{"id":25983,"date":"2022-09-14T19:53:21","date_gmt":"2022-09-14T14:23:21","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=25983"},"modified":"2022-09-14T19:53:25","modified_gmt":"2022-09-14T14:23:25","slug":"inter-2nd-year-maths-2b-definite-integrals-solutions-ex-7d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2b-definite-integrals-solutions-ex-7d\/","title":{"rendered":"Inter 2nd Year Maths 2B Definite Integrals Solutions Ex 7(d)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFine the area of the region enclosed by the given curves.
\ni) y = cos x, y = 1 – \\(\\frac{2x}{\\pi}\\)
\nSolution:
\nEquations of the given curves are
\ny = cos x ………….. (1)
\ny = 1 – \\(\\frac{2x}{\\pi}\\) ………….. (2)
\nEliminating y from eq’s (1) and (2)
\ncos x = 1 – \\(\\frac{2x}{\\pi}\\)
\nWhen x = \\(\\frac{\\pi}{2}\\), cos x = cos\\(\\frac{\\pi}{2}\\) = 0
\n1 – \\(\\frac{2}{\\pi}\\), x = \\(\\frac{2}{\\pi}\\) . \\(\\frac{\\pi}{2}\\) = 1 – 1 = 0
\nWhen x = 0, cos x = cos 0 = 1
\n1 – \\(\\frac{2x}{\\pi}\\) = 1 – 0 = 1
\n\u2234 Point of intersection are A = (\\(\\frac{\\pi}{2}\\), 0) B = [\u03c0 – 1]
\n\"Inter<\/p>\n

Question 2.
\ny =cos x, y = sin 2x, x = 0, x = \\(\\frac{\\pi}{2}\\).
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 3.
\ny = x\u00b3 + 3, y = 0, x = -1, x = 2.
\nSolution:
\nRequired area PABQ
\n\"Inter<\/p>\n

Question 4.
\ny = ex<\/sup>, y = x, x = 0, x = 1.
\nSolution:
\n\"Inter<\/p>\n

Question 5.
\ny = sin x, y = cos x; x = 0, x = \\(\\frac{\\pi}{2}\\).
\nSolution:
\n\"Inter
\nBetween 0 and \\(\\frac{\\pi}{4}\\).
\ncos x > sin x
\nBetween \\(\\frac{\\pi}{4}\\) and \\(\\frac{\\pi}{2}\\)
\ncos x < sin x
\nRequired area
\n\"Inter<\/p>\n

Question 6.
\nx = 4 – y\u00b2, x = 0.
\nSolution:
\nThe given parabola x = 4 – y\u00b2 meets, the x – axis at A(4, 0) and Y – axis at P(0, 2) and Q(0, -2).<\/p>\n

The parabola is symmetrical about X – axis
\n\"Inter
\nRequired area = 2 Area of OAP
\n\"Inter<\/p>\n

Question 7.
\nFind the area enclosed with in the curve |x| + |y| = 1
\nSolution:
\n\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nx = 2 – 5y – 3y\u00b2, x = 0.
\nSolution:
\nSolving the equation of given curves
\n2 – 5y – 3y\u00b2 = 0
\n3y\u00b2 + 5y – 2 = 0
\n(y + 2)(3y – 1) = 0 y = -2 or \\(\\frac{1}{3}\\)
\n\"Inter<\/p>\n

Question 2.
\nx\u00b2 = 4y, x = 2, y = 0.
\nSolution:
\n\"Inter<\/p>\n

Question 3.
\ny\u00b2 = 3x, x = 3.
\nSolution:
\n\"Inter
\nThe parabola is symmetrical about X – axis
\nRequired area = 2\\(\\int_0^3\\)\u221a3. \u221ax dx
\n\"Inter<\/p>\n

Question 4.
\ny = x\u00b2, y = 2x.
\nSolution:
\nGiven equation are y = x\u00b2 ………….. (1)
\ny = 2x …………. (2)
\n\"Inter
\nEliminating y, we get x\u00b2 = 2x
\nx\u00b2 – 2x = 0
\nx(x – 2) = 0
\nx = 0 or x = 2
\ny = 0, or y = 4<\/p>\n

Point of intersection are O(0, 0), A(2, 4)
\n\"Inter<\/p>\n

Question 5.
\ny = sin 2x, y = \u221a3 sin x, x = 0, x = \\(\\frac{\\pi}{6}\\).
\nSolution:
\nGiven equation are y = sin 2x ………… (1)
\ny = \u221a3 sin x …………. (2)
\nsin 2x = \u221a3 sin x
\n2 sin x. cos x = \u221a3 sin x
\nsin x = 0 or 2 cos x = \\(\\frac{\\sqrt{3}}{2}\\)
\nx = 0, cos x = \\(\\frac{\\sqrt{3}}{2}\\) \u21d2 x = \\(\\frac{\\pi}{6}\\)
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 6.
\ny = x\u00b2, y = x\u00b3.
\nSolution:
\nGiven equations are y = x\u00b2 ………….. (1)
\ny = x\u00b3 ………. (2)
\n\"Inter
\nFrom equation (1) and (2) x\u00b2 = x\u00b3
\nx\u00b3 – x\u00b2 = 0
\nx\u00b2(x – 1) = 0
\nx = 0 or 1
\n\"Inter<\/p>\n

Question 7.
\ny = 4x – x\u00b2, y = 5 – 2x
\nSolution:
\n\"Inter
\ny = 4x – x\u00b2 ………… (i)
\ny = 5 – 2x …………..(ii)
\ny = -([x – 2]\u00b2) = 4
\ny – 4 = -(x – 2)\u00b2
\nSolving equations (i) and (ii) we get
\n4x – x\u00b2 = 5 – 2x
\nx\u00b2 – 6x + 5 = 0
\n(x – 5)(x – 1 ) = 0
\nx = 1, 5
\n\"Inter<\/p>\n

Question 8.
\nFind the area in Sq. units bounded by the x – axis, part of the curve y = 1 + \\(\\frac{8}{x^2}\\) and the ordinates x = 2 and x = 4.
\nSolution:
\nGiven equations y = 1 + \\(\\frac{8}{x^2}\\)
\n\"Inter<\/p>\n

Question 9.
\nFind the area of the region bounded by the parabolas y\u00b2 = 4x and x\u00b2 = 4y.
\nSolution:
\nEquations of the given curve are
\ny\u00b2 = 4x
\nx\u00b2 = 4y
\n(\\(\\frac{x^2}{4}\\))\u00b2 = 4x
\n\\(\\frac{x^4}{16}\\) = 4x
\nx4<\/sup> = 64x \u21d2 x4<\/sup> = 0 or x\u00b3 = 64, x = 4
\n\"Inter<\/p>\n

Question 10.
\nFind the area bounded by the curve y = l<\/em>nx the X – axis and the straight line x = e.
\nSolution:
\nEquation of the curve is y = l<\/em>nx
\nx = 1 \u21d2 y = 0
\nThe curve y = l<\/em>nx meets
\nX – axis at C(1, 0)
\n\"Inter
\nRequired area = \\(\\int_1^e\\)lnx dx
\n= (x.lnx)e<\/sup>1<\/sub> – \\(\\int_1^e\\)x.\\(\\frac{1}{x}\\) dx
\n= (e.l<\/em>n e – 1.l<\/em>n 1) – (x)e<\/sup>1<\/sub>
\n= e – (e – 1)
\n= e – e + 1 = 1 Sq.unit.<\/p>\n

III.<\/span><\/p>\n

Question 1.
\ny = x\u00b2 + 1, y = 2x – 2, x = -1, x = 2.
\nSolution:
\nEquation of the curve are
\ny = x\u00b2 + 1 …………. (1)
\ny = 2x – 2 ………….. (2)
\n\"Inter
\nArea between the given curves
\n\"Inter<\/p>\n

Question 2.
\ny\u00b2 = 4x, y\u00b2 = 4(4 – x).
\nSolution:
\nEquations of the curve are y\u00b2 = 4x ………… (1)
\ny\u00b2 = 4(4 – x) …………. (2)
\nEliminating y, we get
\n4x = 4(4 – x)
\n2x = 4 \u21d2 x = 2
\nSubstituting in equation (1), y\u00b2 = 8
\ny = \u00b1 2\u221a2
\n\"Inter
\nPoints of intersection are
\nA(2, 2\u221a2), B(2, -2\u221a2)
\nRequired area is symmetrical about X – axis
\nArea OACB
\n\"Inter<\/p>\n

Question 3.
\ny = 2 – x\u00b2, y = x\u00b2.
\nSolution:
\n\"Inter
\ny = 2 – x\u00b2 …………. (1)
\ny = x\u00b2 …………. (2)
\nx\u00b2 = -(y – 2)
\nFrom equation (2)
\n2 – x\u00b2 = x\u00b2
\n2 = 2x\u00b2 or x\u00b2 = 1
\nx = \u00b11
\nArea bounded by two curve be
\n\"Inter<\/p>\n

Question 4.
\nShow that the area enclosed between the curve y\u00b2 = 12(x + 3) and y\u00b2 = 20(5 – x) is 64\\(\\sqrt{\\frac{5}{3}}\\).
\nSolution:
\nEquation of the curve are
\ny\u00b2 = 12(x + 3) ……….. (1)
\ny\u00b2 = 20(5 – x) ……….. (2)
\nEliminating y
\n12(x + 3) = 20(5 – x)
\n3x + 9 = 25 – 5x
\n8x = 16
\nx = 2
\ny\u00b2 = 12(2 + 3) = 60
\ny = \u221a60 = \u00b12\u221a15
\n\"Inter
\nPoints of intersection are B’ (2, 2\u221a15)
\nB’ (+2, -2\u221a15)
\nThe required area is symmetrical about X – axis
\nArea ABCB’
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 5.
\nFind the area of the region {(x, y)\/x\u00b2 – x – 1 \u2264 y \u2264 -1}
\nSolution:
\n\"Inter<\/p>\n

Question 6.
\nThe circle x\u00b2 + y\u00b2 = 8 is divided into two parts by the parabola 2y = x\u00b2. Find the area of both the parts.
\nSolution:
\nEquations of the curves are
\nx\u00b2 + y\u00b2 = 8 ………… (1)
\n2y = x\u00b2 ………… (2)
\nEliminating y between equations (1) and (2)
\n\"Inter
\nLet x\u00b2 = t
\n4t + t\u00b2 = 32
\nt\u00b2 + 4t – 32 = 0
\n(t + 8)(t – 4) = 0
\nt = -8 (not possible) x\u00b2 = 4 \u21d2 x = \u00b12
\n\"Inter
\nAs curve is symmetric about Y – axis, total area be
\n\"Inter<\/p>\n

Question 7.
\nShow that the area of the region bounded \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1 (ellipse) is \u03c0 ab. also deduce the area of the cricle x\u00b2 + y\u00b2 = a\u00b2.
\nSolution:
\n\"Inter
\nThe ellipse is symmetrical about X and Y axis
\nArea of the ellipse = 4 Area of CAB
\n= 4.\\(\\frac{\\pi}{4}\\) ab
\nEquation of elliple = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1
\n\"Inter
\nSubstituting b = a, we get the circle
\nx\u00b2 + y\u00b2 = a\u00b2
\nArea of the circle = \u03c0a(a) = \u03c0a\u00b2 sq. units.<\/p>\n

Question 8.
\nFind the area of region enclosed by the curves y = sin \u03c0x, y = x\u00b2 – x, x = 2.
\nSolution:
\nRequired area
\n\"Inter
\n\"Inter<\/p>\n

Question 9.
\nLet AOB be the positive quadrant of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1 with OA = a, OB = b. Then show that the area bounded the chord AB and the arc AB of the elliple is \\(\\frac{(\\pi-2) a b}{4}\\).
\nSolution:
\nLet OA = a, OB = b
\nEquation of AB is \\(\\frac{x}{a}+\\frac{y}{b}\\) = 1
\n\\(\\frac{y}{b}\\) = 1 – \\(\\frac{x}{a}\\)
\ny = b(1 – \\(\\frac{x}{a}\\))
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 10.
\nProve that curves y\u00b2 = 4x and x\u00b2 = 4y divide the area of the square bounded by the lines x = 0, x = 4, y = 4 and x = 0 into three equal parts.
\nSolution:
\n\"Inter
\nThe given equations are y\u00b2 = 4x …………. (1)
\nx\u00b2 = 4y …………. (2)
\nThe points of intersection are O(0, 0) A(4, 4)
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Definite Integrals Solutions Exercise 7(d) I. Question 1. Fine the area of the region enclosed by the given curves. i) y = cos … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25983"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=25983"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25983\/revisions"}],"predecessor-version":[{"id":26046,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/25983\/revisions\/26046"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=25983"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=25983"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=25983"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}