Tiêu đề 8. APPLICATIONS OF THE INTEGRALS.pdf ✅

Revision Notes z Area Under Simple Curves: (i) Let us find the area bounded by the curve y = f(x), X-axis and the ordinates x = a and x = b. Consider the area under the curve as composed by large number of thin vertical strips. Let there be an arbitrary strip of height y and width dx. Area of elementary strip dA = y dx, where y = f(x). Total area A of the region between X-axis ordinates x = a, x = b and the curve y = f(x) = sum of areas of elementary thin strips across the region PQML. A y dx f x a b a b = = ∫ ∫ ( ) dx (ii)The area A of the region bounded by the curve x = g(y), Y-axis and the lines y = c and y = d is given by A x dy g y dy c d c d = = ∫ ∫ ( ) (iii)If the curve under consideration lies below X-axis, then f(x) < 0 from x = a to x = b, the area bounded by the curve y = f(x) and the ordinates x = a, x = b and X-axis is negative. But, if the numerical value of the area is to be taken into consideration, then Area = ∫ f x dx a b ( ) (iv) It may also happen that some portion of the curve is above X-axis and some portion is below X-axis as shown in the figure. Let A1 be the area below X-axis and A2 be the area above the X-axis. Therefore, area bounded by the curve y = f(x), X-axis and the ordinates x = a and x = b is given by A = |A1| + |A2| Example-1 Example 1: Find the area enclosed by the ellipse x a y b 2 2 2 2 + = 1. Sol. Let x a y b 2 2 2 2 + = 1 ...(i) The area of the region ABA’B’A bounded by the ellipse = 4 × area of the region AOBA =− ∫ 4 0 y dx a APPLICATIONS OF THE INTEGRALS LEARNING OBJECTIVES After going through this Chapter, the student would be able to learn:  Area under simple curves  Area bounded by a curve and a line 8 CHAPTER
MATHEMATICS, Class-XII Example-1 From eq. (i), we get y = ± − b a a x 2 2 Therefore, The area of the region ABA’B’A bounded by the ellipse = 4 0 b 2 2 a a x dx a ∫ − [As the region AOBA lies in the first quadrant, y is taken as positive] = 4 2 2 2 2 2 1 0 b a x a x a x a a − +         − sin = 4 2 0 2 1 0 2 b 1 a a a × +         −         − sin = 4 2 2 2 b a a × × π = pab Thus, the area of the region ABA’B’A bounded by the ellipse is pab square units. Example-2 Example 2: Find the area of the region bounded by the curve y = x2 and the line y = 4. Sol. Given curve y = x2 is a parabola which is symmetrical about y-axis only. So, by above figure, the required area of the region AOBA = 2 0 4 × ∫ x dy = 2 0 4 × ∫ y dy [Since, x = y ] = 2 3 2 3 2 0 4 ×           y = 2 2 3 4 0 3 2 × −         = 4 3 8 32 3 × = Thus, the area of the region bounded by the curve y = x2 and the line y = 4 is 32 3 square units. KEY-TERMS Arbitrary: An arbitrary constant is a symbol that can be assigned different values. Curve: A curve is a line or shape that has bends or turns and is smoothly drawn in a plane SUBJECTIVE TYPE QUESTIONS Very Short Answer Type Questions (1 mark each) 1. Find the area bounded by y = x2 , the X-axis and the lines x = −1 and x = 1. R [SQP 2020-21] Sol. Area of shaded region =− ∫ 2 2 0 1 x dx = 3 1 0 2 [ ] 3 x = 2 3 sq. unit 1 [Marking Scheme, SQP 2020-21] Commonly Made Error Students fail to identify the figure correctly. Learn to draw the graphs correctly. Answering Tip 2. Find the area bounded by the curve x = 2y + 3, Y-axis and the lines y = 1 and y = –1. R&U [OEB] 3. Find the area lying between the curve y2 = 4x and y = 2x. R [OEB] Sol. The area lying between the curve y2 = 4x and y = 2x is represented by the shaded area OBAO as The points of intersection of the curves are O(0, 0) and A(1, 2). We draw AC perpendicular to X-axis such that coordinate of C is (1, 0). This question is for practice and its solution is available at the end of the chapter
Applications of the Integrals 1 1 0 0 1 1 3 2 2 3 2 0 0 Area of Area of Area of 2 2 2 2 2 4 = 1 3 1 = 3 1 = sq. units 3 OBAO OCABO OCA A x dx x dx x x =∆ −∆ = −       = −               −     + ∫ ∫ 1 4. Find the area lying in the first quadrant and bounded by circle x2 + y2 = 4 and the lines x = 0 and x = 2. R [OEB] Short Answer Type Questions-I (2 marks each) 1. Sketch the region bounded by the lines 2x + y = 8, y = 2, y = 4 and the y-axis. Hence, obtain its area using integration. Ap [Delhi Set-1, 2023] Sol. 2x + y = 8 y = 2 y = 4 1 Required Area = 4 2 xdy ∫ = 8 2 2 4 − ∫ y dy = 4 2 2 4 4 y y     −     = [16 – 4 – 8 + 1] Required Area = 5 unit2 1 [Making Scheme Delhi Set-1, 2023-24] Commonly Made Error Students do not write the unit. Students should write the sq. units with the answer. Answering Tip 2. Find the area of the shaded region between the curve, y = 4 – x2 , 0 ≤ x ≤ 3 and the X-axis. U [CFPQ] Sol. Required area is given by the absolute value of the integral I, where I = 0 2 2 2 3 2 4 4 ∫ ∫ ( ) − + x dx x − − ( )dx 1 = 4 3 4 3 3 0 2 3 2 3 x x x x −         − −         = 8 8 3 0 12 9 8 8 3 −       − − ( ) − − −             = 16 3 3 16 3 − −       = 23 3 square units 1 3. Find the area of the region bounded by the curve y2 = 4x, y-axis and line y = 3. R&U [APQ Set-2, 2023-24] Sol. 1 Required area = 2 3 0 4 y ∫ dy = 3 3 0 12 y    = 27 12 – 0 = 9 4 square units 1 [Marking Scheme APQ, Set-2 2023-24] 4. Find the area of the region bounded by the parabola y2 = 8x and the line x = 2. R&U [SQP 2020-21] Short Answer Type Questions-II (3 marks each) 1. Find the area of the following region using integration: {(x, y) : y2 ≤ 2x and y ≥ x – 4} Ap [Delhi Set-2, 2023] These Questions are for practice and their solutions are given at the end of the chapter.