# Area Under Curve-02-Solved example

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SOLVED SUBJECTIVE EXAMPLES Example 1 : Find the area bounded by y = x |sinx| and x-axis between x = 0, x = 2. Solution : y   x sin x, if sin x  0, i.e., 0  x   x sin x, if sin x  0, i.e.,   x  2  2  Required area =  x sin x dx   (x sin x) dx  x ( cos x)    ( cos x) dx 0  0 y 2 y= x  (x(cos x)) 2   (cos x) dx    sin x   (2  ) sin x 2  4 sq. units 0  Example 2 : O  2 x If the line y = mx divides the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y = 1 + 4x – x2 into two equal parts, then find the value of m. Solution : The given curve is y – 5 = – (x – 2)2 Thus given curve is a parabola with vertex at (2, 5) and axis x = 2 Given that area CBFC = Area CDEBC So area CDEBFC = 2 Area CBFC Area CDEBFC  3/ 2  0 (1 4x  x2 )dx 3/ 2 3  9  9 39  x  2x2    2     sq. units Area CBFC  3/ 2  0 2 0 mxdx  9m 4  4  8 8 So we must have 39  18m 8 8 or m  13 6 Example 3 : Find out the area enclosed by the polynomial function of least degree satisfying  f (x) 1/ x lim 1 x0  Solution : x3  = e and the circle x2 + y2 = 2 above the axis  f (x) 1/ x Since lim 1 x0  x3  exists and f(x) is of least degree, so f(x) must be of the form f(x) = ax4, solving the limit we get a = 1, so f(x) = x4 . Hence we have two curves y = x4 ... (i) y Solving we get,  x 4  x8  x 2  2  0 ... (ii)  (x2 – 1) (x6 + x4 + x2 + 2) = 0  x2 – 1 = 0  x =  1 Required area  2 1  0  x4 dx x x5  1  1  1   3    2 sin1    2        sq. units.  2 2 5  0  2 4 5   5 2  Example 4 :  Find out the area enclosed by y = x2 + cosx and its normal at x = 2 Solution : f(x) = x2 + cos x  f (x) = 2x – sinx in the first quadrant.    f ( )   1 2  Equation of normal at x = 2 is  2  1     y  4   1   x  2      at x axis, y = 0  ( 1)2  x   4 2 Required area = area OAPQO + area of triangle PQR / 2 2 1  ( 1) 2   2   (x  cos x) dx  2  4     4 0   x3  / 2  sin x  0  ( 1)2 8 2 2   3   1 2 3 4 =  1      1  1 8 4 24 32 5  4  3  32 32 24 Example 5 : Find out the ratio of areas in which the function f(x) =  x3 x  100 35  divides the circle   x2 + y2 – 4x + 2y + 1 = 0 ([.] denotes the greatest integer function). Solution : Circle is x2 + y2 – 4x + 2y + 1 = 0 Y (x –2)2+ (y + 1) 2= 4 (2 - 3, 0) (2 + X 3, 0) f(x) = 0 for 0 < x < 4 or, (x – 2)2 + (y + 1)2 = 4 = 22 ... (i) Now for 0  x  4 x3 x  x3 x  0    1      0 100 35 100 35  So, the circle have to find out the ratio in which x axis divides the circle (i). Now, at x-axis, y = 0. So, (x–2)2 = 3 So, the circle cuts the x axis at the points (2 – , 0) and (2 + , 0) 2 3 Let

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### Area Under Curve-02-Solved example

• 1. SOLVED SUBJECTIVE EXAMPLES Example 1 : Find the area bounded by y = x |sinx| and x-axis between x = 0, x = 2. Solution: xsin x, if sin x 0, i.e., 0 x y xsin x,if sin x 0, i.e., x 2              Required area = 2 0 xsin x dx ( xsin x)dx        0 0 x ( cos x) ( cos x)dx        2 2 (x( cos x)) ( cos x)dx          y y= x O x 2  2 0 sin x (2 ) sin x 4           sq. units Example 2 : If the line y = mx divides the area enclosed by the lines x = 0, y = 0, x = 3/2 and the curve y= 1 + 4x – x2 into two equalparts, then find the value ofm. Solution: The given curve is y – 5 = – (x – 2)2 Thus givencurve is a parabola with vertex at (2, 5) and axis x = 2 Given that area CBFC =Area CDEBC So area CDEBFC = 2 Area CBFC Area CDEBFC 3/2 2 0 (1 4x x )dx     3/2 3 2 0 x 3 9 9 39 x 2x 2 3 2 4 8 8              sq. units Area CBFC 3/ 2 0 9m mxdx 4    y = mx y x F x = 3/2 y = 1 + 4x – x 2 B E D C So we must have 39 18m 8 8  or 13 m 6 
• 2. Example 3 : Find out the area enclosed bythe polynomialfunctionofleast degree satisfying 1/ x 3 x 0 f(x) lim 1 x         = e and the circle x2 + y2 = 2 above the axis Solution: Since 1/ x 3 x 0 f(x) lim 1 x         exists and f(x) is of least degree, so f(x) must be of the formf(x) = ax4 , solving the limit we get a = 1, so f(x) = x4 . Hence we have two curves y = x4 ... (i) 2 y 2 x   ... (ii) Solving we get, 2 4 8 2 2 x x x x 2 0        (x2 – 1) (x6 + x4 + x2 + 2) = 0 y = x4 x –1 1 2 y 2 x = -  x2 – 1 = 0  x =  1 Required area   1 2 4 0 2 2 x x dx     1 5 2 1 0 x 2 x x 2 2 x sin 2 2 5 2              1 1 3 2 2 4 5 5 2                    sq. units. Example 4 : Find out the area enclosed byy = x2 + cosx and its normal at x= 2  in the first quadrant. Solution: f(x) = x2 + cos x  f (x)  = 2x – sinx  f ( ) 1 2     Equation ofnormalat x = 2  is 2 1 y x 4 1 2                      x (0, 1)A P Q R O
• 3. at x axis, y = 0  2 ( 1) x 4 2       Required area = area OAPQO + area of triangle PQR /2 2 2 2 0 1 ( 1) (x cos x)dx 2 4 2 2 4                       /2 3 2 0 x ( 1) sin x 3 8                =     3 2 3 4 1 1 1 1 24 8 4 24 32                         = 5 4 3 1 32 32 24       . Example 5 : Find out the ratio of areas in which the function f(x) = 3 x x 100 35          divides the circle x2 + y2 – 4x + 2y + 1 = 0 ([.] denotes the greatest integer function). Solution: Circle is x2 + y2 – 4x + 2y + 1 = 0 X Y (x –2) + (y + 1) = 4 2 2 f(x) = 0 for 0 < x < 4 (2 3,0) - (2 3, + 0) or, (x – 2)2 + (y + 1)2 = 4 = 22 ... (i) Now for 0 x 4   3 3 x x x x 0 1 0 100 35 100 35               So, the circle have to find out the ratio inwhichx axis divides the circle (i). Now, at x-axis, y= 0. So, (x–2)2 = 3
• 4. So, the circle cuts the x axis at the points (2 – 3 , 0) and (2 + 3 , 0) Let   2 3 2 2 3 4 3 3 A 4 (x 2) 1 dx 3           The required ratio is A 4 3 3 4 A 8 3 3        Example 6 : Find the area of the figure enclosed by the curve 5x2 + 6xy + 2y2 + 7x + 6y + 6 = 0 Solution: Equationofcurve canbe re-written as 2y2 + 6(1 + x) y + 5x2 + 7x + 6 = 0 1 3(1 x) (3 x)(x 1) y 2       , 2 3(1 x) (3 x)(x 1) y 2       Therefore the curves (y1 and y2 ) are defined for values of xfor which (3 – x) (x – 1)  0 i.e., 1 x 3   (Actuallythe given equationdenotes an ellipse, because 0   and h2 < ab). Required area willbe given by 3 3 1 2 1 1 A (y y )dx A (3 x)(x 1) dx         y x O y1 y2 1 2 3 –3 –6 Put x = 3 cos2  + sin2  i.e., dx = – 2sin2 d / 2 2 0 A 2 sin 2 d sq. units 2       Example 7 : Find the area enclosed between the curves y = ln(x + e), x = ln 1 y       and x-axis. Solution: Given curves are y= ln(x + e) and x = ln 1 y       or, x = –ln y or, y = e–x or, x 1 y e      
• 5. Clearlythe two curves cut at x = 0 Graph ofcurves (i) and (ii) willbe as showninthe figure. x C A O (1 – e, 0) x = –e Y y = ln(x + e) y = e–x Y’ B x’ Required area = shaded area 1 1 2 0 (x x )dy    1 y 0 [ ln y (e e)]dy      1 y 0 (yln y – y) e ey         1 y 0 y ln y y e ey          y 0 (0 1 e e) (0 0 1 0) lim ylog y 0                  = 1 + 1 = 2 sq. units. Example 8 : Find the area enclosed by the parabola (y – 2)2 = x– 1, the tangent to the parabola at (2, 3) and the x-axis. Solution: Given parabola is (y – 2)2 = x – 1 ... (i) Its axis is y = 2 and vertex is (1, 2). Let P (2, 3)  . From (i), 2(y – 2) dy 1 dx  dy 1 dx 2(y 2)    At dy 1 P(2, 3), dx 2   Equation of tangent at P(2, 3) is y x O y = 3 y = 2 Q(–4, 0) P (2, 3) A (1, 2) R (5, 0) y – 3 = 1 (x 2) 2  or, x 2y 4 0    ... (ii) Line (ii) cuts the x-axis at (–4, 0) and y-axis at (0, 2). Required area, RQPAR = 3 1 2 0 (x x )dy   3 3 2 2 0 0 [(y 2) 1 (2y 4)]dy (y 6y 9)dy           3 3 2 0 y 3y 9y 3            = (9 – 27 + 27) – 0 = 9 sq. units.
• 6. Example 9 : Find the area of the region bounded by the curves y = f(x), y = |g(x)| and the lines x = 0, x = 2, where fand g are continuous functions satisfying f(x+ y) = f(x) + f(y) – 8xyfor all x, y R  and g(x + y) = g(x) + g(y) + 3xy (x + y) for all x, y R  . also f (0) 8 and g (0) 4     Solution: Given, f(x + y) = f(x) + f(y) – 8xyfor all x, y Putting x = 0 and y = 0, we get f(0) = f(0) + f(0)  f(0) = 0 h 0 f (x h) f (x) f (x) lim h      h 0 f (x h) f (x 0) lim h      h 0 f (x) f(h) 8xh f(x) f(0) 8x.0 lim h        h 0 f (h) f (0) lim 8x f (0) 8x h        Thus f (x) 8 8x [ f (0) 8]       Integrating bothsides, we get f(x) = 8x – 4x2 + c ... (2) Putting x = 0, we get f(0) = 0 + c  c = 0 [ f (0) 0]   Hence f(x) = 8x – 4x2 ... (3) Given g(x + y) = g(x) + g(y) + 3xy(x + y) for allx and y Putting x = y= 0, we get g(0) = 0 ... (4) Now h 0 h 0 g(x h) g(x) g(x h) g(x 0) g (x) lim lim h h           2 h 0 g(h) g(0) lim 3x 3xh h            2 2 g (0) 3x 4 3x [ g (0) 4]          Thus g (x )  = – 4 + 3x2 ... (5)  g(x) = – 4x + x3 + k Putting x = 0, we get g(0) = k  k = 0 [ g(0) 0]    g(x) = x3 – 4x ... (6) For points where y= f(x) and y= g(x) intersect,
• 7. 8x – 4x2 = x3 – 4x  x3 + 4x2 – 12 x = 0  x = 0, 2, – 6 Sign scheme for f(x) i.e., for (8x – 4x2 ) is – –ve +ve 0 2 –ve  Y O X y = f(x) (2, 0) y = |g(x)| Sign scheme for g(x) i.e. for x(x2 – 4) is – –ve +ve +ve –2 0 2 –ve  f(x) – |g(x)| = 8x – 4x2 – (4x – x3 ) ( g(x)  0 in [0, 2]) = x3 – 4x2 + 4x = x(x – 2)2  0 Area bounded byy = f(x) and y = |g(x)| between x = 0 and x = 2 2 2 2 3 1 2 0 0 (y y )dx [(8x 4x ) (4x x )]dx         2 3 2 0 4 (x 4x 4x)dx sq. units 3      . Example 10 : Find the area enclosed by the circle x2 + y2 = 4, the parabola y = x2 + x + 1, the curve 2 x x y sin cos 4 4         and the x-axis, (where [x] denotes the integralpart ofx). Solution: Equation ofgiven circle is x2 + y2 = 4 ... (1)  – 2  x  2 and – 2  y  2 Let z = sin2 x x cos 4 4  = 1 – cos2 x x cos 4 4  = 1 + t – t2 , where t = cos x 4 Now – 1 x 1 2 4 2    0 < t  1  5 1 z 4   for x [ 2, 2],   curve y = 2 x x sin cos 4 4        becomes y = 1 ... (2)
• 8. Given parabola is y = x2 + x + 1 ... (3) or, 2 1 3 x y 2 4          ... (4) Its axis is 1 x 2   and vertex is 1 3 , 2 4        . Now, we have to find out the area enclosed by the circle x2 + y2 = 4, parabola 2 3 1 y x , 4 2                line y= 1 and x-axis y x O (0, 2) B A E Q R U S T P D C (2, 0) (–2, 0) – 3 1 1 2 - - T S R U D (–2, 0) ( 3, 0)  (–1, 0) (–1/2, 0) (2, 0) Required area isshown as shaded regionin the figure. Hence required area = area OABCO + area PQRS + area RQEAOR + 2 area CBDC. =   0 2 2 2 1 3 3 1 ( 3 1) 1 (x x 1)dx 2 4 x dx               0 2 3 2 2 1 3 1 x x x x 2 3 1 x 2 4 x 2sin 3 2 2 2                                 1 1 3 2 2 3 1 0 1 2 (0 ) 3 2 2 3                                            5 2 2 3 1 3 6 3       2 1 3 sq. units 3 6           .
• 9. SOLVED OBJECTIVE EXAMPLES Example 1 : Area enclosed by the curve |x –2| + |y + 1| = 1 is equal to (A) 4 sq. units (B) 6 sq. units (C) 2 sq. units (D) 8 sq. units Solution: After shiftingthe origin at the point (2, –1) theequationofcurve becomes, |x| + |y| =1. This curve willrepresent a square as shown inthe adjacent figure. y C O A D y - x = 1 x + y = 1 x + y = –1 x – y = 1 Area ofthis square is clearlyequalto 4 times the area oftriangleOAB. Thus required area = 2 sq. units. Example 2 : Area bounded by the curves y = |x| – 2 and y = 1 – |x–1| is equal to (A) 4 sq. units (B) 6 sq. units (C) 2 sq. units (D) 8 sq. units Solution: Bounded figureABCD is a rectangle. AB 1 1 2    BC 4 4 2 2    A(1, 1) B C D 2 0 1 –1 –2 –2 y = |x| –2 y = 1 – |x –1| Thus, bounded area = ( 2) (2 2) = 4 sq. units. Example 3 : Area bounded bythe curve y= max{sinx, cosx} and x-axis, betweenthe lines x 4   and x = 2 is equalto (A) (4 2 1) 2  sq. units (B) (4 2 1)sq.units  (C) (4 2 1) sq. units 2  (D) None ofthese
• 10. Solution: Bold lines represents the graph ofy= max{sinx, cosx}. Required area, /4     y x O y = sin x y = cos x 5 /4 /4 sin x dx sin x dx         3 /2 2 5 / 4 3 /2 cos xdx cos x dx         (4 2 1) sq.units 2   Example 4 : Area bounded by the parabola y = x2 – 2x + 3 and tangents drawn to it fromthe point P(1, 0) is equalto (A) 4 2 sq. units (B) 4 2 sq. units 3 (C) 8 2 sq. units 3 (D) 16 2 sq. units 3 Solution: Let thedrawntangents be PAand PB.AB isclearlythe chord ofcontact ofpoint P.Thus equation ofAB is 1 . (y 0) 2  = x.1 – (2 + 1) + 3 i.e., y = 4 x coordinates ofpointsAand B willbe given by, x2 – 2x + 3 = 4 i.e., x2 – 2x –1 = 0  x = 1 2  ThusAB = 2 2 units. Hence PAB 1 (2 2).4 4 2 2    sq. units A B x y = 4 P(1, 0) y = x – 2x + 3 2 Now area bounded by lineAB and parabola is equalto 1 2 2 1 2 (4 2 (x 2x 3))dx       = 4 2 3 sq. units. Thus required area = 4 2 8 2 4 2 3 3   sq. units.
• 11. Example 5 : Area bounded bythe curves y= sinx, tangent drawn to it at x = 0 and the line x = 2  , is equalto (A) 2 4 sq. units 2   (B) 2 4 sq. units 4   (C) 2 2 sq.units 4   (D) 2 2 sq. units 2   Solution: The tangent drawnto y= sinxat x = 0 is the line y= x. Clearly the line y= x lies above the graph of y= sinx x 0, 2         . Thus required area / 2 0 (x sin x)dx     / 2 2 2 0 x 4 cos x 2 4               sq. units. Example 6 : IfA(n) represents the area bounded bythe curve y= n. lnx, where n N  and n> 1, the x-axis and the lines x = 1 and x= e, then the value ofA(n) + nA(n–1) is equalto (A) 2 n e 1  (B) 2 n e 1  (C) n2 (D) en2 Solution: A(n) = n e e e 1 1 1 n x dx n n x.x dx n              l l  A(n – 1) = (n – 1)  A(n) + nA(n – 1) = n + n(n – 1) = n2 Example 7 : Value ofthe parametera such that the area bounded byy= a2 x2 + ax + 1, coordinateaxes and the line x = 1, attains it’s least value, is equalto (A) – 1 4 (B) – 1 2 (C) 3 4  (D) –1
• 12. Solution: a2 x2 + ax + 1 is clearlypositive for allreal values of x.Area under consideration 1 2 2 2 0 a a (a x ax 1)dx 1 3 2        2 1 (2a 3a 6) 6    2 1 3 9 18 2 a a 6 6 2 16 16                  = 2 1 3 39 2 a 6 4 8                 which is clearlyminimumfor a = – 3 4 . Example 8 : Area oftheregion which consists ofallthe points satisfying the conditions|x– y| + |x+ y|  8 and xy  2, is equalto (A) 4(7 – ln8) sq. units (B) 4 (9 – ln8) sq. units (C) 2(7 – ln8) sq. units (D) 2 (9 – ln 8) sq. units Solution: The expression|x– y| + |x+ y|  8, represents the interior regionofthe square formedbythe lines x 4, y 4 and xy 2    represents the regionlying inside the hyperbola xy= 2. Required area,   4 4 1/2 1/2 2 2 4 dx 2 4x 2 nx x                 l = 4(7 – 3 ln2) sq. units x = –4 x = 4 y = 4 y = x D A x C B y y = –4 y = –x Example 9 : Apoint Pmoves inxyplaneinsuchawaythat [|x|]+ [|y|]=1, where [.] denotesthe greatest integer function.Areaofthe region representing allpossible positions ofthe point Pis equalto (A) 4 sq. units (B) 16 sq. units (C) 2 2 sq. units (D) 8 sq. units
• 13. Solution: If [|x|] = 1 and [|y|] = 0 then 1 | x | 2, 0 | y | 1     x ( 2, 1] [1, 2), y ( 1,1)        if [|x|] = 0, [|y|] = 1 Then x ( 1, 1), y ( 2, 1] [1,2]       y x –2 2 –2 –1 1 1 –1 O 2 Area ofrequired region = 4(2 – 1) (1 – (–1)) = 8 sq. units. Example 10 : Area enclosed bythe curve y= f(x) defined parametricallyas 2 2 2 1 t 2t x , y 1 t 1 t      is equalto (A) sq.units  (B) / 2 sq. units  (C) 3 sq. units 4  (D) 3 2  sq. units Solution: Clearlyt canbe any real number Let 2 2 1 tan t tan x 1 tan          x cos2   and 2 2tan y sin 2 1 tan        x2 + y2 = 1 Thus required area = 2 .1   sq. units.
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