Using the method of integration find the area of the region boun

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 Multiple Choice QuestionsShort Answer Type

41.

Find the area bounded by the curve y = sin x between x = 0 and x = 2 straight pi.

103 Views

 Multiple Choice QuestionsLong Answer Type

42.

Using integration, find the area of the triangular region whose sides have the equations y = 2 x + 1, y = 3 x + 1 and x = 4.

132 Views

Advertisement

43. Using the method of integration find the area of the region bounded by lines:
2 x + y = 4, 3 x - 2 y = 6 and x - 3 y + 5 = 0.


The equations of the sides are
2 x + y - 4 = 0    ...(1)
3 x - 2 y - 6 = 0  ...(2)
x - 3 y + 5 = 0    ...(3)
Solving (1) and (2), we get
            fraction numerator straight x over denominator negative 6 minus 8 end fraction equals fraction numerator straight y over denominator negative 12 plus 12 end fraction equals fraction numerator 1 over denominator negative 4 minus 3 end fraction space space space space or space space fraction numerator straight x over denominator negative 14 end fraction equals straight y over 0 equals fraction numerator 1 over denominator negative 7 end fraction

therefore space space space space space space space straight x space equals space fraction numerator negative 14 over denominator negative 7 end fraction space equals space 2 comma space space space space straight y space equals space fraction numerator 0 over denominator negative 7 end fraction space equals space 0

therefore space space space space lines space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space intersect space in space straight C left parenthesis 2 comma space 0 right parenthesis

Solving (2) and (3), we get,
                                  fraction numerator straight x over denominator negative 10 minus 18 end fraction equals fraction numerator straight y over denominator negative 6 minus 15 end fraction equals fraction numerator 1 over denominator negative 9 plus 2 end fraction space space space or space space fraction numerator straight x over denominator negative 28 end fraction equals fraction numerator straight y over denominator negative 21 end fraction equals fraction numerator 1 over denominator negative 7 end fraction

therefore space space space space space space space space space straight x equals fraction numerator negative 28 over denominator negative 7 end fraction space equals 4 comma space space space space straight y space equals space fraction numerator negative 21 over denominator negative 7 end fraction space equals space 3

therefore space space space space lines space left parenthesis 2 right parenthesis space and space left parenthesis 3 right parenthesis space intersect space in space straight A left parenthesis 4 comma space 3 right parenthesis

Solving (1) and (3), we get,
            fraction numerator straight x over denominator 5 minus 12 end fraction equals fraction numerator straight y over denominator negative 4 minus 10 end fraction equals space fraction numerator 1 over denominator negative 6 minus 1 end fraction  or  fraction numerator straight x over denominator negative 7 end fraction equals fraction numerator straight y over denominator negative 14 end fraction space equals space fraction numerator 1 over denominator negative 7 end fraction

therefore space space space space space space space space straight x space equals space fraction numerator negative 7 over denominator negative 7 end fraction space equals space 1 comma space space space space space straight y space equals space fraction numerator negative 14 over denominator negative 7 end fraction space equals space 2

therefore space space lines space left parenthesis 1 right parenthesis space and space left parenthesis 3 right parenthesis space intersect space in space straight B left parenthesis 1 comma space 2 right parenthesis
From B. draw BL ⊥ x-axis and from A. draw AM ⊥ x-axis.

Required area = Area of increment ABC = Area of region BLMA - area of increment BLC space - area of increment ACM
equals space integral subscript 1 superscript 4 open parentheses fraction numerator straight x plus 5 over denominator 3 end fraction close parentheses dx space minus space integral subscript 1 superscript 2 left parenthesis 4 minus 2 right parenthesis space dx space minus space integral subscript 2 superscript 4 fraction numerator 3 straight x minus 6 over denominator 2 end fraction dx
equals space 1 third integral subscript 1 superscript 4 left parenthesis straight x plus 5 right parenthesis space dx space minus space integral subscript 1 superscript 2 left parenthesis 4 minus 2 straight x right parenthesis space dx space minus 1 half integral subscript 2 superscript 4 left parenthesis 3 straight x minus 6 right parenthesis space dx
equals space 1 third open square brackets straight x squared over 2 plus 5 straight x close square brackets subscript 1 superscript 4 space minus space open square brackets 4 straight x minus straight x squared close square brackets subscript 1 superscript 2 space minus space 1 half open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction minus 6 straight x close square brackets subscript 2 superscript 4
equals space 1 third open square brackets open parentheses 16 over 2 plus 20 close parentheses minus space open parentheses 1 half plus 5 close parentheses close square brackets space minus space left square bracket left parenthesis 8 minus 4 right parenthesis space minus space left parenthesis 4 minus 1 right parenthesis right square bracket minus 1 half left square bracket left parenthesis 24 minus 24 right parenthesis minus left parenthesis 6 minus 12 right parenthesis right square bracket
equals space 1 third open parentheses 28 minus 11 over 2 close parentheses minus left parenthesis 4 minus 3 right parenthesis minus 1 half left parenthesis 0 plus 6 right parenthesis space equals space 1 third cross times 45 over 2 minus 1 minus 3 space equals space 15 over 2 minus 4 space equals 7 over 2 space sq. space units

161 Views

Advertisement
44.

Using integration, find the area of the region bounded by (2, 5), (4, 7) and (6, 2).

105 Views

Advertisement
45. Using the method of integration, find the area of the triangle ABC, co-ordinates of whose vertices are A (2, 0), B (4, 5), C (6, 3).
154 Views

46.

Using integration, find the area of the region bounded by the triangle whose vertices are (-1, 1), (0, 5) and (3, 2).

286 Views

47. Using integration, find the area of the triangle ABC whose vertices have coordinates A (3, 0), B(4, 6) and C (6, 2).
144 Views

48. Using integration, find the area of the triangle ABC whose vertices are A (3, 0) B (4, 5) and C (5, 1).
182 Views

Advertisement

 Multiple Choice QuestionsShort Answer Type

49.

Using integration, find the area of the region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1).

109 Views

 Multiple Choice QuestionsLong Answer Type

50.

Using integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2).

151 Views

Advertisement