Chapter 8.1 Overview This chapter deals with a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses, and finding the area bounded by the above said curves. 8.1.1 The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) is given by the formula: bb xdx Area = ydx = f () aa 8.1.2 The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d is given by the formula: dd y dy Area = xdy () cc 8.1.3 The area of the region enclosed between two curves y = f (x), y = g (x) and the lines x = a, x = b is given by the formula. b xg () Area = f ()– x dx , where f (x) g (x) in [a, b] a 8.1.4 If f (x) g (x) in [a, c] and f (x) g (x) in [c, b], a < c < b, then c b Area = f () – ( ) xg x dx ( ) – g x f ()x dx a c 8.2 Solved Examples Short Answer (S.A.) Example 1 Find the area of the curve y = sin x between 0 and π. Solution We have π π π–cos xArea OAB = ∫ ydx =∫sin x dx = 0 oo = cos0 – cosπ = 2 sq units. Example 2 Find the area of the region bounded by the curve ay2 = x3, the y-axis and the lines y = a and y = 2a. Solution We have 2a 2a 1 2 Area BMNC = xdy 3 3a y dy a a 1 52a3a3 y3= 5 a 1 5 53a3 2a 3– a3= 5 3 15 5 aa 2–1 ()3= 33 5 232a2.23 –1 = sq units.5 Example 3 Find the area of the region bounded by the parabola y2 = 2x and the straight line x – y = 4. Solution The intersecting points of the given curves are obtained by solving the equations x – y = 4 and y2 = 2x for x and y. We have y2 = 8 + 2y i.e., (y – 4) (y + 2) = 0 which gives y = 4, –2 and x = 8, 2. Thus, the points of intersection are (8, 4), (2, –2). Hence 4 ⎛ 12 ⎞ Area = ∫⎜4 +y –2 y ⎟dy –2 ⎝⎠ 4 y214 y + – y3 = 26 –2 Example 4 Find the area of the region bounded by the parabolas y2 = 6x and x2 = 6y. Solution The intersecting points of the given parabolas are obtained by solving these equations for x and y, which are 0(0, 0) and (6, 6). Hence 3 6 Area OABC = 6 0 6–x 2 6 x dx = 26 32 – 318 x x 0 3 = 26 (6)2 3 – 3(6) 18 = 12 sq units. Example 5 Find the area enclosed by the curve x = 3 cost, y = 2 sint. Solution Eliminating t as follows: x x = 3 cost, y = 2 sint ⇒ = cost ,3 y sin t , we obtain2 x2 y2 = 1,94 which is the equation of an ellipse. From Fig. 8.5, we get 32 the required area = 4 9– x2 dx 03 38 x 29 –1 x = 9– xsin = 6 π sq units.32 230 Long Answer (L.A.) 3x2 Example 6 Find the area of the region included between the parabola y = and the4 line 3x – 2y + 12 = 0. 3x2 Solution Solving the equations of the given curves y = and 3x – 2y + 12 = 0,4 we get 3x2 – 6x – 24 = 0 ⇒ (x – 4) (x + 2) = 0 ⇒ x = 4, x = –2 which give y = 12, y = 3 From Fig.8.6, the required area = area of ABC4 4212 3x 3xdx – dx= 24–2 –2 42⎛ 3x ⎞4 3x3 6x += ⎜⎟ – 4 12⎝⎠–2 −2 Example 7 Find the area of the region bounded by the curves x = at2 and y = 2at between the ordinate coresponding to t = 1 and t = 2. Solution Given that x = at2 ...(i), yy = 2at ...(ii) ⇒ t = putting the2a value of t in (i), we get y2 = 4ax Putting t = 1 and t = 2 in (i), we get x = a, and x = 4a Required area = 2 area of ABCD = 4a 4a 2 ∫ydx = 2 × 2 ∫ax dxaa 4a3 x()2 56 28 a= = a sq units.3 3 a Example 8 Find the area of the region above the x-axis, included between the parabola y2 = ax and the circle x2 + y2 = 2ax. Solution Solving the given equations of curves, we have x2 + ax = 2ax or x = 0, x = a, which give y = 0. y = ± a From Fig. 8.8 area ODAB = ∫ 0 a (2–ax x2– ax )dx Let x = 2a sin2θ. Then dx = 4a sinθ cosθ dθ and x = 0, ⇒θ = 0, x = a ⇒θ = .4 a ax xdx Again, ∫2– 2 0 π ∫ 4 = (a )( cos )d2sin θcosθ4asin θ θθ 0 π π4 2⎛ sin 4 θ⎞4 a2 a2= 1–cos4 θθd =aθ– = .∫()⎜ ⎟ 40 ⎝ 4 ⎠0 Further more, a 32 ⎛⎞ a 2axdx ax 22⎜⎟ = a=∫330 ⎝⎠ 0 π 222 2 Thus the required area = a – a = a2 – sq units.43 43 Example 9 Find the area of a minor segment of the circle x2 + y2 = a2 cut off by the aline x = .2 aSolution Solving the equation x2 + y2 = a2 and x = , we obtain their points of2 aa 3a intersection which are a ,3and ,– .2 222 Hence, from Fig. 8.9, we get a Required Area = 2 Area of OAB = 2 a 2 ⎡ x 22 a2 –1 x ⎤a a – x + sin ⎥= 2 ⎢ 22 aa⎣⎦ 2 ⎡ a2 π a 3 a2 π⎤ .–.a –. ⎥= 2 ⎢ 224 2 26⎣⎦ a`= 2 (6π–3 3– 2π)12 a = 24π–3 3 ) sq units.12 ( Objective Type Questions Choose the correct answer from the given four options in each of the Examples 10 to 12. Example 10 The area enclosed by the circle x2 + y2 = 2 is equal to (A) 4π sq units (B) 22π sq units (C)4π2 sq units (D) 2π sq units 2 Solution Correct answer is (D); since Area = 4 2 – x2 0 2 x = 4 x 2– x2 sin –1 = 2π sq. units.2 20 x2 y2 Example 11 The area enclosed by the ellipse 2 +2 = 1 is equal toab (A) π2ab (B) πab (C) πa2b (D) πab2 Solution Correct answer is (B); since Area = 4 ∫ ab a0 ⎡ a2 x ⎤a4bx 22 –1 a – x + sin = ⎢ ⎥ = πab.a 2 2 a⎣ ⎦0 Example 12 The area of the region bounded by the curve y = x2 and the line y = 16 32 256 64 128 (A) ` (B) (C) (D)333 3 16 Solution Correct answer is (B); since Area = 2 ∫ 0 Fill in the blanks in each of the Examples 13 and 14. Example 13 The area of the region bounded by the curve x = y2, y-axis and the line y = 3 and y = 4 is _______. 37Solution sq. units3Example 14 The area of the region bounded by the curve y = x2 + x, x-axis and the line x = 2 and x = 5 is equal to ________. 297 Solution sq. units6 8.3 EXERCISES Short Answer (S.A.)1. Find the area of the region bounded by the curves y2 = 9x, y = 3x.2. Find the area of the region bounded by the parabola y2 = 2px, x2 = 2py.3. Find the area of the region bounded by the curve y = x3 and y = x + 6 and x = 0.4. Find the area of the region bounded by the curve y2 = 4x, x2 = 4y.5. Find the area of the region included between y2 = 9x and y = x6. Find the area of the region enclosed by the parabola x2 = y and the line y = x + 27. Find the area of region bounded by the line x = 2 and the parabola y2 = 8x8. Sketch the region {(x, 0) : y = 4– x2 } and x-axis. Find the area of the region using integration.9. Calcualte the area under the curve y = 2 x included between the lines x = 0 and x = 1. 10. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8. 11. Draw a rough sketch of the curve y = x –1 in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5. 12. Determine the area under the curve y = a2– x2 included between the lines x = 0 and x = a. 13. Find the area of the region bounded by y = x and y = x. 14. Find the area enclosed by the curve y = –x2 and the straight lilne x + y + 2 = 0. 15. Find the area bounded by the curve y = x , x = 2y + 3 in the first quadrant and x-axis. Long Answer (L.A.) 16. Find the area of the region bounded by the curve y2 = 2x and x2 + y2 = 4x. 17. Find the area bounded by the curve y = sinx between x = 0 and x = 2π. 18. Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration. 19. Draw a rough sketch of the region {(x, y) : y2 ≤ 6ax and x2 + y2 ≤ 16a2}. Also find the area of the region sketched using method of integration. 20. Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7. 21. Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5. 22. Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2π. 23. Draw a rough sketch of the given curve y = 1 + |x +1|, x = –3, x = 3, y = 0 and find the area of the region bounded by them, using integration. Objective Type Questions Choose the correct answer from the given four options in each of the Exercises 24 to 34. π24. The area of the region bounded by the y-axis, y = cosx and y = sinx, 0 ≤ x ≤ is2 (A) 2sq units (B) ( 21+ ) sq units (C) ( 2–1) sq units (D) (22 –1) sq units 25. The area of the region bounded by the curve x2 = 4y and the straight line x = 4y – 2 is 3 579(A) sq units (B) sq units (C) sq units (D) sq units8 888 26. The area of the region bounded by the curve y = 16−x2 and x-axis is (A) 8 sq units (B) 20πsq units (C) 16π sq units (D) 256π sq units 27. Area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32 is (A) 16π sq units (B) 4π sq units (C) 32π sq units (D) 24 sq units 28. Area of the region bounded by the curve y = cosx between x = 0 and x = π is (A) 2 sq units (B) 4 sq units (C) 3 sq units (D) 1 sq units 29. The area of the region bounded by parabola y2 = x and the straight line 2y = x is 4 21(A) sq units (B) 1 sq units (C) sq units (D) sq units3 33 30. The area of the region bounded by the curve y = sinx between the ordinates π x = 0, x = and the x-axis is2 (A) 2 sq units (B) 4 sq units (C) 3 sq units (D) 1 sq units x2 y2 31. The area of the region bounded by the ellipse + = 1 is25 16 (A)20π sq units (B) 20π2 sq units (C)16π2 sq units (D) 25 π sq units 32. The area of the region bounded by the circle x2 + y2 = 1 is (A) 2π sq units (B) π sq units (C) 3π sq units (D) 4π sq units 33. The area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3 is 7 9 1113(A) sq units (B) sq units (C) sq units (D) sq units2 222 34. The area of the region bounded by the curve x = 2y + 3 and the y lines. y = 1 and y = –1 is 3(A) 4 sq units (B) sq units (C) 6 sq units (D) 8 sq units2

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