Area Under the Curve Exercise # 3

EXERCISE # 3 Part-A Subjective Type Questions Q.1 Find area of the loop of the curve y ^ 2 = (x - 2) * (x - 4) ^ 2 Sol. =2 int 0 ^ sqrt 2 (t^ 2 -2),t.2tdt = 4 * integrate (t ^ 4 - 2t ^ 2) dt from 0 to sqrt(2) A = 2 * integrate (x - 4) * sqrt(x - 2) dx from 2 to 4 x - 2 = t ^ 2 = [4/5 * 0.4sqrt(2) - 8/3 * 2sqrt(2)] = 2/15 * 16sqrt(2) Q.2 Find the ratio in which the area enclosed by the curve y = cos x * (0 <= x <= pi / 2) in the first quadrant is divided by the curve y = sin x Sol. pi / 4 A_{1} = integrate (cos x - sin x) dx from 0 to pi / 4 = 1/(sqrt(2)) + 1/(sqrt(2)) - 1 = sqrt(2) - 1 A_{2} = integrate sin x dx from 0 to pi / 4 + integrate cos x dx from pi / 4 to pi / 2 = - (1/(sqrt(2)) - 1) + (1 - 1/(sqrt(2))) = 2 - sqrt(2) A_{1}/A_{2} = (sqrt(2) - 1)/(2 - sqrt(2)) = ((sqrt(2) - 1)(2 + sqrt(2)))/2 = (2sqrt(2) + 2 - 2 - sqrt(2))/2 = 1/(sqrt(2)) Q.3 Find the area of the figure bounded by y = - 3x ^ 2 - |x| + 2 and y = 0 Sol. - 2/3 2/3 A = 2 * integrate (- 3x ^ 2 - x + 2) dx from 0 to 2/3 =-2[x^ 3 ] 0 ^ (2/3) -[x^ 2 ] 0 ^ 2/3 +4[x] 0 ^ 2/3 = - 16/27 - 4/9 + 8/3 = 44/27 Q.4 Find the area bounded by x ^ 2 + y ^ 2 - 2x = 0 and y = sin(pi*x / 2) in the upper half of the circle. Sol. x ^ 2 + y ^ 2 - 2x = 0an * dy = sin((pi*x)/2) Rightarrow (x - 1) ^ 2 + y ^ 2 = 1an * dy = sin((pi*x)/2) (1, 1) (0,0) (1.0) (2,0) unit = pi/2 + 2/pi * \{- 1 - 1\} = pi 2 + 2 pi [cos((pi*x)/2)] 0 ^ 2 required areu= pi/2 - integrate sin((pi*x)/2) dx from 0 to 2 = (pi/2 - 4/pi) * s Q.5 Find the area of the region bounded by the parabola y ^ 3 = 4x and the normal to it at one of the ends of its latus rectum chi(1, 2) (9,-6) Sol. Line y - 2 = - (x - 1) x = 3 - y A = integrate (3 - y) dy from - 6 to 2 - integrate (y ^ 2)/4 dy from - 6 to 2 = 24 + 16 - 56/3 = 64/3 Q.6 Calculate the area enclosed by the curve 4 <= x ^ 2 + y ^ 2 <= 2(|x| + |y|) Sol. Given x ^ 2 + y ^ 2 >= 4 and ! (|x| + |y|) >= 4 Rightarrow x ^ 2 + y ^ 2 >=4 and[x|+|y|>=2 B(0, 2) A(2, 0) C(- 2,0) [ D(0, - 2) Common area Area of square ABCD 1 = (product of diagonal) 2 = 1/2 * (16) = 8 ; . unit. Q.7 Find the area of the region bounded by the axis of x, 0 <= x <= 1 y = arc(cos x) and y = arc(sin x) Sol. y = cos x, y = arcsin(x) x-axis, 0 <= x <= 1 1 / (sqrt(2)) * 1 11√2 required area = sinxdx + + cosx cos xdx 12 1/√2 =[xsinxdx + [x cosx+ (5)dx 1/2 dx + 11-x2 Put I - x ^ 2 = t ^ 2 and solving, we get arca= (sqrt(2) - 1) sq. unit. integrate x/(sqrt(1 - x ^ 2)) dx from 1 / (sqrt(2)) to 1 Q.8 Find the area enclosed by the curve y = d/dx (x * ln(x)) and the coordinate axes. Sol. y = 1 + lambda*nx A= [x*lambda_{1}*x] 0 ^ 1/e = - 1/\mathfrak{e} -lim x 0 * lambda n x |A| = 1/c Q.9 Find the area of curve enclosed by [x] + [y] = 4 in the Ist quadrant. Sol. [x] + [y] = 4 [x] = 4 - [y] when y = 0 [x]=4 Rightarrow 4 <= x < 5 [x]=3 Rightarrow 3 <= x < 4 y = 1 [x]=2 Rightarrow 2 <= x < 3 y = 2 [x]=l Rightarrow 1 <= x < 2 y = 3 [x]=0 Rightarrow 0 <= x < 1 y = 4 5 432 DC AB required area = 5 (area of square ABCD) = 5 * 1 = 5 Q.10 Let f(x) = minimum (ex, 3/2 l+ e ^ (- x) |0<= x<=1| Find the area bounded by y = f(x) x-axis, y-axis and the line x = 1 Sol. t - 1/t - 1 = 0 1 = (1 plus/minus sqrt(5))/2 c ^ x = 1 + e ^ (- x) t ^ 2 - t - 1 = 0 Q.11 Let y ^ z = 4[sqrt(y)] * x and x ^ 2 = 4[sqrt(x)] * y be two curves, where [] represents greatest integer function. Find area bounded by these two curves within the square formed by the lines x = 4 y = 4 x=1. y = 1 . 4 Sol. 12 A = 2 * integrate sqrt(x) dx from 1 to 4 - 1 - integrate (x ^ 2)/4 dx from 2 to 4 = 28/3 - 1 - 56/12 = 11/3 Q.12 Find out the area enclosed by circle |z| = 2 parabola y = x ^ 2 + x + 1 the curve y = [sin^2 (x/4) + cos(x/4)] * and x-axis, where [1 denotes the greatest integer function. Sol. so area enclose by the curves and 1 y=x^ 2 +x+1 Rightarrow (y - 3/4) = (x + 1/2) ^ 2 and y = 1 y = [sin^2 (x/4) + cos(x/4)] |z|=2 Rightarrow x ^ 2 + y ^ 2 = 4 1 < sin^2 (x/4) + cos(x/4) <2 Rightarrow y=1 25-1-1/20 2 required area solving we get sq. sq unit = sqrt(3) * 1 + (sqrt(3) - 1) * 1 + integrate (x ^ 2 + x + 1) dx from - 1 to 0 =(2 sqrt 3 -1)+ [(x ^ 3)/3 + (x ^ 2)/2 + x] -1 ^ 0 integrate sqrt(4 - x ^ 2) dx from sqrt(3) to 2 2 * [x/2 * sqrt(4 - x ^ 2) + 2arcsin(x/2)] ^ 2 area= ((2pi)/3 + sqrt(3) - 1/6) ; Q.13 Compute the area of the loop of the curve y ^ 2 = x ^ 2 * [(1 + x) / (1 - x)] Sol. (2 + (- pi) / 2) units^ 2 Q.14 Find the area included between the curve x ^ 2 * y ^ 2 = a ^ 2 * (y ^ 2 - x ^ 2) and its asymptotes. Sol. y ^ 2 = (u ^ 2 * x ^ 2)/(a ^ 2 - x ^ 2) y = plus/minus (ax)/(sqrt(a ^ 2 - x ^ 2)) =-2a*2 (sqrt(a ^ 2 - x ^ 2)) 0 ^ z A = 4 * integrate (ax)/(sqrt(a ^ 2 - x ^ 2)) dx from 0 to 2 = - 2u * integrate (- 2x)/(sqrt(a ^ 2 - x ^ 2)) dx from 0 to 2 = - 4a(- a) = 4a ^ 2 Part-B Passage based Question Passage I (Q. 15 to 17) Let there are three functions described here: f(x) = \{sin x\} g * (x) = \{cos x\} , h(x) = [x / pi] Where [x] is greatest integral part of x & {x) is the fractional part of x. On the basis of above information, answer the following questions- Q.15 If the area bounded by x-axis, x = 0 x = pi and f(x) is t_{1} and the area bounded by x-axis, x = pi x = 2pi and f(x) is t_{2} 1 then- (A) t_{2}/t_{1} > 1/2 (B) 0 < t_{2}/t_{1} < 1/2 (C) t_{2}/t_{1} > 1 (D) None of these Sol. [A] f(x) = \{sin x\}, g(x) = \{cos x\}, h(x) = \{x / pi\} 2 g(x) f(x) h(x) g(x) f(x) 1/2 h(x) (5pi)/4 (3pi)/2 2pi and 1t_{2} = 2 * integrate (1 + sin x) dx from pi to 3pi / 2 =2[x-cos x] pi ^ 3 pi/2 i_{i} = integrate sin x dx from 0 to pi =[-cos x] 0 ^ pi =1+1=2 = pi/2 - sqrt(2) 12=0.57 (approx.) 11 Clearly = 2[(3pi)/2 + 0 - pi - 1] = pi - 2 t_{2}/t_{1} = (pi - 2)/2 = pi/2 - 1 = 3.14/2 - 1 = 1.57 - 1 t_{2}/t_{1} > 1/2 Q.16 The area bounded by f(x), h(x) x = pi and x = 2pi is (A) 2 (Β) pi - 2 (C) (pi - 2)/2 (D) None Sol. [A] Area bounded by f(x) h(x) x = pi and x = 2pi from Q.1, we have Area bounded by x-axis, x = pi x = 2pi and and area of rectangle made by f(x) is i = pi - 2 x = pi x = 2pi and x-axis is pi*l = pi required area = pi - (pi - 2) = 2 Q.17 The area bounded by f(x) g(x), x-axis, x = pi and x = (3pi)/2 is- (A) pi/2 - 1/(sqrt(2)) (B) pi/2 - sqrt(2) () pi/2 - 1/2 (D) None of these Sol. [B] Required area = (5pi)/4 - 1/(sqrt(2)) - pi + (3pi)/2 - (5pi)/4 - 1/(sqrt(2)) =[x+sin x] pi ^ 5 pi/4 +[1-c os x] 5pi / 4 ^ (3pi / 4) = integrate (1 + cos x) dx from pi to 5pi / 4 + integrate (1 + sin x) dx from 5pi / 4 to 3pi / 2 Passage II (Q. 18 to 20) Five curves defined as follows: C_{2} C_{1} |x - y| <= 1 |x + y| <= 1 C_{4} |y| <= 1/2 |x| <= 1/2 C_{3} C_{5} 3x ^ 2 + 3y ^ 2 = 1 Q.18 The ratio of region bounded by C_{V} C_{2} and C_{3} C_{4} (A) 1.5 (B) 2 (C) 1/2 (D) None Sol. [B] A_{1} = 1 area bounded by C_{1} = area of square ABCD = 2 A_{2} = Area bounded by C_{3} C_{4} Area of square EFGH = 1 Rightarrow A_{1}/A_{2} = 2/1 = 2 C_{2} Q.19 The area bounded by C_{1} and C_{2} which does not contain the area of C_{5} is (A) 2 - pi/4 (B) 2 - pi/6 (C) 2 - pi/3 (D) None of these Sol. [C] area bounded by C_{1} and C_{2} = 2 area bounded by → area bounded by C_{1} and C_{2} which does not contain the area of C_{5} C_{s} = pi/3 = 2 - pi/3 Q.20 That part of area which is bounded by C_{1} and C_{2} but not bounded by C_{3} and C_{4} is- (A) 1 (B) 1/2 (C) 1/3 (D) None Sol. [A] Area bounded by C_{1} and C_{2} = 2 area bounded by C_{3} and C_{4} = 1 area bounded by C_{1} and C_{2} but not bounded by Rightarrow C_{4} = 2 - 1 = 1 C_{3} Passage III (Q. 21 to 23) Let there are two function defined here: f(x) = min(|x - 2|, |x + 2|) and g(x) = min(e ^ x, e ^ (- x)) Now the root of the equation c ^ (- x) + x - 2 = 0 is a, where alpha \in R Q.21 The area bounded by f(x) and x-axis is (A) 1 sq. unit (C) 6 sq. unit (B) 4 sq. unit (D) None of these Sol. [B] 1/2 * 4 * 2 = 4 Q.22 Which statement is correct - (Α) alpha \in (2, 3) (C) alpha \in (0, 2) (Β) alpha \in (- 1, 0) [C] Q.23 The area bounded by f(x) and x = 0 in g(x) (D) None of these Sol. first quadrant is (A) \mathfrak{c} ^ (- alpha) - 1 (C) 1 + e ^ (- alpha) (B) 2 - e ^ (- alpha) Sol. [D] (D) None of these 2 A = integrate (2 - x - c ^ (- x)) dx from 0 to alpha = 2a - 1/2 * a ^ 2 + c ^ (- alpha) - 1 Passage IV (Q. 24 to 26) In the adjacent figure, the graphs of two functions y = f(x) and y = sin x are given. They intersect at origin, A(a, f(a)) B(pi, 0) and C(2pi,0) A_{i} (i = 1, 2, 3) is the area bounded by the curves as shown in the figure respectively for If A_{1} = 1 + (a - 1) * cos a - sin a x \in (0, a) x \in (a, pi) x \in (pi, 2pi) A_{2} B 11 C 2π y = sin x ^ 4 A_{1} y = f(x) y = f(x) A_{3} Q.24 The function f(x) is- (A) x ^ 2 * sin x (B) x sinx (C) 2xsinx (D) x ^ 3 * sin x Sol. [B] From fig, we have diff. with respect to a, we get integrate (sin x - f(x)) dx from 0 to n = A_{1} Rightarrow integrate (sin x - f(x)) dx from 0 to 1 = 1 + (a - 1) * cos a - sin a Rightarrow f(x) = x * sin x Rightarrow f(a) = a * sin a sin a - f(a) = cos a - (a - l) * sin a - cos a Q.25 Value of A_{2} is- (Α) (pi - 1) units^ 2 (Β) (pi / 2 - 1) units^ 2 (C) (π-sin1-1) units² (D) π/2units Sol. [C] f(x) = x * sin x and y = sin x = [- x * cos x] 1 ^ pi +[sin x] 1 ^ pi +[cos x] 1 ^ pi A_{2} = integrate (x * sin x - sin x) dx from 1 to pi = pi + cos 1 - sin 1 - 1 - cos 1 = (pi - sin 1 - l) unit^ 2 Rightarrow (a, f(a)) = (1, a * sin a) Q.26 Value of A_{3} is- (Α) (2pi - 1) units^ 2 (C) (3pi - 2) units^ 2 (Β) (3pi - sin 2) units^ 2 (D) (pi - 2) units^ 2 Sol. [C] A_{3} = |integrate (sin x - x * sin x) dx from pi to 2pi| =[-cos x] pi ^ 2 pi +[cos x] pi ^ 2 pi - integrate cos x dx from pi to 2pi | =| - 1 - 1 + 2pi + pi -[sin x] pi ^ 2 pi | = (3pi - 2) unit^ 2 Passage V (Q. 27 to 29) Area enclosed by curve y = f(x) and y = x ^ 2 + 2 between the abscissa x = 2 and x = alpha is given as (a' - 4a ^ 2 + 8) sq. unit. It is known that curve y = f(x) lies below the parabola y = x ^ 2 + 2 Q.27 Area enclosed by curve y = f(x) with x-axis, x = 1 is x = 0 (A) 8/3 (B) 16/3 (C) 16/7 (D) 4/3 Sol. [B] Diff. w.r.to a f(x) = - 2x ^ 2 + 8x + 2 f(alpha) = - 2alpha ^ 2 + 8alpha + 2 Now alpha ^ 3 - 4alpha ^ 2 + 8 = integrate (x ^ 2 + 2 - f(x)) dx from 2 to a 3alpha ^ 2 - 8alpha = alpha ^ 2 + 2 - f(alpha) A = integrate (- 2x ^ 2 + 8x + 2) dx from 0 to 1 = - 2/3 + 4 + 2 = 16/3 Q.28 If f(x) lies above x-axis in then (q + p) is equal to x \in (p, q) (A) 2 (B) 3 (C) 4 (D) 8 Sol. [C] p + q = - (- 8)/2 = 4 2x ^ 2 - 8x + 2 = 0 < q ^ p Q.29 Value of area bounded by line y = x + 2 and y = f(x) x = 2 and x = 4 is (A) 36/5 (8) 7/5 (C) 123/13 (D) None of these Sol. [D] (7/11) * 11/2 4 2x ^ 2 - 7x = 0 - 2x ^ 2 + 8x + 2 = x + 2 x(2x - 7) = 0 = 33/8 + |(- 2/3 * (64 - 343/8)) + 4(16 - 49/4) + 2(4 - 7/2)| s = 1/2 * x * 3/2 * x * 11/2 + |integrate (- 2x ^ 2 + 8x + 2) dx from 1/2 to 4| = 33/8 + [- 338/24 + 15 + 1] = 33/8 + 46/24 = 145/24