U Ye Min Htoo(YTU) Grade 12 Mathematics CHAPTER 11 Application of Integration Rules for Definite Integration Suppose f and g are continuous on [a, b] and k ∈ R. Z b Z b 1. kf (x) dx = k f (x) dx. a Z b [f (x) ± g(x)] dx = 2. a Z b f (x) dx ± a Z a 3. 0 Z b g(x) dx. a 10 1 a f (x) dx = 0. a Z a f (x) dx = − f (x) dx. a b 5. Z b Z c Z b 09 45 1 4. 23 Z b f (x) dx f (x) dx + f (x) dx = where c ∈ (a, b). c a a Example 1. H Evaluate the following definite integrals. Z 2 Z π 2 3 (a) x dx (b) sin x dx −1 0 Z π | cos x| dx Z 1 (c) M −1 UY π 4 Solution 3 2 Z 2 8 (−1) 23 (−1)3 x 2 x dx = (a) − = − = 3. = 3 −1 3 3 3 3 −1 Z π 2 (b) π sin x dx = [− cos x]02 = − cos 0 Z 1 (c) √ Z 1 1 (x + 2) 2 dx x + 2 dx = −1 π + cos 0 = 0 + 1 = 1. 2 −1 3 1 2 = (x + 2) 2 3 −1 1 √ x + 2 dx (d) U Ye Min Htoo(YTU) = Grade 12 Mathematics 3 2 3 (3) 2 − (1) 2 3 2 √ = (3 3 − 1). 3 Z π (d) π 4 | cos x| dx 0 When π4 ≤ x ≤ π2 , | cos x| = cos x. Z π Z π Z π 2 | cos x| dx = cos x dx + π 4 23 π 4 10 1 When π2 ≤ x ≤ π, | cos x| = − cos x. (− cos x) dx π 2 π 4 2 09 45 1 π 2 = sin x π − sin x π π π π − sin π − sin = sin − sin 2 4 2 √ H 2 2 M =2− UY Example 2. Z 2 Evaluate the integral √ x 2 − x dx. 1 Solution Method 1 Z 2 √ x 2 − x dx 1 Let u = 2 − x. Then du = −dx. 2 U Ye Min Htoo(YTU) Z Grade 12 Z √ x 2 − x dx = − R =− Mathematics √ (2 − u) u du 1 3 2u 2 − u 2 du 3 5 = − 43 u 2 + 25 u 2 3 5 2 3 5 4 2 x 2 − x dx = − (2 − x) 2 + (2 − x) 2 3 5 1 1 Z 2 √ 09 45 1 14 = 0 − − 34 + 25 = 15 Method 2 Let u = 2 − x. Then du = −dx. 1→0 Z 2 M u UY 1→2 H Changing the variable from x to u, we get x 23 Next 10 1 0 = − 34 (2 − x) 2 + 25 (2 − x) 2 √ x 2 − x dx = − Z 0 1 1 3 2u 2 − u 2 du 1 Z 1 = 1 2 2u − u 3 2 du 0 1 4 3 2 5 = u2 − u2 3 5 0 = 14 15 3 U Ye Min Htoo(YTU) Grade 12 Mathematics Exercise 11.1 1. Evaluate the following definite integrals. Z 1 Z 1 √ ex (a) dx (b) x x + 2 dx x 0 e +1 −1 Z 4 (d) 2 Z 2 x+5 dx 2 x − 6x + 5 (e) e −x Z 6 (c) 0 1 x−1 3 3 dx Z π cos 2x dx (f) tan 2x dx π 2 0 0 Solution; Z 1 0 ex dx = ln |ex + 1|10 x e +1 10 1 (a) 23 = ln |e1 + 1| − ln |e0 + 1| Z 1 (b) 09 45 1 = ln(e + 1) − ln 2 e+1 = ln 2 √ x x + 2 dx −1 M When x = −1, u = 1. H Let u = x + 2. Then du = dx and x = u − 2. UY When x = 1, u = 3. 4 U Ye Min Htoo(YTU) Z 1 Grade 12 Z 3 √ x x + 2 dx = −1 Mathematics √ (u − 2) u du 1 Z 3 1 (u − 2)u 2 du = Z1 3 3 1 u 2 − 2u 2 du 1 3 2 5 4 3 = u2 − u2 5 3 1 2 5 4 3 2 5 4 3 2 2 2 2 = (3) − (3) − (1) − (1) 5 3 5 3 √ √ 2 4 2 4 = ·9 3− ·3 3 − − 5 3 5 3 √ √ 2 4 18 3 −4 3− + = 5√ √ 5 3 18 3 − 20 3 2 4 − + = 5 5 3 √ 2 3 2 4 =− − + 5√ 5 3 2( 3 + 1) 4 =− + 3 √5 −6( 3 + 1) + 20 = √15 20 − 6 3 − 6 = 15 √ 14 − 6 3 = 15 (c) 0 Let 1 x−1 3 3 dx UY Z 6 M H 09 45 1 23 10 1 0 = u = 13 x − 1 du 1 = dx 3 3du = dx Z 1 x−1 3 3 Z dx = u3 · 3du = 3 u4 4 5 U Ye Min Htoo(YTU) 3 = 4 Z 6 1 x−1 3 1 x−1 3 Mathematics 4 3 " 4 # 6 3 1 dx = x−1 4 3 0 4 4 3 1 3 1 = ×6−1 − ×0−1 4 3 4 3 3 3 = (2 − 1)4 − (0 − 1)4 4 4 3 3 = − 4 4 10 1 0 0 Grade 12 =0 Z 6 0 09 45 1 23 Another Method 1 x−1 3 3 1 dx = 4 UY M H 3 = 4 Z 4 (d) 2 x+5 = 1 x−1 3 4 1 6 × 1 4 6 0 3 0 3 = (2 − 1)4 − (0 − 1)4 4 3 = [1 − 1] 4 =0 x+5 dx x2 − 6x + 5 x2 − 6x + 5 1 x−1 3 x+5 A B = + (x − 1)(x − 5) x−1 x−5 x+5 A(x − 5) + B(x − 1) = x2 − 6x + 5 (x − 1)(x − 5) x + 5 = A(x − 5) + B(x − 1) 6 U Ye Min Htoo(YTU) Grade 12 Mathematics When x = 1, 1 + 5 = A(1 − 5) 6 = −4A A=− 3 2 When x = 5, 5 + 5 = B(5 − 1) 5 2 10 1 B= 0 10 = 4B Z Z x+5 3 4 1 5 4 1 dx = − dx + dx x2 − 6x + 5 2 2 x−1 2 2 x−5 4 4 5 3 = − ln |x − 1| + ln |x − 5| 2 2 2 2 3 3 5 5 = − ln |4 − 1| + ln |2 − 1| + ln |4 − 5| − ln |2 − 5| 2 2 2 2 3 5 = − ln 3 − ln 3 2 2 1 = −4 ln 3 = ln 4 3 09 45 1 Z 4 23 5 − 32 x+5 2 = + ∴ 2 x − 6x + 5 x−1 x−5 Z 2 e−x cos 2x dx UY (e) M H 2 0 Let u = e−x , dv = cos 2x dx Z Z du −x = −e , dv = cos 2x dx dx 1 du = −e−x dx, v = sin 2x 2 Z e −x 1 cos 2x dx = e−x sin 2x − 2 1 1 = e−x sin 2x + 2 2 Z Z 1 sin 2x(−e−x dx) 2 e−x sin 2x dx 7 U Ye Min Htoo(YTU) Z Let I = Grade 12 Mathematics e−x cos 2x dx 1 1 Then I = e−x sin 2x + 2 2 Let u = e−x , Z Z du = −e−x dx e−x sin 2x dx · · · (1) dv = sin 2x dx Z dv = sin 2x dx 0 1 du = −e−x dx v = − cos 2x 2 Z 1 −x 1 sin 2x dx = − e cos 2x − − cos 2x (−e−x dx) 2 2 Z 1 −x 1 = − e cos 2x − e−x cos 2x dx · · · (2) 2 2 09 45 1 23 e −x 10 1 Z Substituting (2) into (1), we get UY M H Z 1 −x 1 1 −x 1 −x I = e sin 2x + − e cos 2x − e cos 2x dx 2 2 2 2 Z 1 −x 1 −x 1 I = e sin 2x − e cos 2x − e−x cos 2x dx 2 4 4 1 1 1 −x I=e sin 2x − cos 2x − I 2 4 4 5 1 I = e−x (2 sin 2x − cos 2x) 4 4 1 −x I = e (2 sin 2x − cos 2x) 5 Z ∴ 1 e−x cos 2xdx = e−x (2 sin 2x − cos 2x) 5 Z 2 e 0 −x 2 1 −x cos 2x dx = e (2 sin 2x − cos 2x) 5 0 1 −2 1 0 = e (2 sin 4 − cos 4) − e (0 − 1) 5 5 1 1 = 2 (2 sin 4 − cos 4) + 5e 5 8 U Ye Min Htoo(YTU) Z π 6 (f) Z π 6 tan 2x dx = π 8 π 8 1 =− 2 Z π 6 π 8 Grade 12 Mathematics sin 2x dx cos 2x −2 sin 2x dx cos 2x π 1 6 = − ln | cos 2x| π 2 8 ! 23 1 2 − ln ln 2 2 09 45 1 = √ 10 1 √ ! 1 2 ln − ln 2 2 1 =− 2 1 = 2 0 1 1 π π + ln cos = − ln cos 2 3 2 4 1 √ ln 2 2 √ = ln 4 2 0 M H 2. Find the integral of Z 2π cos mx sin nx dx UY for (a) m = n and (b) m ̸= n. 9 U Ye Min Htoo(YTU) Grade 12 Mathematics Solution; (a) For m = n, Z 2π Z 2π cos mx sin nx dx = 0 cos mx sin mx dx 0 09 45 1 =0 (b) For m ̸= n, Z 2π Z 1 2π 2 cos mx sin nx dx cos mx sin nx dx = 2 0 Z 1 2π = [sin(m + n)x − sin(m − n)x] dx 2 0 2π 1 cos(m + n)x cos(m − n)x = − + 2 m+n m−n 0 1 1 cos(m + n)2π cos(m − n)2π 1 = + + − − − 2 m+n m−n m+n m−n UY M H 0 23 10 1 0 Z 1 2π = 2 sin mx cos mx dx 2 0 Z 1 2π = sin 2mx dx 2 0 2π cos 2mx 1 − = 2 2m 0 1 cos 4mπ 1 = − − − 2 2m 2m 1 1 1 − = − − 2 2m 2m =0 3. Evaluate the following integrals. Z 4 √ | 2x − 2| dx (a) 0 Z 2π Z 1 |e − 1| dx x (b) −1 | sin x| dx (c) 0 Solution; 10 U Ye Min Htoo(YTU) Z 4 (a) Grade 12 Mathematics √ | 2x − 2| dx For the intersection point of x-axis and the curve, 0 √ 2x − 2 = 0 √ 2x = 2 x=2 √ √ When 0 ≤ x ≤ 2, | 2x − 2| = −( 2x − 2) √ √ When 2 ≤ x ≤ 4, | 2x − 2| = ( 2x − 2) q 0 |ex − 1| dx 1 2 10 1 Z 2 09 45 1 23 √ H Z 4 q 1 | 2x − 2| dx = (−( 2x + 2)dx + ( 2x 2 − 2)dx 0 0 2 " #2 " √ 3 #4 3 √ x2 2x 2 + 2x + = − 2 − 2x 3 2 2 0 2 ! √ " # 8 8 16 2 = − +4 −0 + −8 − −4 3 3 3 √ 8 16 2 8 =− +4+ −8− +4 3√ 3 3 16 2 16 = − 3 3 16 √ = ( 2 − 1) 3 Z 4 (b) −1 M Z 1 UY For the intersection point of x-axis and the curve, ex − 1 = 0 ex = 1 x=0 When −1 ≤ x ≤ 0, |ex − 1| = −(ex − 1) When 0 ≤ x ≤ 1, |ex − 1| = ex − 1 11 U Ye Min Htoo(YTU) Grade 12 Z 1 Z 0 |e − 1| dx = − x −1 Mathematics Z 1 (e − 1)dx + (ex − 1)dx x −1 0 09 45 1 x=π 23 sin x = 0 10 1 (c) For the intersection point of x-axis and the curve, 0 = − [(ex − x)]0−1 + [(ex − x)]10 = − (e0 − 0) − (e−1 + 1) + (e1 − 1) − (e0 − 0) 1 = − 1 − − 1 + [e − 1 − 1] e 1 = +e−2 e When 0 ≤ x ≤ π, | sin x| = sin x When π ≤ x ≤ 2π, | sin x| = − sin x Z 2π H Z π Z 2π | sin x| dx = UY M 0 (− sin x) dx sin x dx + 0 = (− cos x) π π + cos x 0 2π π = [(− cos π) − (− cos 0)] + [cos 2π − cos π] = [1 + 1] + [1 − (−1)] =4 12 U Ye Min Htoo(YTU) Grade 12 Mathematics Example 3. Find the area enclosed between the graph of (a) f (x) = 3 and the x-axis over [1, 4], 2 (b) f (x) = 3x + 2 and the x-axis over − , 0 , 3 (c) f (x) = −x2 + 4 and the x-axis over [−2, 2]. 10 1 0 Solution Z 4 23 (a) 1 1 = 12 − 3 09 45 1 4 3 dx = 3x = 9 unit2 M H (b) 0 3x2 + 2x (3x + 2) dx = 2 − 23 − 32 2 3 2 2 =0− − −2 − 2 3 3 2 = unit2 3 UY Z 0 13 U Ye Min Htoo(YTU) Grade 12 Mathematics (c) 2 x3 (4 − x ) dx = 4x − 3 −2 −2 8 8 = 8− − −8 + 3 3 2 = 10 unit2 3 Z 2 2 Example 4. (a) Compute the definite integral 0 10 1 Z 5 0 2 (x2 − 2x) dx. (b) Find the total 23 2 area enclosed between the graph of f (x) = x − 2x and the x-axis 5 over 0, . 2 09 45 1 Solution (a) 3 π2 x 2 (x − 2x) dx = −x 3 0 0 1 π 3 π 2 = − 3 2 2 1 = −1 24 2 UY M 2 H Z 5 (b) For the intersection point of x-axis and the curve, f (x) = 0 14 U Ye Min Htoo(YTU) Grade 12 Mathematics x2 − 2x = 0 x(x − 2) = 0 x = 0, 2 10 1 0 Thus the graph of f (x) cuts x-axis at 0 and 2, we divide the domain in two intervals: the 5 interval [0, 2] and the interval 2, . 2 2 3 x 2 −x (x − 2x) dx = 3 0 0 8 = −4 −0 3 4 =− 3 Z 2 09 45 1 23 2 H On [0, 2], the function f ≤ 0, so the value of integration f gives negative. 3 25 x 2 (x − 2x) dx = −x 3 2 2 125 25 8 = − − −4 24 4 3 7 = 24 M Z 5 UY 2 2 5 On 2, , the function f ≥ 0, so the value of integration f gives positive. 2 4 7 5 Area = − + = 1 unit2 3 24 8 15 U Ye Min Htoo(YTU) Grade 12 Mathematics Example 5. Find the area enclosed between the graph of f (x) = cos x and the lines (a) x = 0 and x = (b) x = π 2 π 3π and x = 2 2 (c) x = 0 and x = 3π 2 cos x dx. 10 1 2 (d) Find the definite integral 0 Z 3π UY (a) M Solution; H 09 45 1 23 0 Z π 2 cos x dx = sin x π 2 0 0 π = sin − sin 0 2 =1 Area = 1 unit2 . (b) 16 U Ye Min Htoo(YTU) Grade 12 Mathematics Z 3π 2 π 2 3π 2 cos x dx = sin x π 2 = sin 3π π − sin 2 2 = −2 0 Area = | − 2| = 2 unit2 . 3π is 3 unit2 . 2 10 1 (c) Area between the graph f (x) = cos x and the lines x = 0 and x = 09 45 1 23 (d) Z 3π 2 cos x dx = sin x 3π 2 0 0 3π = sin − sin 0 2 M Example 6. H = −1 UY The shaded area is 16 unit2 . Find the value of k. Solution 17 U Ye Min Htoo(YTU) Grade 12 Mathematics Z 4 √ 2k 3 4 k x dx = x2 3 0 0 2k 3 · 42 = 3 16k = 3 16k = 16, so k = 3. 3 Example 7. M Solution H 09 45 1 23 10 1 Find the area between y = x2 − 2x + 2 and y = 2x − 1. 0 By the problem, UY To find the x-coordinate of intersection points, x2 − 2x + 2 = 2x − 1 x2 − 4x + 3 = 0 (x − 1)(x − 3) = 0 x = 1, 3 18 U Ye Min Htoo(YTU) Grade 12 Mathematics Z 3 [(2x − 1) − (x2 − 2x + 2)] dx Area = Z1 3 (−x2 + 4x − 3) dx 1 3 3 x 2 = − + 2x − 3x 3 1 4 = 0− − 3 1 = 1 unit2 3 Example 8. 10 1 π 5π ≤x≤ . 4 4 H 09 45 1 23 Find the area between y = sin x and y = cos x for 0 = Solution UY M From the figure, we see that sin x ≥ cos x when π 5π <x< . 4 4 Z 5π 4 Area = π 4 (sin x − cos x) dx 5π = [− cos x − sin x] π4 4 5π π π 5π − sin − − cos − sin = − cos 4 4 4 4 √ √ ! √ √ ! 2 2 2 2 = + − − − 2 2 2 2 √ = 2 2 unit2 19 U Ye Min Htoo(YTU) Grade 12 Mathematics Exercise 11.2 UY M H 09 45 1 23 10 1 0 1. Find the shaded areas. Solution: 20 U Ye Min Htoo(YTU) Grade 12 Mathematics (a) For the intersection point for x-axis and the curve, f (x) = 0 x2 − 2x − 3 = 0 (x + 1)(x − 3) = 0 x = −1, 3 3 3 x 2 2 (x − 2x − 3) dx = − x − 3x 3 −1 −1 3 3 (−1)3 2 2 = −3 −3·3 − − (−1) − 3(−1) 3 3 1 = (9 − 9 − 9) − − − 1 + 3 3 2 = −9 − 1 3 2 = −10 3 23 10 1 0 Z 3 2 2 = 10 unit2 3 3 (b) To find the x-coordinate of intersection point 09 45 1 The shaded area = −10 cos x = sin x M H 1= tan x = 1 π x= 4 Z π UY Z π 4 shaded area = sin x cos x (cos x − sin x) dx + 0 2 π 4 (sin x − cos x) dx π π = [sin x + cos x]04 + [− cos x − sin x] π2 i h 4 h π π π π π π i − (0 + 1) + − cos − sin − − cos − sin = sin + cos 4# " 2 2 4 4 "√ 4 √ √ √ !# 2 2 2 2 = + − 1 + (0 − 1) − − − 2 2 2 2 √ √ = 2−1−1+ 2 √ =2 2−2 √ = 2( 2 − 1) unit2 21 U Ye Min Htoo(YTU) Grade 12 Mathematics (c) 4 0 x x3 2 (x + x − 2x) dx = + −x 4 3 2 2 16 8 =0− + −4 4 3 48 − 32 − 48 =0− 12 32 = 12 8 = 3 4 1 Z 1 x x3 3 2 2 (x + x − 2x) dx = + −x 4 3 0 0 1 1 = + −1 −0 4 3 3 + 4 − 12 = 12 5 =− 12 Z 0 0 2 09 45 1 23 10 1 3 5 8 5 32 + 5 37 1 8 + − = + = = =3 unit2 . 3 12 3 12 12 12 12 (d) To find the x-coordinate of intersection points. x2 = x + 2 x2 − x − 2 = 0 (x + 1)(x − 2) = 0 UY M H The shaded area = x = −1, 2 22 U Ye Min Htoo(YTU) Grade 12 Mathematics Z 2 −1 (x + 2 − x2 ) dx x2 = √ 09 45 1 x4 = x x 23 (e) To find the x-coordinate of intersection points. 10 1 2 2 x x3 = + 2x − 2 3 −1 4 8 1 1 = +4− − −2+ 2 3 2 3 10 7 = + 3 6 27 = 6 1 = 4 unit2 2 x4 − x = 0 x(x3 − 1) = 0 UY M H x = 0, 1 Z 1 √ x − x2 dx The shaded area = Z0 1 √ = x − x2 dx 0 " 3 #1 x2 x3 = 3 − 3 2 0 " 3 #1 2x 2 x3 = − 3 3 0 2 1 − −0 = 3 3 1 = unit2 3 23 0 The shaded area = U Ye Min Htoo(YTU) Grade 12 Mathematics (f) To find the x-coordinate of the intersection points. sin 2x = sin x 2 sin x cos x = sin x 2 sin x cos x − sin x = 0 sin x(2 cos x − 1) = 0 π x = 0, , π 3 Z π Z π 3 (sin x − sin 2x) dx (sin 2x − sin x) dx + The shaded area = π 3 π cos 2x cos 2x = − + cos x + − cos x + 2 2 π 0 3 1 2π π 1 = − cos − − cos 0 + cos 0 + cos 2 3 3 2 π 1 2π 1 + − cos π + cos 2π − − cos + cos 2 3 2 3 1 1 1 1 1 1 = + − − +1 + 1+ − − − 4 2 2 2 2 4 1 1 1 3 1 = − +1+ + + 4 2 2 2 4 3−2+4+2+2+1 = 4 10 = 4 5 = unit2 2 10 1 π3 0 0 M H 09 45 1 23 UY 2. Find the area under each curve between the given x-values: (a) f (x) = ln x from x = 1 to x = e. (b) f (x) = 12 − 3x2 (c) f (x) = xex 2 (d) f (x) = sin x from x = −1 to x = 2. from x = 0 to x = 1. from x = 0 to x = (e) f (x) = (1 + cos x) sin x 3π . 2 from x = 0 to x = π. Solution. 24 U Ye Min Htoo(YTU) (a) f (x) = ln x Grade 12 Mathematics from x = 1 to x = e. Z e The area = ln x dx 1 = [x ln x − x]e1 = (e ln e − e) − (1 ln 1 − 1) = (e − e) − (0 − 1) = 1 unit2 0 from x = −1 to x = 2 Z 2 The area = −1 (12 − 3x2 ) dx 23 = [12x − x3 ]2−1 10 1 (b) f (x) = 12 − 3x2 09 45 1 = (24 − 8) − (−12 + 1) = 16 + 11 = 27 unit2 H from x = 0 to x = 1 UY Let u = x2 du = 2x dx 1 du = x dx 2 2 M (c) f (x) = xex Z Z x2 xe dx = 1 eu du 2 1 = eu 2 1 2 = ex Z2 1 2 xex dx 0 1 1 x2 = e 2 0 1 = (e − 1) unit2 2 The area = (d) f (x) = sin x from x = 0 to x = 3π 2 25 U Ye Min Htoo(YTU) Grade 12 Mathematics For the intersection points of x-axis and sin x sin x = 0 x=π Z π sin x dx = [− cos x]π0 = − cos π + cos 0 = 2 0 Z 3π 3π sin x dx = [− cos x]π2 = − cos 0 π Area = |2| + | − 1| = 3 unit2 (e) f (x) = (1 + cos x) sin x 3π + cos π = −1 2 from x = 0 to x = π. 23 Z π 10 1 2 (1 + cos x) sin x dx The area = Z π 09 45 1 0 (sin x + cos x sin x) dx Z π 1 = sin x + (2 sin x cos x) dx 2 0 Z π 1 = sin x + sin 2x dx 2 0 π 1 = − cos x − cos 2x 4 0 1 1 = − cos π − cos 2π − − cos 0 − cos 0 4 4 1 1 = 1− − −1 − 4 4 3 5 = + 4 4 = UY M H 0 = 2 unit2 3. Z 1 √ (a) Compute x 1 − x2 dx. −1 √ (b) Find the area enclosed between the graph of f (x) = x 1 − x2 and the x-axis over [−1, 1]. 26 Grade 12 Mathematics 0 U Ye Min Htoo(YTU) 10 1 Solution Z 1 √ (a) x 1 − x2 dx 23 −1 du = −2x dx 1 du = x dx 2 09 45 1 Let u = 1 − x2 . 1 x 1 − x2 dx = u − du 2 Z 1 =− u1/2 du 2 Z 1 √ 1 1 x 1 − x2 dx = − (1 − x2 )3/2 −1 3 −1 1 = − [0 − 0] 3 √ Z UY M H Z =0 Z 1 √ (b) x 1 − x2 dx −1 27 √ U Ye Min Htoo(YTU) Grade 12 Mathematics Z 1 √ 1 1 x 1 − x2 dx = − (1 − x2 )3/2 −1 3 −1 1 = − [1 − 0] 3 1 =− 3 The enclosed area = − 13 + 13 = 23 unit2 . 10 1 0 4. Find the area enclosed between the graphs of y = x3 − 3x and y = x for 0 ≤ x ≤ 2. Z 2 23 Solution. (x − (x3 − 3x)) dx The enclosed area = 09 45 1 Z0 2 (x − x3 + 3x) dx = Z0 2 (4x − x3 ) dx 0 2 x4 2 = 2x − 4 0 = (8 − 4) − 0 = 4 unit2 UY M H = 5. Find the area enclosed of the graphs of ln(3x − 2) and y = x2 − 2x + 1 for 1 ≤ x ≤ 2. Solution. Z ln(3x − 2) dx Let u = ln(3x − 2), dv = dx du 1 = × 3, dx 3x − 2 du = 3 dx, 3x − 2 R dv = R dx v=x 28 U Ye Min Htoo(YTU) Grade 12 Mathematics Z Z 3 x· dx 3x − 2 Z 2 dx = x ln(3x − 2) − 1+ 3x − 2 2 = x ln(3x − 2) − x + ln |3x − 2| 3 Z 3 x (x2 − 2x + 1) dx = − x2 + x 3 10 1 0 ln(3x − 2) dx = x ln(3x − 2) − Z 2 [ln(3x − 2) − (x2 − 2x + 1)] dx Z 2 Z1 2 (x2 − 2x + 1) dx ln(3x − 2) dx − = 1 1 2 3 2 2 x 2 = x ln(3x − 2) − x + ln |3x − 2| − −x +x 3 3 1 1 2 8 1 = 2 ln 4 − 2 − − (0 − 1 − 0) − −4+2 − −1+1 3 3 3 2 1 4 = ln 4 − 2 + 1 − + 3 3 3 4 4 = ln 4 − 3 3 4 = (ln 4 − 1)unit2 3 M H 09 45 1 23 The enclosed area = UY 6. The shaded area is 8a unit2 . Find the value of a. Solution. 29 U Ye Min Htoo(YTU) Grade 12 Mathematics 3 a x (x + 1) dx = +x 3 −a 3 −a a (−a)3 = +a − + (−a) 3 3 3 a3 a +a− − −a = 3 3 3 3 a a = +a+ +a 3 3 2a3 + 2a = 3 Z a 10 1 0 2 23 By the problem: 09 45 1 2a3 + 2a = 8a 3 2a3 + 6a = 24a 2a3 − 18a = 0 a3 − 9a = 0 H a(a2 − 9) = 0 M a(a + 3)(a − 3) = 0 UY a = 0 (impossible) or a = −3 (impossible) ∴a=3 30 or a=3
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