Page 1 E 1 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s 1 . AREA UNDER THE CURVES : ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and O x=a x=b dx y=ƒ (x) x y x = b is given by A = b a y dx ? , where y = ƒ (x) lies above the x-axis and b > a. Here vertical strip of thickness dx is considered at distance x. (b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider O x y a b the magnitude only, i.e. A = b a y dx ? (c) If curve crosses the x-axis at x = c, then A = c b a c y dx ydx ? ? ? O x y x=a x=b c (d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : O x y y=b dy y=a Area bounded by the curve, y-axis and the two abscissae at y = a & y = b is written as b a A xdy ? ? . Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Find the area bounded by y = sec 2 x, x = 6 ? , x = 3 ? & x-axis Solution : Area bounded = / 3 / 6 ydx ? ? ? = / 3 2 / 6 sec xdx ? ? ? = / 3 / 6 [tan x] ? ? = tan 3 ? – tan 6 ? = 3 – 1 3 = 2 3 sq.units. Illustration 2 : Find the area in the first quadrant bounded by y = 4x 2 , x = 0, y = 1 and y = 4. Solution : Required area = 4 1 x ? dy = 4 1 y 2 ? dy = 4 3 / 2 1 1 2 y 2 3 ? ? ? ? ? ? y=1 y=4 x=0 Y O X = 1 3 [4 3/2 – 1] = 1 3 [8 – 1] = 7 3 = 2 3 ? sq.units. Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = 4 ? and 3 x 4 ? ? Solution : Required area = / 2 3 / 4 / 4 / 2 sin 2xdx sin 2xdx ? ? ? ? ? ? ? = / 2 3 / 4 / 4 / 2 cos2x cos2x 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?/4 3 /4 ? ?/2 = 1 1 [ 1 0] (0 ( 1)) 2 2 ? ? ? ? ? ? = 1 sq. unit AREA UNDER THE CURVE JEEMAIN.GURU Page 2 E 1 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s 1 . AREA UNDER THE CURVES : ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and O x=a x=b dx y=ƒ (x) x y x = b is given by A = b a y dx ? , where y = ƒ (x) lies above the x-axis and b > a. Here vertical strip of thickness dx is considered at distance x. (b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider O x y a b the magnitude only, i.e. A = b a y dx ? (c) If curve crosses the x-axis at x = c, then A = c b a c y dx ydx ? ? ? O x y x=a x=b c (d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : O x y y=b dy y=a Area bounded by the curve, y-axis and the two abscissae at y = a & y = b is written as b a A xdy ? ? . Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Find the area bounded by y = sec 2 x, x = 6 ? , x = 3 ? & x-axis Solution : Area bounded = / 3 / 6 ydx ? ? ? = / 3 2 / 6 sec xdx ? ? ? = / 3 / 6 [tan x] ? ? = tan 3 ? – tan 6 ? = 3 – 1 3 = 2 3 sq.units. Illustration 2 : Find the area in the first quadrant bounded by y = 4x 2 , x = 0, y = 1 and y = 4. Solution : Required area = 4 1 x ? dy = 4 1 y 2 ? dy = 4 3 / 2 1 1 2 y 2 3 ? ? ? ? ? ? y=1 y=4 x=0 Y O X = 1 3 [4 3/2 – 1] = 1 3 [8 – 1] = 7 3 = 2 3 ? sq.units. Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = 4 ? and 3 x 4 ? ? Solution : Required area = / 2 3 / 4 / 4 / 2 sin 2xdx sin 2xdx ? ? ? ? ? ? ? = / 2 3 / 4 / 4 / 2 cos2x cos2x 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?/4 3 /4 ? ?/2 = 1 1 [ 1 0] (0 ( 1)) 2 2 ? ? ? ? ? ? = 1 sq. unit AREA UNDER THE CURVE JEEMAIN.GURU 2 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Do yourself - 1 : (i) Find the area bounded by y = x 2 + 2 above x-axis between x = 2 & x = 3. (ii) Using integration, find the area of the curve 2 y 1 x ? ? with co-ordinate axes bounded in first quadrant. (iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?. (iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x = 1 2 ? and x=1. 2 . AREA ENCLOSED BETWEEN TWO CURVES : O y y =ƒ (x) 1 y =g(x) 2 x 1 x 2 x ( a ) Area bounded by two curves y = ƒ (x) & y = g(x) such that ƒ (x) > g(x) is 2 1 x 1 2 x A (y y )dy ? ? ? 2 1 x x A [ƒ(x) g(x)]dx ? ? ? (b ) In case horizontal strip is taken we have y x =g(y) 2 x =ƒ (y) 1 x y 2 y 1 2 1 y 1 2 y A (x x )dy ? ? ? 2 1 y y A [ƒ(y) g(y)]dy ? ? ? (c) If the curves y 1 = ƒ (x) and y 2 = g(x) intersect at x = c, then required area x=a x=b c y x y =g(x) 2 y =ƒ (x) 1 A = c b a c (g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ? ? ? = b a ƒ(x) g(x) dx ? ? Note : Required area must have all the boundaries indicated in the problem. Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 and x-axis Solution : To determine the sign, we follow the usual rule of change of sign. y = +ve for x > 3 y = –ve for 2 < x < 3 y = +ve for 1 < x < 2 y = –ve for x < 1. 3 0 | y| dx ? = 1 0 | y| dx ? + 2 1 | y| dx ? + 3 2 | y| dx ? 1 2 3 E D F C B O Y A (0, –6) X = 1 0 – y dx ? + 2 1 y dx ? + 3 2 –y dx ? Now let F(x) = ? (x – 1) (x – 2) (x – 3) dx = ? (x 3 – 6x 2 + 11x – 6) dx = 1 4 x 4 – 2x 3 + 11 2 x 2 – 6x. ? F(0) = 0, F(1) = – 9 4 , F(2) = –2, F(3) = – 9 4 . Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2 3 4 sq.units. JEEMAIN.GURU Page 3 E 1 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s 1 . AREA UNDER THE CURVES : ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and O x=a x=b dx y=ƒ (x) x y x = b is given by A = b a y dx ? , where y = ƒ (x) lies above the x-axis and b > a. Here vertical strip of thickness dx is considered at distance x. (b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider O x y a b the magnitude only, i.e. A = b a y dx ? (c) If curve crosses the x-axis at x = c, then A = c b a c y dx ydx ? ? ? O x y x=a x=b c (d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : O x y y=b dy y=a Area bounded by the curve, y-axis and the two abscissae at y = a & y = b is written as b a A xdy ? ? . Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Find the area bounded by y = sec 2 x, x = 6 ? , x = 3 ? & x-axis Solution : Area bounded = / 3 / 6 ydx ? ? ? = / 3 2 / 6 sec xdx ? ? ? = / 3 / 6 [tan x] ? ? = tan 3 ? – tan 6 ? = 3 – 1 3 = 2 3 sq.units. Illustration 2 : Find the area in the first quadrant bounded by y = 4x 2 , x = 0, y = 1 and y = 4. Solution : Required area = 4 1 x ? dy = 4 1 y 2 ? dy = 4 3 / 2 1 1 2 y 2 3 ? ? ? ? ? ? y=1 y=4 x=0 Y O X = 1 3 [4 3/2 – 1] = 1 3 [8 – 1] = 7 3 = 2 3 ? sq.units. Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = 4 ? and 3 x 4 ? ? Solution : Required area = / 2 3 / 4 / 4 / 2 sin 2xdx sin 2xdx ? ? ? ? ? ? ? = / 2 3 / 4 / 4 / 2 cos2x cos2x 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?/4 3 /4 ? ?/2 = 1 1 [ 1 0] (0 ( 1)) 2 2 ? ? ? ? ? ? = 1 sq. unit AREA UNDER THE CURVE JEEMAIN.GURU 2 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Do yourself - 1 : (i) Find the area bounded by y = x 2 + 2 above x-axis between x = 2 & x = 3. (ii) Using integration, find the area of the curve 2 y 1 x ? ? with co-ordinate axes bounded in first quadrant. (iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?. (iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x = 1 2 ? and x=1. 2 . AREA ENCLOSED BETWEEN TWO CURVES : O y y =ƒ (x) 1 y =g(x) 2 x 1 x 2 x ( a ) Area bounded by two curves y = ƒ (x) & y = g(x) such that ƒ (x) > g(x) is 2 1 x 1 2 x A (y y )dy ? ? ? 2 1 x x A [ƒ(x) g(x)]dx ? ? ? (b ) In case horizontal strip is taken we have y x =g(y) 2 x =ƒ (y) 1 x y 2 y 1 2 1 y 1 2 y A (x x )dy ? ? ? 2 1 y y A [ƒ(y) g(y)]dy ? ? ? (c) If the curves y 1 = ƒ (x) and y 2 = g(x) intersect at x = c, then required area x=a x=b c y x y =g(x) 2 y =ƒ (x) 1 A = c b a c (g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ? ? ? = b a ƒ(x) g(x) dx ? ? Note : Required area must have all the boundaries indicated in the problem. Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 and x-axis Solution : To determine the sign, we follow the usual rule of change of sign. y = +ve for x > 3 y = –ve for 2 < x < 3 y = +ve for 1 < x < 2 y = –ve for x < 1. 3 0 | y| dx ? = 1 0 | y| dx ? + 2 1 | y| dx ? + 3 2 | y| dx ? 1 2 3 E D F C B O Y A (0, –6) X = 1 0 – y dx ? + 2 1 y dx ? + 3 2 –y dx ? Now let F(x) = ? (x – 1) (x – 2) (x – 3) dx = ? (x 3 – 6x 2 + 11x – 6) dx = 1 4 x 4 – 2x 3 + 11 2 x 2 – 6x. ? F(0) = 0, F(1) = – 9 4 , F(2) = –2, F(3) = – 9 4 . Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2 3 4 sq.units. JEEMAIN.GURU E 3 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves y = 2 x , y = 2x – x 2 . Solution : Figure is self-explanatory y = 2 x , (x – 1) 2 = – (y – 1) y=2x–x 2 x=2 R(2,4) M(2,0) y=2 x Q (0,1) O The required area = 2 1 2 0 (y y ) ? ? dx where y 1 = 2 x and y 2 = 2x – x 2 = 2 x 2 0 (2 2x x )dx ? ? ? = 2 x 2 3 0 2 1 x x ln 2 3 ? ? ? ? ? ? ? ? = 4 8 4 ln 2 3 ? ? ? ? ? ? ? ? – 1 ln 2 = 3 4 – ln 2 3 sq.units. Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y 2 , x = 1 – 3y 2 . Solution : Solving the equations x = –2y 2 , x = 1 – 3y 2 , we find that ordinates of the points of intersection of the two curves as y 1 = – 1, y 2 = 1. The points are (–2, –1) and (–2, 1). 1 Y x=1–3y 2 P 1 1 X –1 –2 –1 O P 2 x=–2y 2 (–2, 1) (–2, –1) The required area 2 1 1 2 0 (x x ) ? ? dy = 2 1 2 2 0 [(1 3y ) (–2y )]dy ? ? ? =2 1 2 0 (1 y )dy ? ? = 2 1 3 0 y y 3 ? ? ? ? ? ? ? = 4 3 sq.units. Do yourself - 2 : (i) Find the area bounded by y x ? and y = x. (ii) Find the area bounded by the curves x = y 2 and x = 3 – 2y 2 . (iii) Find the area of the region bounded by the curves x = 1 2 , x = 2, y = logx and y = 2 x . 3 . CURVE TRACING : The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be extremely useful to quickly and correctly evaluate the area under the curves. ( a ) Symmetry : The symmetry of the curve is judged as follows : (i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x. (ii) If all the powers of x are even, the curve is symmetrical about the axis of y. (iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y. (iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is symmetrical about y = x. (v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is symmetry in opposite quadrants. (b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents. (c) Find the points where the curve crosses the x ?axis & also the y ?axis. (d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when x ? ? or ? ?. Illustration 7 : Find the area of a loop as well as the whole area of the curve a 2 y 2 = x 2 (a 2 – x 2 ). Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0) Area of a loop = 2 a 0 y dx ? = 2 a 2 0 x a x dx a ? ? ? Y O A (a,0) X X' A' (–a,0) = – a 2 2 0 1 a x (–2x)dx a ? ? = – a 2 2 3 / 2 0 1 2 2 (a x ) a 3 3 ? ? ? ? ? ? ? ? a 2 Total area = 2 × 2 3 a 2 = 4 3 a 2 sq.units. s. JEEMAIN.GURU Page 4 E 1 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s 1 . AREA UNDER THE CURVES : ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and O x=a x=b dx y=ƒ (x) x y x = b is given by A = b a y dx ? , where y = ƒ (x) lies above the x-axis and b > a. Here vertical strip of thickness dx is considered at distance x. (b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider O x y a b the magnitude only, i.e. A = b a y dx ? (c) If curve crosses the x-axis at x = c, then A = c b a c y dx ydx ? ? ? O x y x=a x=b c (d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : O x y y=b dy y=a Area bounded by the curve, y-axis and the two abscissae at y = a & y = b is written as b a A xdy ? ? . Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Find the area bounded by y = sec 2 x, x = 6 ? , x = 3 ? & x-axis Solution : Area bounded = / 3 / 6 ydx ? ? ? = / 3 2 / 6 sec xdx ? ? ? = / 3 / 6 [tan x] ? ? = tan 3 ? – tan 6 ? = 3 – 1 3 = 2 3 sq.units. Illustration 2 : Find the area in the first quadrant bounded by y = 4x 2 , x = 0, y = 1 and y = 4. Solution : Required area = 4 1 x ? dy = 4 1 y 2 ? dy = 4 3 / 2 1 1 2 y 2 3 ? ? ? ? ? ? y=1 y=4 x=0 Y O X = 1 3 [4 3/2 – 1] = 1 3 [8 – 1] = 7 3 = 2 3 ? sq.units. Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = 4 ? and 3 x 4 ? ? Solution : Required area = / 2 3 / 4 / 4 / 2 sin 2xdx sin 2xdx ? ? ? ? ? ? ? = / 2 3 / 4 / 4 / 2 cos2x cos2x 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?/4 3 /4 ? ?/2 = 1 1 [ 1 0] (0 ( 1)) 2 2 ? ? ? ? ? ? = 1 sq. unit AREA UNDER THE CURVE JEEMAIN.GURU 2 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Do yourself - 1 : (i) Find the area bounded by y = x 2 + 2 above x-axis between x = 2 & x = 3. (ii) Using integration, find the area of the curve 2 y 1 x ? ? with co-ordinate axes bounded in first quadrant. (iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?. (iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x = 1 2 ? and x=1. 2 . AREA ENCLOSED BETWEEN TWO CURVES : O y y =ƒ (x) 1 y =g(x) 2 x 1 x 2 x ( a ) Area bounded by two curves y = ƒ (x) & y = g(x) such that ƒ (x) > g(x) is 2 1 x 1 2 x A (y y )dy ? ? ? 2 1 x x A [ƒ(x) g(x)]dx ? ? ? (b ) In case horizontal strip is taken we have y x =g(y) 2 x =ƒ (y) 1 x y 2 y 1 2 1 y 1 2 y A (x x )dy ? ? ? 2 1 y y A [ƒ(y) g(y)]dy ? ? ? (c) If the curves y 1 = ƒ (x) and y 2 = g(x) intersect at x = c, then required area x=a x=b c y x y =g(x) 2 y =ƒ (x) 1 A = c b a c (g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ? ? ? = b a ƒ(x) g(x) dx ? ? Note : Required area must have all the boundaries indicated in the problem. Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 and x-axis Solution : To determine the sign, we follow the usual rule of change of sign. y = +ve for x > 3 y = –ve for 2 < x < 3 y = +ve for 1 < x < 2 y = –ve for x < 1. 3 0 | y| dx ? = 1 0 | y| dx ? + 2 1 | y| dx ? + 3 2 | y| dx ? 1 2 3 E D F C B O Y A (0, –6) X = 1 0 – y dx ? + 2 1 y dx ? + 3 2 –y dx ? Now let F(x) = ? (x – 1) (x – 2) (x – 3) dx = ? (x 3 – 6x 2 + 11x – 6) dx = 1 4 x 4 – 2x 3 + 11 2 x 2 – 6x. ? F(0) = 0, F(1) = – 9 4 , F(2) = –2, F(3) = – 9 4 . Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2 3 4 sq.units. JEEMAIN.GURU E 3 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves y = 2 x , y = 2x – x 2 . Solution : Figure is self-explanatory y = 2 x , (x – 1) 2 = – (y – 1) y=2x–x 2 x=2 R(2,4) M(2,0) y=2 x Q (0,1) O The required area = 2 1 2 0 (y y ) ? ? dx where y 1 = 2 x and y 2 = 2x – x 2 = 2 x 2 0 (2 2x x )dx ? ? ? = 2 x 2 3 0 2 1 x x ln 2 3 ? ? ? ? ? ? ? ? = 4 8 4 ln 2 3 ? ? ? ? ? ? ? ? – 1 ln 2 = 3 4 – ln 2 3 sq.units. Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y 2 , x = 1 – 3y 2 . Solution : Solving the equations x = –2y 2 , x = 1 – 3y 2 , we find that ordinates of the points of intersection of the two curves as y 1 = – 1, y 2 = 1. The points are (–2, –1) and (–2, 1). 1 Y x=1–3y 2 P 1 1 X –1 –2 –1 O P 2 x=–2y 2 (–2, 1) (–2, –1) The required area 2 1 1 2 0 (x x ) ? ? dy = 2 1 2 2 0 [(1 3y ) (–2y )]dy ? ? ? =2 1 2 0 (1 y )dy ? ? = 2 1 3 0 y y 3 ? ? ? ? ? ? ? = 4 3 sq.units. Do yourself - 2 : (i) Find the area bounded by y x ? and y = x. (ii) Find the area bounded by the curves x = y 2 and x = 3 – 2y 2 . (iii) Find the area of the region bounded by the curves x = 1 2 , x = 2, y = logx and y = 2 x . 3 . CURVE TRACING : The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be extremely useful to quickly and correctly evaluate the area under the curves. ( a ) Symmetry : The symmetry of the curve is judged as follows : (i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x. (ii) If all the powers of x are even, the curve is symmetrical about the axis of y. (iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y. (iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is symmetrical about y = x. (v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is symmetry in opposite quadrants. (b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents. (c) Find the points where the curve crosses the x ?axis & also the y ?axis. (d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when x ? ? or ? ?. Illustration 7 : Find the area of a loop as well as the whole area of the curve a 2 y 2 = x 2 (a 2 – x 2 ). Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0) Area of a loop = 2 a 0 y dx ? = 2 a 2 0 x a x dx a ? ? ? Y O A (a,0) X X' A' (–a,0) = – a 2 2 0 1 a x (–2x)dx a ? ? = – a 2 2 3 / 2 0 1 2 2 (a x ) a 3 3 ? ? ? ? ? ? ? ? a 2 Total area = 2 × 2 3 a 2 = 4 3 a 2 sq.units. s. JEEMAIN.GURU 4 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Illustration 8 : Find the whole area induded between the curve x 2 y 2 = a 2 (y 2 – x 2 ) and its asymptotes. Solution : (i) The curve is symmetric about both the axes (even powers of x & y) (ii) Asymptotes are x = ± a a 0 A 4 ydx ? ? y x=–a x=a dx x a 2 2 0 ax 4 dx a x ? ? ? a 2 2 0 4a a x ? ? ? = 4a 2 Illustration 9 : Find the area bounded by the curve xy 2 = 4a 2 (2a–x) and its asymptote. Solution : (i) The curve is symmetrical about the x-axis as it contains even powers of y. (ii) It passes through (2a,0). (iii) Its asymptote is x = 0, i.e., y-axis. y x (2a,0) 2a 2a 0 0 2a x A 2 ydx 2 2a dx x ? ? ? ? ? Put x = 2a sin 2 ? A = 16a 2 / 2 2 0 cos d ? ? ? ? = 4 ?a 2 4 . IMPORTANT POINTS : ( a ) Since area remains invariant even if the co-ordinate axes are shifted, hence shifting of origin in many cases proves to be very convenient in computing the area. Illustration 10 : Find the area enclosed by |x – 1| + |y + 1| = 1. Solution : Shift the origin to (1, –1). (0,1) (0,–1) (1,0) (–1,0) 2 X = x – 1 Y = y + 1 |X| + |Y| = 1 Area = 2 2 ? = 2 sq. units Illustration 11 : Find the area of the region common to the circle x 2 + y 2 + 4x + 6y – 3 = 0 and the parabola x 2 + 4x = 6y + 14. Solution : Circle is x 2 + y 2 + 4x + 6y – 3 = 0 ? (x + 2) 2 + (y + 3) 2 = 16 x (0,4) (–2 3,2) ? (2 3,2) ? Shifting origin to (–2,–3). X 2 + Y 2 = 16 equation of parabola ? ? (x + 2) 2 = 6(y + 3) ? X 2 = 6Y Solving circle & parabola, we get X = ± 2 3 Hence they intersect at ? ? 2 3,2 ? & ? ? 2 3,2 2 4 2 0 2 A 2 6Y dY 16 Y dY ? ? ? ? ? ? ? ? ? ? ? 4 2 3 / 2 2 1 0 2 2 1 16 Y 2 6 Y Y 16 Y sin 3 2 2 4 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 3 16 3 3 ? ? ? ? ? ? ? ? ? ? ? sq. units JEEMAIN.GURU Page 5 E 1 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s 1 . AREA UNDER THE CURVES : ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and O x=a x=b dx y=ƒ (x) x y x = b is given by A = b a y dx ? , where y = ƒ (x) lies above the x-axis and b > a. Here vertical strip of thickness dx is considered at distance x. (b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider O x y a b the magnitude only, i.e. A = b a y dx ? (c) If curve crosses the x-axis at x = c, then A = c b a c y dx ydx ? ? ? O x y x=a x=b c (d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : O x y y=b dy y=a Area bounded by the curve, y-axis and the two abscissae at y = a & y = b is written as b a A xdy ? ? . Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Find the area bounded by y = sec 2 x, x = 6 ? , x = 3 ? & x-axis Solution : Area bounded = / 3 / 6 ydx ? ? ? = / 3 2 / 6 sec xdx ? ? ? = / 3 / 6 [tan x] ? ? = tan 3 ? – tan 6 ? = 3 – 1 3 = 2 3 sq.units. Illustration 2 : Find the area in the first quadrant bounded by y = 4x 2 , x = 0, y = 1 and y = 4. Solution : Required area = 4 1 x ? dy = 4 1 y 2 ? dy = 4 3 / 2 1 1 2 y 2 3 ? ? ? ? ? ? y=1 y=4 x=0 Y O X = 1 3 [4 3/2 – 1] = 1 3 [8 – 1] = 7 3 = 2 3 ? sq.units. Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = 4 ? and 3 x 4 ? ? Solution : Required area = / 2 3 / 4 / 4 / 2 sin 2xdx sin 2xdx ? ? ? ? ? ? ? = / 2 3 / 4 / 4 / 2 cos2x cos2x 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?/4 3 /4 ? ?/2 = 1 1 [ 1 0] (0 ( 1)) 2 2 ? ? ? ? ? ? = 1 sq. unit AREA UNDER THE CURVE JEEMAIN.GURU 2 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Do yourself - 1 : (i) Find the area bounded by y = x 2 + 2 above x-axis between x = 2 & x = 3. (ii) Using integration, find the area of the curve 2 y 1 x ? ? with co-ordinate axes bounded in first quadrant. (iii) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2 ?. (iv) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x = 1 2 ? and x=1. 2 . AREA ENCLOSED BETWEEN TWO CURVES : O y y =ƒ (x) 1 y =g(x) 2 x 1 x 2 x ( a ) Area bounded by two curves y = ƒ (x) & y = g(x) such that ƒ (x) > g(x) is 2 1 x 1 2 x A (y y )dy ? ? ? 2 1 x x A [ƒ(x) g(x)]dx ? ? ? (b ) In case horizontal strip is taken we have y x =g(y) 2 x =ƒ (y) 1 x y 2 y 1 2 1 y 1 2 y A (x x )dy ? ? ? 2 1 y y A [ƒ(y) g(y)]dy ? ? ? (c) If the curves y 1 = ƒ (x) and y 2 = g(x) intersect at x = c, then required area x=a x=b c y x y =g(x) 2 y =ƒ (x) 1 A = c b a c (g(x) ƒ(x))dx (ƒ(x) g(x))dx ? ? ? ? ? = b a ƒ(x) g(x) dx ? ? Note : Required area must have all the boundaries indicated in the problem. Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and x = 3 and x-axis Solution : To determine the sign, we follow the usual rule of change of sign. y = +ve for x > 3 y = –ve for 2 < x < 3 y = +ve for 1 < x < 2 y = –ve for x < 1. 3 0 | y| dx ? = 1 0 | y| dx ? + 2 1 | y| dx ? + 3 2 | y| dx ? 1 2 3 E D F C B O Y A (0, –6) X = 1 0 – y dx ? + 2 1 y dx ? + 3 2 –y dx ? Now let F(x) = ? (x – 1) (x – 2) (x – 3) dx = ? (x 3 – 6x 2 + 11x – 6) dx = 1 4 x 4 – 2x 3 + 11 2 x 2 – 6x. ? F(0) = 0, F(1) = – 9 4 , F(2) = –2, F(3) = – 9 4 . Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2 3 4 sq.units. JEEMAIN.GURU E 3 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves y = 2 x , y = 2x – x 2 . Solution : Figure is self-explanatory y = 2 x , (x – 1) 2 = – (y – 1) y=2x–x 2 x=2 R(2,4) M(2,0) y=2 x Q (0,1) O The required area = 2 1 2 0 (y y ) ? ? dx where y 1 = 2 x and y 2 = 2x – x 2 = 2 x 2 0 (2 2x x )dx ? ? ? = 2 x 2 3 0 2 1 x x ln 2 3 ? ? ? ? ? ? ? ? = 4 8 4 ln 2 3 ? ? ? ? ? ? ? ? – 1 ln 2 = 3 4 – ln 2 3 sq.units. Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y 2 , x = 1 – 3y 2 . Solution : Solving the equations x = –2y 2 , x = 1 – 3y 2 , we find that ordinates of the points of intersection of the two curves as y 1 = – 1, y 2 = 1. The points are (–2, –1) and (–2, 1). 1 Y x=1–3y 2 P 1 1 X –1 –2 –1 O P 2 x=–2y 2 (–2, 1) (–2, –1) The required area 2 1 1 2 0 (x x ) ? ? dy = 2 1 2 2 0 [(1 3y ) (–2y )]dy ? ? ? =2 1 2 0 (1 y )dy ? ? = 2 1 3 0 y y 3 ? ? ? ? ? ? ? = 4 3 sq.units. Do yourself - 2 : (i) Find the area bounded by y x ? and y = x. (ii) Find the area bounded by the curves x = y 2 and x = 3 – 2y 2 . (iii) Find the area of the region bounded by the curves x = 1 2 , x = 2, y = logx and y = 2 x . 3 . CURVE TRACING : The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be extremely useful to quickly and correctly evaluate the area under the curves. ( a ) Symmetry : The symmetry of the curve is judged as follows : (i) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x. (ii) If all the powers of x are even, the curve is symmetrical about the axis of y. (iii) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y. (iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is symmetrical about y = x. (v) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is symmetry in opposite quadrants. (b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents. (c) Find the points where the curve crosses the x ?axis & also the y ?axis. (d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when x ? ? or ? ?. Illustration 7 : Find the area of a loop as well as the whole area of the curve a 2 y 2 = x 2 (a 2 – x 2 ). Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0) Area of a loop = 2 a 0 y dx ? = 2 a 2 0 x a x dx a ? ? ? Y O A (a,0) X X' A' (–a,0) = – a 2 2 0 1 a x (–2x)dx a ? ? = – a 2 2 3 / 2 0 1 2 2 (a x ) a 3 3 ? ? ? ? ? ? ? ? a 2 Total area = 2 × 2 3 a 2 = 4 3 a 2 sq.units. s. JEEMAIN.GURU 4 E NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Illustration 8 : Find the whole area induded between the curve x 2 y 2 = a 2 (y 2 – x 2 ) and its asymptotes. Solution : (i) The curve is symmetric about both the axes (even powers of x & y) (ii) Asymptotes are x = ± a a 0 A 4 ydx ? ? y x=–a x=a dx x a 2 2 0 ax 4 dx a x ? ? ? a 2 2 0 4a a x ? ? ? = 4a 2 Illustration 9 : Find the area bounded by the curve xy 2 = 4a 2 (2a–x) and its asymptote. Solution : (i) The curve is symmetrical about the x-axis as it contains even powers of y. (ii) It passes through (2a,0). (iii) Its asymptote is x = 0, i.e., y-axis. y x (2a,0) 2a 2a 0 0 2a x A 2 ydx 2 2a dx x ? ? ? ? ? Put x = 2a sin 2 ? A = 16a 2 / 2 2 0 cos d ? ? ? ? = 4 ?a 2 4 . IMPORTANT POINTS : ( a ) Since area remains invariant even if the co-ordinate axes are shifted, hence shifting of origin in many cases proves to be very convenient in computing the area. Illustration 10 : Find the area enclosed by |x – 1| + |y + 1| = 1. Solution : Shift the origin to (1, –1). (0,1) (0,–1) (1,0) (–1,0) 2 X = x – 1 Y = y + 1 |X| + |Y| = 1 Area = 2 2 ? = 2 sq. units Illustration 11 : Find the area of the region common to the circle x 2 + y 2 + 4x + 6y – 3 = 0 and the parabola x 2 + 4x = 6y + 14. Solution : Circle is x 2 + y 2 + 4x + 6y – 3 = 0 ? (x + 2) 2 + (y + 3) 2 = 16 x (0,4) (–2 3,2) ? (2 3,2) ? Shifting origin to (–2,–3). X 2 + Y 2 = 16 equation of parabola ? ? (x + 2) 2 = 6(y + 3) ? X 2 = 6Y Solving circle & parabola, we get X = ± 2 3 Hence they intersect at ? ? 2 3,2 ? & ? ? 2 3,2 2 4 2 0 2 A 2 6Y dY 16 Y dY ? ? ? ? ? ? ? ? ? ? ? 4 2 3 / 2 2 1 0 2 2 1 16 Y 2 6 Y Y 16 Y sin 3 2 2 4 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 4 3 16 3 3 ? ? ? ? ? ? ? ? ? ? ? sq. units JEEMAIN.GURU E 5 NODE6\E\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#08\Eng\01. Area Under the curve.p65 J E E - M a t h e m a t i c s Do yourself : 3 (i) Find the area inside the circle x 2 –2x + y 2 – 4y + 1 = 0 and outside the ellipse x 2 –2x+4y 2 –16y+13= 0 (b ) If the equation of the curve is in parametric form, then t t dx A y .dt dt ? ? ? ? ? ? or t t dy x .dt dt ? ? ? ? ? , where ? ? ?? ? are values corresponding to values of x and ? & ? are values corresponding to values of y. Illustration 12 : Find the area bounded by x-axis and the curve given by x = asint, y = acost for 0 ? t ? ? ?. Solution : Area = 0 dx y .dt dt ? ? = 0 a cos t(a cos t)dt ? ? = 2 2 0 0 a a sin 2t (1 cos2t)dt t 2 2 2 ? ? ? ? ? ? = 2 2 a a 2 2 ? ? ? Alternatively, Area = 0 dy x .dt dt ? ? = 0 a sin t( a sin t)dt ? ? ? = 2 0 a (cos2t 1)dt 2 ? ? ? = 2 0 a sin 2t t 2 2 ? ? ? 2 a 2 ? ? Illustration 13 : Find the area of the figure bounded by one arc of the cycloid x = a(t – sint), y = a(1 – cost) and the x-axis. Solution : To find the points where an arc cuts x-axis a(1 – cost) = 0 ? t = 0, ? Area = 2 2 0 0 dx y dt a (1 cos t) dt dt ? ? ? ? ? ? 2 0 3 sin 2t a t 2 sin t 2 4 ? ? ? ? 2 2 3 3 a a 2 2 ? ? ? ? ? ? ? ? ? ? Do yourself - 4 : (i) Find the area of the loop of the curve : (a) x = 3t 2 , y = 3t – t 3 (b) x = t 2 – 1, y = t 3 – t (c) If y = ƒ (x) is a monotonic function in (a, b), then the area bounded by the ordinates at x = a, x = b, y = ƒ (x) and y = ƒ (c) [where c ? (a, b)] is minimum when a b c 2 ? ? . Proof : Let the function y = ƒ (x) be monotonically increasing. y=ƒ (c) x y=ƒ (x) x=a x=c x=b y O Required area A = c b a c [ƒ(c) ƒ(x)]dx [ƒ(x) ƒ(c)]dx ? ? ? ? ? For minimum area, dA 0 dc ? ? [ƒ '(c).c ƒ(c) ƒ '(c)a ƒ(c)] [ ƒ(c) ƒ '(c).b ƒ '(c).c ƒ(c)] 0 ? ? ? ? ? ? ? ? ? ? a b ƒ '(c) c 0 2 ? ? ? ? ? ? ? ? ? ? c = a b 2 ? ( ??ƒ '(c) ? 0) JEEMAIN.GURURead More

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