{"id":20558,"date":"2020-01-22T11:02:58","date_gmt":"2020-01-22T05:32:58","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20558"},"modified":"2021-07-02T15:28:00","modified_gmt":"2021-07-02T09:58:00","slug":"2nd-puc-maths-question-bank-chapter-8-miscellaneous-ex","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-8-miscellaneous-ex\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 8 Application of Integrals Miscellaneous Exercise"},"content":{"rendered":"

Students can Download Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Miscellaneous Exercise<\/h2>\n

Question 1.
\nFind area under gives curves and line:
\n(i) y = x2<\/sup>, x = I, x = 2 and x – axis
\n(ii) y = x2<\/sup>, x = 1, x = 5 and x – axis
\nAnswer:
\n(i) y = x2<\/sup> is upward parabola
\nArea = shaded one)
\n\"2nd<\/p>\n

\"2nd
\n\"KSEEB
\n(ii) y = x4<\/sup> is upward parabola
\nArea = shaded one)
\n\"2nd
\n\"2nd<\/p>\n

Question 2.
\nFind the area between the curves y = x and y = x2<\/sup>.
\nAnswer:
\ny = x is line with pts (0, 0) (-1, 1) (1, 1) (2, 2) etc
\ny = x2<\/sup> is parabola with vertex (0,0) up ward parabola
\nThey meet at pts (0, 0) and (1, 1)
\n\"2nd
\n\u21d2 y = x putting in y = x2
\n<\/sup>x = x2<\/sup> \u21d2 x = 0 \u21d2 y = 0
\nx =1 \u21d2 y = 1
\nArea is (shaded one) = (Area under line x = 0 to x = 1)
\n– (Area under parabola x = 0 to x = 1)
\n\"2nd<\/p>\n

Question 3.
\nFind the area of the region lying in the first quadrant and bounded by
\ny = 4x2<\/sup>, x = 0, y = 1 and y = 4.
\nAnswer:
\n(1) y = 4x2<\/sup> is an upward parabola with vertex (0, 0)
\n\"2nd<\/p>\n

x = 0 is line with pts (0,1) (0,0) etc….
\ny = 1 is line with pts (0, 1) (1, 1) etc….
\ny = 4 is line with pts (0,4) (1, 4) etc….
\nparabola meets at pts (0,0) with line x = 0
\n(1\/2,1) with line y = 1 (1,4) with line y = 4
\nHere we need to do along \u2018y’ axis as we don\u2019t know eq: of line with respect to \u2018x\u2019. Required area (shaded one) = Area under parabola from y = 1 to y = 4<\/p>\n

\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nSketch the graph of y = |x+3| and evaluate \\(\\int _{ -6 }^{ 0 } |x+3|dx\\)
\nAnswer:
\ny = |x+3|
\npoints are\u00a0\u00a0\u00a0 x = 0, y = 3 ; x = 1, y = 4 etc
\nx = -1, y = 2 ; x = -4, y = (-1) =\u00a0\u00a0 1
\nso point are (0, 3), (1,4), (2, 5), (-1,2), (-2, 1) (-3, 0), (-4,1) etc.
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 5.
\nFind the area bounded by the curve y = sin x between x = 0 and x = 2\u03c0
\nAnswer:
\nSin curve is
\n\"2nd
\ny = |sin x| is a sin curve
\nx = 0 is a line pts (0,1) (0, 2) etc
\nx = 2 \u03c0 is line through pts 2 \u03c0
\nHere in this we take n multiples in \u2018x\u2019 axis and 1 and -1 in \u2018y\u2019 axis
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nFind the area enclosed between the parabola y2<\/sup> = 4ax and the line y = mx.
\nAnswer:
\ny2<\/sup> = 4ax is right sided parabola vertex (O, O). y
\ny = mx is straight line through (O, O)
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 7.
\nFind the area enclosed by the parabola 4y = 3x2<\/sup> and the line 2y = 3x + 12.
\nAnswer:
\n4y = 3x2<\/sup> is parabola upward
\nline 2y = 3x +12
\n\"2nd<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n2<\/td>\n-2<\/td>\n<\/tr>\n
y<\/td>\n6<\/td>\n0<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

They meet at pts (4,12) and (-2,3)
\n4y = 3x2<\/sup>
\nputting 2y = 3x +12 in it
\n2(3x +12) = 3x2
\n\"2nd<\/sup>
\nRequired area (shaded) = Area under line 2y = 3x + 12 from x = -2 to x = 4
\n– Area under parabola 4y = 3x2<\/sup> from x = -2 to x = 4
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nFind the area of the smaller region bounded by the ellipse
\n\\(\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1 \\text { and the line } \\frac{x}{3}+\\frac{y}{2}=1\\)
\nAnswer:
\n\"2nd
\n\"2nd
\nRequired area (shaded area) = Area under ellipse from x = 0 to x = 3)
\n-(Area under line from x = 0 to x = 3)
\n\"2nd
\n\"2nd<\/p>\n

Question 9.
\nFind the area of the smaller region bounded by the ellipse
\n\\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\text { and the line } \\frac{x}{a}+\\frac{y}{b}=1\\)
\nAnswer:
\n\"2nd<\/p>\n

\"2nd
\nRequired area (shaded area) = Area under ellipse from x = 0 to x = 3)
\n-(Area under line from x = 0 to x = 3)
\n\"2nd
\n\"2nd
\nOR
\nRequired area (Shaded) = Area of ellipse in 1st quadrant – Area of \u0394 ABC
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nFind the area of the region enclosed by the parabola x2<\/sup> = y, the line y = x + 2 and the
\nx – axis.
\nAns:
\nx2<\/sup> = y is an upward parabola
\nline y = x +2
\n\"2nd
\nThey meet at pts (2, 4) and (-1, 1)
\nBy solving the equation
\nx2<\/sup> = y, y = x + 2
\n\u21d2 x2<\/sup>– x – 2 = 0
\n\u21d2 x = 2 or -1
\ny = 4 or 1
\n\\(Required\\quad area=\\int _{ -1 }^{ 2 } Area\\quad of\\quad line-\\int _{ -1 }^{ 2 } Area\\quad of\\quad parabola\\)
\n\"2nd
\n\"2nd<\/p>\n

Question 11.
\nUsing the method of integration find the area bounded by the curve |x| + |y| = 1.
\n[Hint: The required is bounded by lines x + y = 1, – x + y = 1 and -x -y = 1].
\nAnswer:
\n|x|+ |y| = 1
\n\"2nd
\nso points are (0, 1) (0, -1) (1, 0) (-1, 0)
\nbut |x| + |y| =\u00a0 1 can be written as: follows: \u21d2 -x + y = 1 (2nd<\/sup> quadrant); x + y =1
\n(1st<\/sup> quadrant); x – y = 1 (4th<\/sup> qudrant); -x -y = 1 (3rd<\/sup> quadrant) .
\n|x|+ |y|=1 in different quadrants
\n\"2nd
\nNot by Integration
\nso area required (shaded)
\n= 4 x Area of any one part \u0394
\n= 4 x Area \u0394 ABC
\n\\(=4 \\times \\frac{1}{2} \\times(\\mathrm{AB}) \\times|\\mathrm{BC}|\\)
\n\\(=4 \\times \\frac{1}{2} \\times 1 \\times 1=2 \\mathrm{sq.units.}\\)
\n\"2nd
\nBy integration
\nArea required
\n= 4 x Area of \u0394ABC
\n= 4 x Area under line AC (x + y = 1)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nFind the area bounded by curves {(x, y): y \u2265 x2<\/sup> and y = |x| }.
\nAnswer:
\ny \u2265 x2<\/sup> and y \u2264 |x|
\n\"2nd
\nconsider then as curves y = x2<\/sup> (upward parabola vertex (0, 0)
\ny= |x| (line)
\nAs y \u2265 x2<\/sup>, so area above parabola
\ny \u2264 |x|, so area below modulus
\n\u2234 required area is shaded one
\n= 2x [Area under line in 1st quadrant)
\n– (Area under parabola in 1 st quadrant)]
\n\"2nd<\/p>\n

Question 13.
\nUsing the method of integration find the area of the triangle ABC, coordinates of whose
\nvertical are A(2,0), B(4 ,5) and C (6 , 3)
\nAnswer:
\nBy 2 points form, equation of the line is
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

Question 14.
\nUsing the method of integration find the area of the region bounded by lines:
\n2x + y = 4,3x – 2y = 6 and x – 3y + 5 = 0
\nAnswer:
\n\"2nd
\n\"2nd
\nRequired area (shaded) = Area under line AB – Area under AC -Area under BC
\n\\(=\\int_{1}^{4} \\text { line } \\mathrm{AB}-\\int_{1}^{2} \\text { line } \\mathrm{AC}-\\int_{2}^{4} \\text { line } \\mathrm{BC}\\)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nFind the area of the region {(x, y) : y2<\/sup>< 4x, 4×2 + 4y2<\/sup>< 9}
\nAnswer:
\nCurve y2<\/sup> = 4x is a parabola right sided
\nCurve 4x2<\/sup> + 4y2<\/sup> = 9 is a circle with radius \\(\\frac{3}{2}\\) and centre (0, 0)
\n\"2nd
\nThey meet at points
\n\\(\\left(\\frac{1}{2}, \\sqrt{2}\\right)\\)
\nThey meet at points
\ny2<\/sup> = 4x in 4x2<\/sup> + 4y2<\/sup> = 9
\n\u21d2 4x2<\/sup> + 4 (4x) = 9
\n\u21d2 4x2<\/sup>+16x – 9 = 0
\n\\(\\Rightarrow x=\\frac{1}{2}, \\frac{-9}{2} \\text { but } x \\ 0, \\text { so } \\frac{-9}{2} \\text { not possible }\\)
\nSo, required area (shaded) = 2 x half of the area we get above \u2018x\u2019 axis
\n= 2 x [Area under curve y2<\/sup>= 4x + Area under curve 4x2<\/sup> + 4y2<\/sup> = 9]
\n\"2nd
\n\"2nd<\/p>\n

Choose the correct answer in the following Exercise from 16 to 20.<\/p>\n

Question 16.
\nArea bounded by the curve y = x3<\/sup>, the x-axis and the ordinates x = -2 and x = 1 is
\n(A) -9
\n(B)\\( \\frac{-15}{4}\\)
\n(C)\\(\\frac{15}{4}\\)
\n(D)\\(\\frac{17}{4}\\)
\nAnswer:
\ny = x3
\n\"2nd<\/sup>
\nRequired area (shaded)
\n\"2nd
\n= Area under curve y = x3 <\/sup>with respect to ‘x’ axis from x = -2 to x = 1
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nThe area bounded by the curve y = x |x| , x – axis and the ordinates x = -1 and x = 1 is given by
\n(A) 0
\n(B)\\( \\frac{1}{3}\\)
\n(C)\\( \\frac{2}{3}\\)
\n(D)\\( \\frac{4}{3}\\)
\n[Hint: y = x2<\/sup> if x > 0 and y = – x2<\/sup> if x < 0].
\nAnswer:
\n\"2nd
\nNOTE: As explained in summary of this chapter
\nSo correct answer is (C) \\( \\frac{2}{3}\\)<\/p>\n

Question 18.
\nThe area of the circle x2<\/sup> + y2<\/sup> = 16 exterior to the parabola y2<\/sup> = 6x is
\n(A) \\(\\frac{4}{3}(4 \\pi-\\sqrt{3})\\)
\n(B) \\(\\frac{4}{3}(4 \\pi+\\sqrt{3})\\)
\n(C) \\(\\frac{4}{3}(8 \\pi-\\sqrt{3})\\)
\n(D) \\(\\frac{4}{3}(8 \\pi+\\sqrt{3})\\)
\nAnswer:
\n\"2nd
\nx2<\/sup> + y2<\/sup> = 16 is circle with radius 4 centre (0, 0)
\ny2<\/sup> = 6x is is right side parabola, vertex (0, 0)
\nThey both meet at pts:
\nx2<\/sup> + y2<\/sup> = 16, But given y2<\/sup> = 6x
\nso x2<\/sup> + 6x – 16 = 0 \u21d2 x = -8, but x <0 is eliminated
\nx = 2, \\(y=2 \\sqrt{3}\\)
\nRequired area (shaded) = Area of circle – shaded area (as ext. to parabola in circle)
\n= \u03c0r2<\/sup> – shaded area
\nShaded area = 2 x [(Area of parabola from x = 0 to x = 2) +
\n(Area of circle from x = 2 to x = 4)]
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB
\nQuestion 19.
\nThe area bounded by the y – axis, y = cos x and y = sin x when \\(0 \\leq x \\leq \\frac{\\pi}{2} \\text { is }\\)
\n(A)\\(2(\\sqrt{2-1})\\)
\n(B)\\(\\sqrt{2}-\\)
\n(C)\\(\\sqrt{2}+1\\)
\n(D)\\(\\sqrt{2}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

2nd PUC\u00a0 Maths Application of Integrals Miscellaneous Exercise Additional Questions and Answers<\/p>\n

Question 1.
\nFind the area of ABC, where vertices are A (4, 1) B (6, 6), C (8, 4) (CBSE 2010)
\nAnswer:
\n\"2nd<\/p>\n

Question 2.
\nFind the area of the circle 4x2<\/sup> + 4y2<\/sup> = 9, which is interior to the parabola x2<\/sup> = 4y.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 3.
\nSketch the graph of y=|x+3| and evaluate x-axis and between x = – 6 and x = 0.
\n(CBSE – 2011)
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nFind the area of the region {(x, y) : x2<\/sup> + y2<\/sup> = 4, x + y \u2265 2}\u00a0 (CBSE 2012)
\nAnswer:
\n\"2nd
\nCircle x2<\/sup> + y2<\/sup> = 4 his centre (0, 0) and radius 2
\nline x + y = 2 pts are (1,1) (2, 0) (0, 2) etc………
\ncircle and line intersect at pts.
\nIt can be found by solving the equation
\nx2<\/sup> + y2<\/sup> = 4 put (y = 2 – x)
\n\u21d2 x2<\/sup> + (2 -x)2<\/sup> = 4 \u21d2 2x2<\/sup> – 4x + 0 = 0
\n\u21d2 2x2<\/sup> – 4x = 0 \u21d2 x2<\/sup> – 2x = 0
\n\u21d2 (x -2) x = 0\u00a0 \u21d2 x = 0 or x = 2
\nso y = 2 – x , x = 0 , y = 2\u00a0\u00a0\u00a0 (0, 2)
\nx = 2 , y = 0 , (2,0)
\nso they meet at (0, 2) and (2, 0)
\n\"2nd
\nRequired area is shaded area:\u00ad-
\n(Area under circle x2<\/sup> + y2<\/sup> = 4 from x = 0 to x = 2)
\n– (Area under line x+y = 2 from x = 0 to x = 2)
\n\"2nd
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 8 Application of Integrals Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Miscellaneous Exercise Question 1. Find …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20558"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20558"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20558\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20558"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20558"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20558"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}