{"id":20458,"date":"2020-01-21T12:58:20","date_gmt":"2020-01-21T07:28:20","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20458"},"modified":"2021-07-02T15:28:10","modified_gmt":"2021-07-02T09:58:10","slug":"2nd-puc-maths-question-bank-chapter-8-ex-8-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-8-ex-8-1\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.1"},"content":{"rendered":"

Students can Download Maths Chapter 8 Application of Integrals Ex 8.1 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 Ex 8.1<\/h2>\n

2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.1<\/h3>\n

Question 1.
\nFind the area of the region bounded by the curve y2<\/sup> = x and the lines x = 1, x = 4 and the x-axis.
\nAnswer:
\nWe know that y2<\/sup> = x represents right hand parabola with vertex (0,0)
\nRequired area is shaded area:-
\n\"2nd
\nIf not mentioned above\/below\/any quadrant, so we should take above and below
\n\\(\\Rightarrow 2 \\times \\frac{14}{3}=\\frac{28}{3} \\text { sq. units. }\\)<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the area of the region bounded by y2<\/sup> = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
\nAnswer:
\nWe know y2<\/sup> = 9x is right handed parabola with vertex (0,0)
\n\"2nd<\/p>\n

Question 3.
\nFind the area of the region bounded by x2<\/sup> = 4y, y = 2, y = 4 and the y – axis in the first quadrant.
\nAnswer:
\nWe know x2<\/sup> = 4y is upward parabola with vertex (0,0).
\nRequired area is shaded area:-
\n\"2nd
\n\"2nd<\/p>\n

Question 4.
\nFind the area of the region bounded by the ellipse
\n\\(\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1\\)
\nAnswer:
\nWe know curve is ellipse vertex (0,0) major axis is \u2018x\u2019 and major ‘y’
\nArea of given area is = 4 x (Area of shaded)
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nFind the area of the region bounded by the ellipse
\n\\(\\frac{x^{2}}{4}+\\frac{y^{2}}{9}=1\\)
\nAnswer:
\nWe know curve is ellipse vertex (0,0) major axis is \u2018y\u2019 and major ‘x’
\nArea of given area is = 4 x (Area of shaded)
\n\\(=4 \\times \\int_{0}^{2}|y| d x\\)
\n\"2nd
\n\"2nd<\/p>\n

Question 6.
\nFind the area of the region in the first quadrant enclosed by x-axis, line \\(x=\\sqrt{3}\\) y and the circle x2<\/sup> + y2<\/sup> = 4.
\nAnswer:
\n\"2nd
\n\"2nd
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nFind the area of the smaller part of the circle x2<\/sup> + y2<\/sup> = a2<\/sup> cut off by the line \\(x=\\frac{a}{\\sqrt{2}} \\)
\nAnswer:
\nCircle x2<\/sup> + y2<\/sup> = a2<\/sup> is circle with centre (0,0) and radius
\n\"2nd
\n\"2nd<\/p>\n

Question 8.
\nThe area between x = y2<\/sup> and x = 4 is divided into two equal parts by the line x = a, find the value of a.
\nAnswer:
\n(1) x = y2<\/sup> is right hand parabola with vertex (0, 0) \u21d2 y2<\/sup> = x
\nAs area between y2<\/sup> = x, x = 4 is divided into 2 parts by x = a,
\n\"2nd
\nso,(area under y2<\/sup> = x, x = 0 to x = a) =(area under y2<\/sup> = x, x = a, x = 4)
\n\u21d2area (A) = area (B)
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nFind the area of the region bounded by the parabola y = x2<\/sup> and y = |x|
\nAnswer:
\n(1) As we know x2<\/sup> = y is a upward parabola with vertex (0,0)
\n(2) line \u21d2 y = |x|
\nSo if y = 1,x = 1 (1,1)
\ny = -1,x = 1 (-1, 1)
\ny = 2, x = 2 (2, 2)
\ny = 2, x = 2 (-2,2)
\n\"2nd
\nArea required is the shaded area
\n= 2 x (area of one half (lets take it of 1st<\/sup> quadrant))
\n= 2 x [(area under line y = |x|) – area under parabola y = x]
\n\"2nd<\/p>\n

Question 10.
\nFind the area bounded by the curve x2<\/sup> = 4y and the line x = 4y – 2.
\nAnswer:
\n(1) Curve x2<\/sup> = 4y is upward parabola vertex (0, 0)
\n(2) Line x = 4y – 2 (put 4y = x2<\/sup> in it)
\n\u21d2 x = x2 <\/sup>– 2 \u21d2 x2<\/sup>– x – 2 = 0
\n\u21d2 x = 2, -1
\n\"2nd
\nSo when x = 2 , y = 1 (2,1)
\n\\(x=-1, \\quad y=\\frac{1}{4} \\quad\\left(-1, \\frac{1}{4}\\right)\\)
\nSo, Area required is shaded area:-
\n= [Area under line x = 4y – 2 from (-1 to 2) (x axis) – [Area under x2<\/sup> = 4y from
\n-1 to 2 (x axis)]
\n\\(\\int_{-1}^{2}\\left(\\frac{x+2}{4}\\right) d x-\\int_{-1}^{2} \\cfrac{x^{2}}{4} d x\\)
\nAs solving along ‘x’ axis ,so we take values in terms of x to solve
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nFind the area of the region bounded by the curve y2<\/sup> = 4x and the line x = 3.
\nAnswer:
\n(1) Curve y2<\/sup> = 4x is right handed parabola with vertex (0,0) y
\n(2) Line x = 3, so points are (3,0) (3, 1) (3, 2)
\n\\(\\Rightarrow y^{2}=12 \\Rightarrow y=\\pm 2 \\sqrt{3}\\)
\n\"2nd
\n(3) But as we solve with respect to \u2018x\u2019 axis, so need only \u2018x\u2019 values:-
\n\u21d2Required area is shaded area = 2 X area of upper half of shaded region<\/p>\n

\"2nd<\/p>\n

Choose the correct answer in the following Exercise 12 and 13<\/p>\n

Question 12.
\nArea lying in the first quadrant and bounded by the circle x2<\/sup> + y2<\/sup> = 4 and the lines x = 0 and x = 2 is
\n(A) \u03c0
\n(B)\\(\\frac{\\pi}{2}\\)
\n(C)\\(\\frac{\\pi}{3}\\)
\n(D)\\(\\frac{\\pi}{4}\\)
\nAnswer:
\n(1) Circle is x2<\/sup> + y2<\/sup> = 4 with centre (0, 0) and radius 2.
\nse required area is shaded area of circle from
\n\"2nd
\n\"2nd
\nso answer is (A) \u03c0 unit ……
\nOR
\nAs circle of radius 2. so area of circle = \u03c0r2<\/sup>
\n\u03c0 x 22 <\/sup>= 4\u03c0
\nAs we need area of only 1 quadrant
\n\\(\\Rightarrow \\frac{1}{4} \\times \\text { area of circle }=\\frac{1}{4} \\times 4 \\pi=\\pi \\mathrm{sq.} \\text { units. }\\)
\noption (A)<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nArea of the region bounded by the curve y2<\/sup> = 4x, y – axis and the line y = 3 is
\n(A) 2
\n(B) \\(\\frac{9}{4}\\)
\n(C) \\(\\frac{9}{3}\\)
\n(D) \\(\\frac{9}{2}\\)
\nAnswer:
\n(1) Curve y2<\/sup> = 4x is right handed parabola
\n(2) line y = 3 \u21d2 points in line are (0,3), (1,3), (2,3),(3, 3), (-1,3) (-2, 3) etc
\n\"2nd
\n(3) Pt where line meets y2<\/sup> = 4x is solved by . solving the equation
\nas we know y = 3 \\(\\text { so } x=\\frac{3^{2}}{4} \\Rightarrow \\frac{9}{4}\\)
\n\\(\\text { so pt is }\\left(\\frac{9}{4}, 3\\right)\\)
\n(If we solve along \u2018y\u2019 axis)
\n[Here to solve along \u2018x\u2019 axis, it is tough as to get equation of line in terms of \u2018x\u2019 is tough.
\n(4) So area is area under parabola from y = 0 to y = 3.
\n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 8 Application of Integrals Ex 8.1 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 Ex 8.1 2nd PUC Maths …<\/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\/20458"}],"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=20458"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20458\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20458"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20458"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20458"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}