Chapter 7 Area Under Curves PDF

60 374 Area Under Curves 7.1 Introduction. E3 We know the methods of evaluating definite integrals. These integrals are used in evaluating certain types of bounded regions. For evaluation of bounded regions defined by given functions, we shall also require to draw rough sketch of the given function....

60 374 Area Under Curves 7.1 Introduction. E3 We know the methods of evaluating definite integrals. These integrals are used in evaluating certain types of bounded regions. For evaluation of bounded regions defined by given functions, we shall also require to draw rough sketch of the given function. The process of drawing rough sketch of a given function is called curve sketching. ID 7.2 Procedure of Curve Sketching. (1) Symmetry: (i) Symmetry about x-axis: If all powers of y in equation of the given curve are even, then it U is symmetric about x-axis i.e., the shape of the curve above x-axis is exactly identical to its shape below x-axis. D YG For example, y 2  4 ax is symmetric about x-axis. (ii) Symmetry about y-axis: If all power of x in the equation of the given curve are even, then it is symmetric about y-axis For example, x 2  4 ay is symmetric about y-axis. (iii) Symmetry in opposite quadrants or symmetry about origin: If by putting –x for x and –y for y, the equation of a curve remains same, then it is symmetric in opposite quadrants. (iv) U For example, x 2  y 2  a 2 and xy  a 2 are symmetric in opposite quadrants. Symmetry about the line y  x : If the equation of a given curve remains unaltered ST by interchanging x and y then it is symmetric about the line y  x which passes through the origin and makes an angle of 45 0 with the positive direction of x-axis. (2) Origin: If the equation of curve contains no constant terms then it passes through the origin. Find whether the curve passes through the origin or not. For examples, x 2  y 2  4 ax  0 passes through origin. (3) Points of intersection with the axes: If we get real values of x on putting y  0 in the equation of the curve, then real values of x and y  0 give those points where the curve cuts the x-axis. Similarly by putting x  0, we can get the points of intersection of the curve and y-axis. Area Under Curves 375 For example, the curve x2 y2   1 intersect the axes at points ( a, 0 ) and (0,  b). a2 b 2 dy  0 , at these points the tangent to the curve dx dx  0. At these points the tangent to the curve x-axis. Find the points at which dy is parallel to 60 (4) Special points: Find the points at which is parallel to y-axis. which determine the region of the curve. For example for the curve xy 2  a 2 (a  x )  y  a E3 (5) Region: Write the given equation as y  f (x ) , and find minimum and maximum values of x ax x ID Now y is real, if 0  x  a , So its region lies between the lines x = 0 and x = a (6) Regions where the curve does not exist: Determine the regions in which the curve does U not exists. For this, find the value of y in terms of x from the equation of the curve and find the value of x for which y is imaginary. Similarly find the value of x in terms of y and determine the values of y for which x is imaginary. The curve does not exist for these values of x and y. D YG For example, the values of y obtained from y 2  4 ax are imaginary for negative value of x, so the curve does not exist on the left side of y-axis. Similarly the curve a 2 y 2  x 2 (a  x ) does not exist for x  a as the values of y are imaginary for x  a. 7.3 Sketching of Some Common Curves. (1) Straight line: The general equation of a straight line is ax  by  c  0. To draw a straight U line, find the points where it meets with the coordinate axes by putting y = 0 and x = 0 respectively in its equation. By joining these two points, we get the sketch of the line. ST (2) Region represented by a linear inequality: To find the region represented by linear inequalities ax  by  c and ax  by  c, we proceed as follows. (i) Convert the inequality into equality to obtain a linear equation in x, y. (ii) Draw the straight line represented by it. (iii) The straight line obtained in (ii) divides the xy-plane in two parts. To determine the region represented by the inequality choose some convenient points, e.g. origin or some point on the coordinate axes. If the coordinates of a point satisfy the inequality, then region containing the point is the required region, otherwise the region not 2 containing the point is the required region. 1 (3) Circle: The equation of a circle having centre at (0,0) and radius r is given by x 2  y 2  r 2. The equation of a circle having centre at (h, k) –2 –1 1 –1 – 2 2 376 Area Under Curves and radius r is given (x  h)2  (y  k )2  r 2. by The general equation of a circle is x 2  y 2  2 gx  2 fy  c  0. This represents the circle whose centre is at (-g,-f) and radius equal to g 2  f 2  c. 60 The figure of the circle x 2  y 2  (2)2 is given. Here centre is (0,0) and radius is 2. E3 (i) y 2  4 ax : For this parabola Directrix (4) Parabola: There are four standard forms of parabola with vertex at origin and the axis along either of coordinate axis. Y (b) Focus: ( a, 0 ) (a) Vertex: (0,0) X ' A Z (c) Directrix: x  a  0 (d) Latus rectum: 4 a (e) Axis y  0 (f) Symmetry : It is symmetric about x-axis. ID Y´ (ii) x  4 ay : For this parabola 2 Y (0, a) A (b) Focus: (0,  a) U (a) Vertex: (0,0) (c) Directrix: y  a  0 D YG (d) Latus rectum: 4 a (e) Axis x = 0 X' Z (f) Symmetry: It is symmetric about y-axis at the origin and major and minor axes along the coordinate axes is 2 x y  2  1 Here a  b. The figure of the ellipse is given. 2 a b Directri x Y' X Y (5) Ellipse: The standard equation of the ellipse having its centre 2 X S (a,0) o) B (0,b) (- a, 0) X A (a, 0) X A B (0, – b) U Y´ Sketch the region bounded by 3 x  4 y  12 Solution: Converting the inequality into equation we get 3 x  4 y  12. This ST Example: 1 line meets x-axis at (4,0) and y-axis at (0,3). Joining these two points we obtain the straight line represented by 3 x  4 y  12. This Y X´ O straight line divides the plane in two parts. One part contains the origin the other does not contain the origin. Clearly, (0,0) satisfies the inequality 3 x  4 y  12. So, the region represented by (0, 3) 3x + 4y = 12 (4, 0) X Y' 3x  4y  12 is the region containing the origin as shown in figure. Y Example: 2 Sketch the parabola y  8 x. Solution: y2  8 x 2 y 2  4(2)x Here vertex is (0,0) and focus is (2,0) X' ´ (0, 0) (2, 0) X Y' ´ Yaxis X(1,0)axis (0,0) (½, – ¼) Area Under Curves 377 Example: 3 Sketch the graph for y  x 2  x. Solution: We note the following points about the curve. (i) The curve does not have any kind of symmetry. 60 (ii) The curve passes through the origin and the tangent at the origin is obtained by equating the lowest degree term to zero. The lowest degree term is x  y. Equation it to zero, we get x  y  0 as the equation of tangent at the origin. E3 (iii) Putting y=0 in the equation of curve, we get x 2  x  0  x  0, 1. So, the curve crosses x-axis at (0,0) and (1,0). Putting x=0 in the equation of the curve, we obtain y=0. So, the curve meets y-axis at (0,0) only. dy d 2y  2 x  1 and 2 dx dx 2 ID (iv) y  x 2  x  dy 1 0 x  , dx 2 At x  1 1 d 2y is point of local minima. ,  0 , So x  2 2 2 dx U Now, dy 1 1 1  0  2 x  1  0  x  , So the curve increases for all x  and decreases for all x  dx 2 2 2 D YG (v) 7.4 Area of Bounded Regions. (1) The area bounded by a cartesian curve y = f(x), x-axis and ordinates x = a and x = b is given by  b Area  y dx  a b  f (x )dx Y a y dx O x=a x=b X ST U y = f(x) (2) If the curve y = f(x) lies below x-axis, then the area bounded by the curve y  f (x ), the x- axis and the ordinates x =a and x =b is negative. So, area is given by b  y dx a (3) The area bounded by a cartesian curve x =f(y), y-axis and abscissa y = c and y = d is given by Y Area =  d c x dy  d  f (y)dy c y=d x dy x = f(y) y=c O X 378 Area Under Curves (4) If the equation of a curve is in parametric form, say x = f(t), y = g(t) then the area  y dx  a  t2 t1 g(t) f ' (t) dt where t1 and t 2 are the values of t respectively corresponding to the values 60  b of a and b of x. 7.5 Sign convention for finding the Areas using Integration. Case I: In the expression E3 While applying the discussed sign convention, we will discuss the three cases. b  f (x ) dx if b  a and f (x )  0 for all a  x  b, then this integration a Case II: If in the expression b  f (x )dx a ID will give the area enclosed between the curve f(x), x-axis and the line x = a and x = b which is positive. No need of any modification. if b > a and f (x )  0 for all a  x  b, then this integration Case III. If in the expression U will calculate to be negative. But the numerical or the absolute value is to be taken to mean the area enclosed between the curve y = f(x), x-axis and the lines x = a and x = b. b  f (x )dx a where b  a but f(x) changes Y D YG its sign a numbers of times in the interval a  x  b, then we must divide the region [a, b] in such a way that we clearly get the points lying between [a, b] where f(x) changes its sign. For the region a c d e f b X where f(x)>0 we just integrate to get the area in that region and then add the absolute value of the integration calculated in the region where f(x)