Area Under the Curve Exercise # 1

AREA UNDER THE CURVE EXERCISE # 1 Questions based on Q.1 Area bounded by curve and x-axis Q.4   Area bounded by y = sec2 x, x = , x = and 6 3 (A) x- axis is2 (A) (B) 3 Sol. 3 2 (C) 2 3 Sol.  sec 2 x dx = [tan x ] /6 3 = 1  3 (C) 4 3 (D) 8 3 [C] y = 4x – x2 – 3  (x – 3) (x – 1) = 0  x = 1, 3 /3 3 /6 A= 2 e  1 e 1 sq. units (B) sq. units 2 2 =1+ 2 (C) e  + 1 sq. units (D) e  – 1 sq. units [B] 2 2   2 A  x e x dx 4 1 = 3 3 The area of the region bounded by the curve y = sin x and the x- axis in [–] is(A) 4 (B) 8 (C) 12 (D) 2 Sol. [A]   Area = 2 sin x dx Put x2 = t 0 2   3) dx Q.5 0 1 2 2 3 2 A  (4x  x   x3 1  3x  = 18 – 9 – 9 – 2 + = 2 x 2  +3 3 3   1 3 Area bounded by the curve y = xe x , x- axis and the ordinates x = 0, x =  is(A) Sol. 2 3 (B) 1 2 Q.3 1 3  x2 – 4x + 3 = 0 A= Sol. (D) [A] /3 Q.2 2 3 The area bounded by the x-axis and the curve y = 4x – x2 – 3 is- e t dt  0 e 2  1 = 2 [ cos x ]0 = 2[1 + 1] = 4 2 The area bounded by the curve y = f(x), x-axis and the ordinates x = 1 and x = b is (b – 1) sin (3b + 4), then f(x) equals(A) (x –1) cos (3x + 4) (B) sin (3x + 4) (C) sin (3x + 4) + 3(x –1) cos (3x + 4) (D) None of these Questions based on Q.6 Sol. (B) 5 7 (C) 7 3 (D) 7 5 [C] 4   f (x) dx = (b – 1) sin (3b + 4) = 1  f(x) = sin (3x+ 4) + 3(x –1) cos (3x + 4) 3 7 1 A y dy 21 b diff. with respect. to b, we have f(b) = sin (3b + 4) + 3(b –1) cos (3b + 4) Area in 1st quadrant bounded by y = 4x2, x = 0, y = 1 and y = 4 is(A) [C] Given that Area bounded by curve and y-axis Q.7   1 3/ 2 y 3 4 1 = 7 3 The area between the curves x = 2 – y – y2 and y-axis, is- (A) 9 Sol. (B) 9 2 (C) 9 4 [B] y2 + y – 2 = 0 (y + 2) (y –1) = 0  y = 1, –2 =2 Q.10 1 required area =  x dy 2 Sol. =  (2  y  y 2 ) dy 2 Questions based on Q.8 Sol.  4a  2 x3    4a . x 3 / 2   =1 3 12a   0  Q.11  = 0 0 +   1 3 x 3 Sol. 2 0 3 8 4 n 2 3 Sol. (C) 4 3 (D) 5 3     2x  4  4x    dx  3  4 1 0 2 3 [A] From curve y2 = 4x and line 2x – 3y + 4 = 0 we get x coordinate of intersection point which is x = 1, x = 4 1 2 8 1 (A) (B) (C) 3 3 3 [C] Parabola is x2 = y – 1 Tangent to it from origin is y = 2x and y = – 2x Intersection point of x2 = y – 1 and y = 2x is (1, 2)  (x (B) 4 1  2  = 2. x 3 / 2  ( x 2  4x ) 3  3 1 Area bounded by y = x2 + 1 and the tangents to it drawn from the origin, is- Area = 2 1 3 Area = 3 4  = n 2 3 Q.9 3 4 The area bounded by the curve y2 = 4x and the line 2x – 3y + 4 = 0, is(A) 0 2 2 (4a ) 3 (4a)2 – =1 3 12a a = A  (2 x  2x  x 2 ) dx    4a 2 – x2 3 4  x 2    4ax  dx = 1  4a  0 Area of figure bounded by straight lines x = 0, x = 2 and the curves y = 2x , y = 2x – x2 is4 3 3 (A) 3log2e – (B) – n 2 3 4 3 4 3 3 (C) + (D) + n 2 4 n 2 3 [B] 2 (D) 2    4ax  x  dx  4a  0 Given that Area = 1 Area bounded by two curves  2x    n 2  (C) 4 3 [D] Curve is y2 = 4ax and x2 = 4ay intersection point is (0, 0) and (4a, 4a) So Area = 1 1 4 8 9 – +4+ – = 2 3 2 3 2 =  (B) 4 3 4a 1  y 2 y3  = 2 y    2 3   2 =2–  2 The value of a for which the area between the curves y2 = 4ax and x2 = 4ay is 1 sq. unit, is(A) 1 1  ( x  1) 3  2 ( x  1) dx = 2    3   0 3 0 1 (D) 3 2  1  2x ) dx = 10 (D) 3 32 32 4 5 1     3 3 3 3 3 Q.12 The area bounded by the curve y = ex & the lines y = |x –1|, x = 2 is given by(A) e2 + 1 (B) e2 –1 (C) e2 – 2 (D) None of these Sol. [C] y = ex, y = | x – 1 |, x = 2 area of the part surrounding by C1, C2 and the yS axis, then lim a2 1 (B) 2 a 0 (A) 4 (0,1) (2,1) Sol. (0, 0) (1,0) (2,0) equals- e dx  1 a (0,e ) (0,1) = e2 – 2 Sol. The average value of f(x) = sec2 x from x = 0 to  x = , is4   2 4 (A) (B) (C) (D)   4 2 [D] a/2  (e Area = a x a/2  e x ) dx  (e a  x  e x ) 0 0 S = ea – 2 ea/2 + 1  Lim a 0 Questions based on The area of the region(s) enclosed by the curves y = x2 and y = (A) 1/3 Sol. a/2 sec2 x dx  0 0 4 / 4 4 4  tan x   0   Q.14 C2 / 4 1 (B) 2/3 Q.16 | x | is- (C) 1/6 (D) 1 [B] y = x2 and y2 = | x | Sol. S a = Lim 2 e a  2e a / 2  1 a 0 a 2  1 4 Parametric Curves The area bounded by the curve x = a cos3t, y = a sin3t is(A) a 2 32 (B) a 2 32 (C) 3a 2 16 (D) None of these [A] x = a cos3t, y = a sin3t  x2/3 + y2/3 = a2/3 (0, a) (1,0) 1 2 x3  Area = 2 ( x  x 2 ) dx  2  x 3 / 2   3   3 0 0 1  2 1 2 =2     3 3 3 (a, 0) a  Area = 4 y dx Q.15 1 4 C1 x 0 Q.13 (D) [D] C1 : y = ex and C2 : y = ea–x 2 Required area = (C) 0 Let 'a' be a positive constant number. Consider two curves C1 : y = ex, C2 : y = ea–x. Let S be the 0  y = a sin3t and dx = –3 a cos2t sin t dt 0 0  Area = = – [ x  e log( x  e)] (a sin 3 t ) (3a cos2 t sin t ) dt 1 e =e+1–e=1 /2 /2  sin = 3a2 4 t cos2 t dt Q.19 0 y = nx & y = (nx)2 is- (3.1) . 1  3a 2 3a2 . .  6.4.2 2 32 [C] The area bounded by the curves x = a ( – sin), y = n x y = a(1 – cos ), 0 2, is- Intersection at x = 1, and x = e (A) a2 sq. units (B) 2a2 sq. units (C) 3a2 sq. units Sol. (B) e –1 (D) 1 (A) e + 1 (C) 3 – e Sol. Q.17 The area of the figure bounded by the curves  (n x)  (n x)  dx e Area = (D) 4a2 sq. units 2 1 [C] e 1  A  a (1  cos ) . a(1 – cos ) d   2 n x dx e e 1 = 1 – e + 2 x n x  x  e 1 2  (1  cos 2 1 0 2  = x n x  x   x (n x ) 2 = a y = (n x)2 and 2   2 cos ) d = 1 – e + 2 (e – e + 1) = 3 – e 0  = a2 [ x ] 02    2  1 1  x  sin 2   0   2 2 0  Q.20 The area enclosed by the curve y2 + x4 = x2, is(A) = a (2 + ) 2 Sol. = 3a2 Questions based on Q.18 Sol. 2 3 (B) 4 3 (C) 8 3 (D) 10 3 [B] Miscellaneous The area enclosed between the curve y = loge (x + e) and the coordinate axes is(A) 4 (B) 3 (C) 2 (D) 1 –1 1 [D] log (x + e) 1 Total Area = 4  0 –e 1–e O 1 y dx = 4 x 1  x 2 dx 0 Solving we get Area = 4 3 0 Area =  log (x + e) dx Q.21 1e 0 = [ x log ( x  e)] 1e 0 =–  1 e x ee dx xe 0  Re(z) =  x  e dx 1e Let z be a complex number such that x x 2  4 , and Im (z) = y  4 satisfying |z| = 10 . Area enclosed by the set of points (x, y) on the complex plane, is(A) 8 6 (B) 4 6 (C) Sol. 20 10 3 40 10 3 (D) (0, 3) [A] x2 + 4 + y – 4 = 10 x2 = – (y – 10) (4, 0) (0,10) 9  10  6 Area in first quadrant 10 6 y4 6  6  A = 2  y dx  4  6  = 2 (10  x 2 )dx = 8 6 0  0     Q.22 Sol. Area of the curve y = (7 – x) (5 + x) above x-axis and between the ordinates x = –5 and x = 1, is- Q.24 = ab 1  ab 4 2 = 12 1   . 12 = 3(–2) 4 2 The area bounded by y = 2 – |2 – x| and y = 2 (A) 9 (B) 18 (C) 15 (D) None of these [A] y2 = – (x – 7) (x + 5) – + –  4  3 n 3 2 3 (C) + n3 2 4  3 n 3 2 1 (D) + n3 2 (A) (B) [B]  True / False type questions 7 1 Area = is- Sol. –5 3 , |x| Q.25 35  2x  x 2 dx 5 The area of the figure bounded by the curves y = |x –1| and y = 3 – |x| is 4 sq. unit. Sol. 1 =  36  ( x  1) 2 dx (0,3) 5 (–1,2) 1 x 1 x 1  36  =  sin 1  36  ( x  1) 2  2 6 2   5 y+x=1 = 9 Q.23 (1,0) The area bounded in the first quadrant 2 between the ellipse 3x + 4y = 12, is(A) 6( –1) (C) 3( –1) Sol. Ellipse A= 2 2 . 2 =4 2 x y + = 1 and the line 16 9 Q.26 (B) 3(–2) (D) None of these Area of the region bounded by y = {x} and 1 2x – 1 = 0, y = 0 is , ({ } stands for fraction 4 part) Sol. [B] x 2 y2 =1  16 9 x y line  = 1 4 3 (2,1) y = x+3 1/2 1/2 A= / 2 1 8 Sol. A=  sin x dx =  cos x / 2 1 0 0 Q.27 The positive value of 'a' for which area covered by figure y = sin ax, y = 0, x = Sol.   and x = is 3a a 1 equal to 3, is . 2 /a 1 A = – cos a x   / 3a a A=  a= 3 1 = 2A 2 when A = 3 Area = 2 Q.32 The area bounded by x-axis, x = 0, x = 2e and f(x) = ||x – e| –e | is……………….. e Sol. O –e Sol. A / 4 A=2  The area bounded by f(x) = min. (|x|, |x –1|) and x-axis is…………………. O (1/2, 0)B (1, 0) Sol. Area= Area of OAB 1  1 1 = 1   = 2  2 4 The area bounded by x-axis, x = f(x) = {sin x} is…………………  and 2  1   1  1 = 2 1  = –2   2  2   =2– / 4 sin y dy = –2 cos y 0 A Q.30 Area bounded by the curve y = sin–1 x, y = cos–1 x and x = 0 is equal to …………. B (2e, e) area = area of OAB 1 = .2e.e = e2 2 Q.29 Sol. Area enclosed by the curve |x – 2| + |y + 1| = 1 is equal to …………………. | x – 2 | + | y + 1| = 1 |X|+|Y|=1 1 1 3 1    a 2  2a  Fill in the blanks type questions Q.28 Q.31 2 0 EXERCISE # 1 Q.No. Ans. Q.No. Ans. 1 A 21 A 2 B 22 A 3 C 23 B 4 C 24 B 5 A 6 C 7 B 8 B 9 C 10 D 11 A 12 C 13 D 14 B 15 D 16 A 17 C 18 D 19 C 25. True 26. False 27. True 28. e2 sq. unit 29. 1/4 sq. unit 30. 1 sq. unit 31. 2 sq. unit 32. (2 – 20 B 2 ) sq. unit