) = 12 -85x +4x-5, x20, find of food doc. x² Given that is constant √415x values for A and find these and ful3-4) dx = A² show that there are two possible (3-4)= (3x²0²5-Adx 6-SA=A² A²+SA-6=0 (A-1) (A-6)-O A-1 or A-6 (x-6)(x-4)=8 point (is at (6,17) x=6,x=4 +60=9x²² - 8x²4/² + 4x-S дохі 9x²-2²-8x²4/²+4x-5dx= 3x²² -8x²*² ² + 4x² - 5x =-9x² 1-16x³/2+2x²-5x+r✓ -1 3/2 2 21: f'(x) = (2-x²)³ XL = - = [6x05-Ax]{ = (6 (8) ⁰.5-A (9)) - (6(4)⁰. S-A (G₂)) -(18-3A)-(12-4A) 1x40 a) show that f'(x) = 8x=²-12 + Ax² + Bx", where A and Bare constants to be found. f'(x) = (2-x4) (2-x²4(2-x²). x² = (²-x²) (4-4x²+x²) 8-8x²+2x4-4x² + 4x²4-26 2² 8-12x²+6x4-x6 22 1'60=8-12x²+6x²-x6 22 x² x -8x ²12+6x²-x²4 A=6✓ B=-1 2² 246) Find "G + 6 = 8x²-12 +6x²-x² 1"(x) = (28x²+² + (2) 6x² - (Wx³ = -16x +12x-4x²³ c) Given that the point (2.9) lies on the core with equation y=f(c), find f(z). dx = 18x² 9-86-2 -12(-2) +2(-2)³-1/5 (-25+C 4+24-16 +32/5 +c 8x12x+6²-²+C f(x) = -8₂²² - 12x + 2x² - 4/5+²³+c = 16--8x12x²+2x²-4/5x²³_47/5/ 22. The finite region S, which is shown shaded, is bounded by the x-axis and the arve with equation y = 3-sx-2x? The curve meets the x-axis at points A and B. a) Find the coordinates of point A and point B. y = 3-5x-2x² y = (2x-1) (x+3) x-4/21x=-3 point Ais at (-3,0)✓ point Bis at (1/20) b) Find the area of the region S. P²¹/23-5x-2x² dx A = B = [3x - 5/2x²-4/3x²14! = (3(4/2)-5/2 (4/2)²-2/3 (4/2³) - (36-3) - 5/2 (-3)²-2/3 (-3³) 9/24-(-2712) -343/24 units ², 9=92/3+( C=-47/5 23. The graph shows a sketch of part of the curve ( with equation y=(x-4) (2x+3). The curve ( crosses the 2-axis at the points A and B. a) Write down the x-coordinates of A and B. y=(x-4) (2x+3) point A is ( x=4₁ x= -3/2 point Bis (4,0)✓ b) The finite region R, shawl shaded, is banded by Use integration to find the area of R. A= √22²-5x-12 dx = 3x3-5122². = (1/3(4) ³ - 3/2 (4)2-1264)) - (436-3/12) ³-562 (-3/2)²-12(-362) =(-136/3) -(81/8) -1334/24 : Area of finite region R = 1331/24 units ² Cand the x-asis. 24: The graph shows asketch of part of the curve ( with equation y=xlx-3)(x+₂). The curve crosses the I-abis at the origin O and the points of and B. Write y= x(x-3)(x+2) of point A is (20) point Bis (30) x=0, x=3₁x==2 b) The finite region shown shaded is bounded by the curve and the x-axis. ase integration to find the total area of this region. A = =10x² x-x²-6x dx + x²-x²-6x dar J-2 - 253/12 units ² do = [/x-43x²³-3x²10₂ + [4/4x4-4/3x³-3x²] ³ ((1/4(0) 4-4/3(0)³-3/019) - (1/4(-284-4/36-2)8-3(-2)²)) (414(34 4/3 (2) ³ - 2 2²)-(1/4(041/3(0) ³ - 4.3/13(0) ³ - X(014) (0)-(163))+((-63/4) -(0)) A = 16/3 + 63/4