{"id":4782,"date":"2023-12-22T10:18:42","date_gmt":"2023-12-22T04:48:42","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=4782"},"modified":"2023-12-23T17:41:06","modified_gmt":"2023-12-23T12:11:06","slug":"samacheer-kalvi-12th-maths-solutions-chapter-9-ex-9-8","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-maths-solutions-chapter-9-ex-9-8\/","title":{"rendered":"Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8"},"content":{"rendered":"

You can Download\u00a0<\/span>Samacheer Kalvi 12th Maths Book Solutions<\/a>\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/span><\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8<\/h2>\n

Question 1.
\nFind the area of the region bounded by 3x – 2y + 6 = 0, x = -3, x = 1 and x-axis.
\nSolution:
\n\"Samacheer<\/p>\n

Question 2.
\nFind the area of the region bounded by 2x – y + 1 = 0, y = – 1, y = 3 and y – axis.
\nSolution:
\n2x – y + 1 = 0
\n\"Samacheer
\nTo find further limit put x = 0, we get y = 1
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nFind the area of the region bounded by the curve 2 + x – x2<\/sup> + y = 0, x – axis, x = – 3 and x = 3.
\nSolution:
\n\"Samacheer<\/p>\n

Question 4.
\nFind the area of the region bounded by the line y = 2x + 5 and the parabola y = x2<\/sup> – 2x.
\nSolution:
\nTo find point of intersection of the curves
\ny = 2x + 5 and y = x2<\/sup> – 2x we get (-1, 3) and (5, 15)
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 5.
\nFind the area of the region bounded between the curves y = sin x and y = cos x and the lines x = 0 and x = \u03c0.
\nSolution:
\nTo find the points of intersection,
\n\"Samacheer<\/p>\n

Question 6.
\nFind the area of the region bounded by y = tan x, y = cot x and the line x = 0, x = \\(\\frac{\\pi}{2}\\), 0
\nSolution:
\nTo find the points of intersection of these two curves between 0 to \\(\\frac{\\pi}{2}\\) is \\(\\frac{\\pi}{4}\\)
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nFind the area of the region bounded by parabola y2<\/sup> = x and the line y = x – 2
\nSolution:
\nTo find the points of intersection solve the two equations y2<\/sup> = x and y = x – 2
\n\"Samacheer<\/p>\n

Question 8.
\nFather of a family wishes to divide his square field bounded by x = 0, x = 4 , y = 4 and y = 0 along the curve y2<\/sup> = 4x and x2<\/sup> = 4y into three equal parts for his wife, daughter and son. Is it possible to divide? If so, find the area to be divided among them.
\nSolution:
\nTo find the points of intersection of the two curves, y2<\/sup> = 4x and x2<\/sup> = 4y are (0, 0) and (4, 4).
\nArea of the square field = 4 \u00d7 4 = 16 sq. units
\n\"Samacheer
\nSo, the remaining each of the two parts must be \\(\\frac{16}{3}\\) sq.units.
\n\u2234 Yes, It is possible to divide the square field into three equal parts.<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nThe curves = (x – 2)2<\/sup> + 1 has a minimum point at P. A point Q on the curve is such that the slope of PQ is 2. Find the area bounded by the curve and the chord PQ.
\nSolution:
\ny = (x – 2)2<\/sup> + 1
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer
\n\u2234 x = 2 is a minimum point
\n\u2234 The point P is (2, 1)
\nBut slope of PQ is 2
\n\u2234 Equation of the chord PQ
\ny – y1<\/sub> = m(x – x1<\/sub>)
\ny – 1 = 2 (x – 2)
\ny – 1 = 2x – 4
\ny = 2x – 3
\nOn solving the curve and line we get the point Q(4, 5)
\n\"Samacheer<\/p>\n

Question 10.
\nFind the area of the region common to the circle x2<\/sup> + y2<\/sup> = 16 and the parabola y2<\/sup> = 6x.
\nSolution:
\nTo find points of intersection of x2<\/sup> + y2<\/sup> = 16 and y2<\/sup> = 6x are (2, \\(2 \\sqrt{3})\\)) and (2, –\\(2 \\sqrt{3})\\))
\nDue to symmetrical property,
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8 Additional Problems<\/h3>\n

Question 1.
\nFind the area of the region enclosed by y2<\/sup> = x and y = x – 2.
\nSolution:
\nThe points of intersection of the parabola y2<\/sup> = x and the line y = x – 2 are (1, -1) and (4, 2)
\n\"Samacheer
\nTo compute the region [shown in figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect toy no splitting is necessary.
\n\"Samacheer<\/p>\n

Question 2.
\nFind the area bounded by the curve y = x3<\/sup> and the line y = x.
\nSolution:
\nThe line y = x lies above the curve y = x3<\/sup> in the first quadrant and y = x3<\/sup> lies above the line y = x in the third quadrant. To get the points of intersection, solve the curves y = x3<\/sup>, y = x \u21d2 x3<\/sup> = x. We get x = {0, \u00b1 1}
\n\"Samacheer<\/p>\n

Question 3.
\nFind the area of the loop of the curve 3ay2<\/sup> = x (x – a)2<\/sup>.
\nSolution:
\nPut y = 0; we get x = 0, a
\nIt meets the x – axis at x = 0 and x = a
\n\u2234 Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nFind the area between the line y = x + 1 and the curve y = x2<\/sup> – 1.
\nSolution:
\nTo get the points of intersection of the curves we should solve the equations y = x +1 and y = x2<\/sup> – 1.
\nwe get, x2<\/sup> – 1 = x + 1
\nx2<\/sup> – x – 2 = 0 \u21d2 (x – 2)(x + 1) = 0
\nx = – 1 or x = 2
\n\u2234 The line intersects the curve at x = – 1 and x = 2.
\n\"Samacheer
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download\u00a0Samacheer Kalvi 12th Maths Book Solutions\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.8 Question 1. Find the area of the region bounded by 3x – 2y + 6 = …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4782"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=4782"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4782\/revisions"}],"predecessor-version":[{"id":60218,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/4782\/revisions\/60218"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=4782"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=4782"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=4782"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}