Tiêu đề AREAS BOUNDED BY CURVES (Q).pdf ✅

\\ Chapter – 1 Definite integral is used to evaluate areas bounded by curves. To do problems under this heading, one must be able to draw a rough figure of the curve when the equation is given. Some rules about drawing curves are given below. Familiar curves like lines, circles and conics are not discussed here. Guidelines (i) Check whether the curve is symmetrical about the x-axis or not. The curve is symmetrical about the x-axis, if its equation is unchanged when y is replaced by y. (ii) The curve is symmetrical about the y-axis if its equation is unchanged when x is replaced by x. (iii) Put y = 0 in the equation of the curve. This will give the points where it cuts the x-axis (iv) Put x = 0 in the equation of the curve. This will give the points where it cuts the y-axis. (v) The curve is symmetrical about the line y = x if its equation does not change when x and y are interchanged. (vi) Find the turning points of the graph by equating 0 dx dy  (vii) Find the intervals of curve in which it increases and decreases if required. (viii) Use periodicity wherever possible. (ix) Check behaviour at x    and y   . Illustration 1 Question: Trace the curve y 2 (2a  x) = x 3 , a > 0. Solution: Note that the curve passes through the origin and is symmetrical about the x-axis. a x x y   2 3 2 Y X O 2a L.H.S. is positive. If x is negative or if x is greater than 2a, R.H.S. becomes negative. Hence the curve lies only in the interval 0 to 2a. When x  2a, y  . Therefore the line x = 2a is an asymptote for the curve. A rough Figure is shown. APPLICATION OF INTEGRATION TO AREAS 1 THEORY CONTENT OF AREAS BOUNDED BY CURVES
Illustration 2 Question: Trace the curve   x x x y    1 1 2 2 Solution: The curve passes through the origin and is symmetrical about the x-axis. It intersects the x-axis at x = 1 and x = 0. If x < 1 or x > 1 the curve is non-existent. As x  1, y  ± a rough diagram is shown below. Y X 1 O +1 The curve has a loop between – 1 and 0. Four cases are discussed below: Case I : PQ is an arc of a curve whose equation is y = f(x). We have an area bounded by PQ on one side; by the x-axis on another and the two parallel lines x = a and x = b (shown by PL and QM), a < b. P Y Q O L M X x = a x = b The area PLMQ y dx f x dx b a x b x a ( )       Case II: PQ is an arc of a curve whose equation is y = f(x) or x = f(y). In this case y-axis is one boundary and the other two are the lines y = c and y = d. P Y M Q X L O y = c y = d The area LPQM x dy f y dy d c y d y c ( )       In this case the integration is with respect to y. ESTIMATION OF AREAS 2
Case III: The figure encloses an area between two curves one of which is represented by PQ with equation y = f(x) and the other by AB with the equation y = g(x). Y O Q B A P y  f(x) 2 y         y1 y  g(x) X Fig. 3 x = a x = b y1 Area      b a PABQ y1 y 2 dx where ( ) y1  f x and ( ) y 2  g x      b a f(x) g(x) dx Case IV: The figure represents the region bounded by a closed curve ACQBP. The area of the region bounded by a closed curve ACQBP is   1 2 1 2 y y dx , y y b a    The values of y1 and y2 are obtained by solving the equation of the curve as a quadratic in y whose larger root y1 and smaller root y2 are functions of x. a and b are the coordinates of the points of contact of tangents drawn parallel to the y-axis. Illustration 3 Question: Find the area of the ellipse 1 0 2 2 2 2    b y a x Solution: The ellipse is symmetrical about both axes and hence the area enclosed = 4 (area of the quadrant) Y O Q A B P 2 y    X Fig. 4 C x  a x  b       1 2 y y O B A (a, 0) P (x, y) B’ A’
  a y dx 0 4 dx a x b a    0 2 2 4 1 =   a a a x a x a x a b a x dx a b 0 1 2 2 2 0 2 2 sin 2 2 4 4               =          4 4 2 a a b = ab sq. units Illustration 4 Question: Find the area included between the parabolas y 2 = 4ax and x 2 = 4ay. Solution: The two parabolas intersect at O(0,0) and A (4a, 4a). The area included between the two curves = area OQAP.      x a x y y dx 4 0 1 2 ( )            a dx a x a x 4 0 2 4 2 a a x a x 4 0 3 3 / 2 3 12 2 2           a a a a 12 64 8. 3 4 3 3 / 2   3 16 2 a  sq.units. Note: Sometimes it is better to use the formula  d c x dy instead of  b a y dx in the computation of area to simplify calculations, as the following illustration shows. Illustration 5 Question: Find the area of the segment cut off from the parabola y 2 = 2x by the line y = 4x –1. Solution: The line y = 4x – 1 intersects the parabola y 2 = 2x at A and B Y O X Q P A x 4ay 2  y 4ax 2 