Page 1 APPLICATION OF INTEGRALS 359 Fig 8.1 ? One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. â€“ BIRKHOFF ? 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter , we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f (x). Chapter 8 APPLICATION OF INTEGRALS A.L. Cauchy (1789-1857) 2019-20 Page 2 APPLICATION OF INTEGRALS 359 Fig 8.1 ? One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. â€“ BIRKHOFF ? 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter , we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f (x). Chapter 8 APPLICATION OF INTEGRALS A.L. Cauchy (1789-1857) 2019-20 360 MATHEMATICS Fig 8.2 This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express A = A () b b b a a a d ydx f x dx == ?? ? The area A of the region bounded by the curve x = g (y), y-axis and the lines y = c, y = d is given by A = () d d c c xdy g y dy = ?? Here, we consider horizontal strips as shown in the Fig 8.2 Remark If the position of the curve under consideration is below the x-axis, then since f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e., () b a f x dx ? . Fig 8.3 Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig 8.4. Here, A 1 < 0 and A 2 > 0. Therefore, the area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by A = |A 1 | + A 2 . 2019-20 Page 3 APPLICATION OF INTEGRALS 359 Fig 8.1 ? One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. â€“ BIRKHOFF ? 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter , we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f (x). Chapter 8 APPLICATION OF INTEGRALS A.L. Cauchy (1789-1857) 2019-20 360 MATHEMATICS Fig 8.2 This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express A = A () b b b a a a d ydx f x dx == ?? ? The area A of the region bounded by the curve x = g (y), y-axis and the lines y = c, y = d is given by A = () d d c c xdy g y dy = ?? Here, we consider horizontal strips as shown in the Fig 8.2 Remark If the position of the curve under consideration is below the x-axis, then since f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e., () b a f x dx ? . Fig 8.3 Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig 8.4. Here, A 1 < 0 and A 2 > 0. Therefore, the area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by A = |A 1 | + A 2 . 2019-20 APPLICATION OF INTEGRALS 361 Example 1 Find the area enclosed by the circle x 2 + y 2 = a 2 . Solution From Fig 8.5, the whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both x-axis and y-axis] = 0 4 a ydx ? (taking vertical strips) = 22 0 4 a a x dx - ? Since x 2 + y 2 = a 2 gives y = 22 ax ±- As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle = 2 2 2 â€“1 0 4 sin 2 2 a x ax ax a ? ? -+ ? ? ? ? = = 2 2 4 22 a a ?? p ?? =p ? ?? ? ? ?? ? Fig 8.5 Fig 8.4 2019-20 Page 4 APPLICATION OF INTEGRALS 359 Fig 8.1 ? One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. â€“ BIRKHOFF ? 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter , we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f (x). Chapter 8 APPLICATION OF INTEGRALS A.L. Cauchy (1789-1857) 2019-20 360 MATHEMATICS Fig 8.2 This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express A = A () b b b a a a d ydx f x dx == ?? ? The area A of the region bounded by the curve x = g (y), y-axis and the lines y = c, y = d is given by A = () d d c c xdy g y dy = ?? Here, we consider horizontal strips as shown in the Fig 8.2 Remark If the position of the curve under consideration is below the x-axis, then since f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e., () b a f x dx ? . Fig 8.3 Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig 8.4. Here, A 1 < 0 and A 2 > 0. Therefore, the area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by A = |A 1 | + A 2 . 2019-20 APPLICATION OF INTEGRALS 361 Example 1 Find the area enclosed by the circle x 2 + y 2 = a 2 . Solution From Fig 8.5, the whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both x-axis and y-axis] = 0 4 a ydx ? (taking vertical strips) = 22 0 4 a a x dx - ? Since x 2 + y 2 = a 2 gives y = 22 ax ±- As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle = 2 2 2 â€“1 0 4 sin 2 2 a x ax ax a ? ? -+ ? ? ? ? = = 2 2 4 22 a a ?? p ?? =p ? ?? ? ? ?? ? Fig 8.5 Fig 8.4 2019-20 362 MATHEMATICS Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle = 0 4 a xdy ? = 22 0 4 a a y dy - ? (Why?) = 2 22 1 0 4 sin 2 2 a a y y ay a - ? ? -+ ? ? ? ? = = 2 2 4 22 a a p =p Example 2 Find the area enclosed by the ellipse 22 22 1 xy ab += Solution From Fig 8.7, the area of the region ABA'B'A bounded by the ellipse = in 4 , 0, area of the region AOBA the first quadrant bounded by thecurve x axis and theordinates x x a ? ? ? ? - == ? ? (as the ellipse is symmetrical about both x-axis and y-axis) = 0 4 (taking verticalstrips) a ydx ? Now 22 22 xy ab + = 1 gives 22 b y ax a =± - , but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is = 22 0 4 a b a x dx a - ? = 2 2 2 â€“1 0 4 sin 22 a bx a x ax a a ? ? -+ ? ? ? ? (Why?) = 2 1 4 0 sin 1 0 22 ba a a - ? ? ? ? ×+ - ? ? ? ? ? ? ? ? ? ? = 2 4 22 ba ab a p =p Fig 8.6 Fig 8.7 2019-20 Page 5 APPLICATION OF INTEGRALS 359 Fig 8.1 ? One should study Mathematics because it is only through Mathematics that nature can be conceived in harmonious form. â€“ BIRKHOFF ? 8.1 Introduction In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in the applications of mathematics to many real life problems. The formulae of elementary geometry allow us to calculate areas of many simple figures. However, they are inadequate for calculating the areas enclosed by curves. For that we shall need some concepts of Integral Calculus. In the previous chapter, we have studied to find the area bounded by the curve y = f (x), the ordinates x = a, x = b and x-axis, while calculating definite integral as the limit of a sum. Here, in this chapter , we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses (standard forms only). We shall also deal with finding the area bounded by the above said curves. 8.2 Area under Simple Curves In the previous chapter, we have studied definite integral as the limit of a sum and how to evaluate definite integral using Fundamental Theorem of Calculus. Now, we consider the easy and intuitive way of finding the area bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b. From Fig 8.1, we can think of area under the curve as composed of large number of very thin vertical strips. Consider an arbitrary strip of height y and width dx, then dA (area of the elementary strip)= ydx, where, y = f (x). Chapter 8 APPLICATION OF INTEGRALS A.L. Cauchy (1789-1857) 2019-20 360 MATHEMATICS Fig 8.2 This area is called the elementary area which is located at an arbitrary position within the region which is specified by some value of x between a and b. We can think of the total area A of the region between x-axis, ordinates x = a, x = b and the curve y = f (x) as the result of adding up the elementary areas of thin strips across the region PQRSP. Symbolically, we express A = A () b b b a a a d ydx f x dx == ?? ? The area A of the region bounded by the curve x = g (y), y-axis and the lines y = c, y = d is given by A = () d d c c xdy g y dy = ?? Here, we consider horizontal strips as shown in the Fig 8.2 Remark If the position of the curve under consideration is below the x-axis, then since f (x) < 0 from x = a to x = b, as shown in Fig 8.3, the area bounded by the curve, x-axis and the ordinates x = a, x = b come out to be negative. But, it is only the numerical value of the area which is taken into consideration. Thus, if the area is negative, we take its absolute value, i.e., () b a f x dx ? . Fig 8.3 Generally, it may happen that some portion of the curve is above x-axis and some is below the x-axis as shown in the Fig 8.4. Here, A 1 < 0 and A 2 > 0. Therefore, the area A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given by A = |A 1 | + A 2 . 2019-20 APPLICATION OF INTEGRALS 361 Example 1 Find the area enclosed by the circle x 2 + y 2 = a 2 . Solution From Fig 8.5, the whole area enclosed by the given circle = 4 (area of the region AOBA bounded by the curve, x-axis and the ordinates x = 0 and x = a) [as the circle is symmetrical about both x-axis and y-axis] = 0 4 a ydx ? (taking vertical strips) = 22 0 4 a a x dx - ? Since x 2 + y 2 = a 2 gives y = 22 ax ±- As the region AOBA lies in the first quadrant, y is taken as positive. Integrating, we get the whole area enclosed by the given circle = 2 2 2 â€“1 0 4 sin 2 2 a x ax ax a ? ? -+ ? ? ? ? = = 2 2 4 22 a a ?? p ?? =p ? ?? ? ? ?? ? Fig 8.5 Fig 8.4 2019-20 362 MATHEMATICS Alternatively, considering horizontal strips as shown in Fig 8.6, the whole area of the region enclosed by circle = 0 4 a xdy ? = 22 0 4 a a y dy - ? (Why?) = 2 22 1 0 4 sin 2 2 a a y y ay a - ? ? -+ ? ? ? ? = = 2 2 4 22 a a p =p Example 2 Find the area enclosed by the ellipse 22 22 1 xy ab += Solution From Fig 8.7, the area of the region ABA'B'A bounded by the ellipse = in 4 , 0, area of the region AOBA the first quadrant bounded by thecurve x axis and theordinates x x a ? ? ? ? - == ? ? (as the ellipse is symmetrical about both x-axis and y-axis) = 0 4 (taking verticalstrips) a ydx ? Now 22 22 xy ab + = 1 gives 22 b y ax a =± - , but as the region AOBA lies in the first quadrant, y is taken as positive. So, the required area is = 22 0 4 a b a x dx a - ? = 2 2 2 â€“1 0 4 sin 22 a bx a x ax a a ? ? -+ ? ? ? ? (Why?) = 2 1 4 0 sin 1 0 22 ba a a - ? ? ? ? ×+ - ? ? ? ? ? ? ? ? ? ? = 2 4 22 ba ab a p =p Fig 8.6 Fig 8.7 2019-20 APPLICATION OF INTEGRALS 363 Alternatively, considering horizontal strips as shown in the Fig 8.8, the area of the ellipse is = 4 0 xdy b ? = 4 22 0 a b b y dy b - ? (Why?) = = 2 â€“1 4 0 sin 1 0 22 ab b b ? ? ? ? ×+ - ? ? ? ? ? ? ? ? ? ? = 2 4 22 ab ab b p =p 8.2.1 The area of the region bounded by a curve and a line In this subsection, we will find the area of the region bounded by a line and a circle, a line and a parabola, a line and an ellipse. Equations of above mentioned curves will be in their standard forms only as the cases in other forms go beyond the scope of this textbook. Example 3 Find the area of the region bounded by the curve y = x 2 and the line y = 4. Solution Since the given curve represented by the equation y = x 2 is a parabola symmetrical about y-axis only, therefore, from Fig 8.9, the required area of the region AOBA is given by 2 0 4 xdy ? = area of theregion BONBbounded bycurve, axis 2 and thelines 0and = 4 y yy - ? ? ? ? = ? ? = 4 0 2 ydy ? = 2 2 3 3 2 4 × ? ? ? ? ? ? ? ? y 0 4 32 8 33 = ×= (Why?) Here, we have taken horizontal strips as indicated in the Fig 8.9. Fig 8.8 Fig 8.9 2019-20Read More

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!