This question is similar to Chapter 8 Class 12 Application of Integrals - Examples

Please check the question here 

https://www.teachoo.com/3357/732/Example-13---Find-area-bounded-by-y--cos-x--x--0--2pi/category/Examples/

Question 32

Draw the rough sketch of the curve y=20 cos 2x; (where π/6≤x≤π/3)

Using integration, find the area of the region bounded by the curve y=20 cos2x from the ordinates x=π/6 to x=π/3 and the x-axis.

 

 

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Transcript

Question 32 Draw the rough sketch of the curve 𝑦=20 cos 2π‘₯; (where πœ‹/6≀π‘₯β‰€πœ‹/3) Using integration, find the area of the region bounded by the curve y=20 cos2x from the ordinates π‘₯=πœ‹/6 to π‘₯=πœ‹/3 and the π‘₯-axis.Now, 𝑦=20 cos 2π‘₯; Since πœ‹/6≀π‘₯β‰€πœ‹/3, We find value of y at key points At x = 𝝅/πŸ” 𝑦=20 cos 2(πœ‹/6) = 20 cos πœ‹/3 = 20 Γ—1/2 = 𝟏𝟎 At x = 𝝅/πŸ‘ 𝑦=20 cos 2(πœ‹/3) = 20 cos 2πœ‹/3 = 20 cos(πœ‹βˆ’πœ‹/3) = 20 Γ— βˆ’ cos πœ‹/3 = 20 Γ—(βˆ’1)/2 = βˆ’πŸπŸŽ At x = 𝝅/πŸ’ 𝑦=20 cos 2(πœ‹/4) = 20 cos πœ‹/2 = 20 Γ—0 = 𝟎 Thus, graph of 𝑦=20 cos 2π‘₯ is Now, Area Required = Area ADB + Area BEC + Area DEF Area ADB Area ADB = ∫_(πœ‹/6)^(πœ‹/( 4))▒〖𝑦 𝑑π‘₯γ€— 𝑦→20 cos⁑2π‘₯ = ∫_(πœ‹/6)^(𝝅/( πŸ’))β–’γ€–πŸπŸŽ π’„π’π’”β‘πŸπ’™ 𝒅𝒙〗 = 20[sin⁑2π‘₯/2]_(πœ‹/6)^(πœ‹/4) =10[sin⁑2(πœ‹/4)βˆ’sin⁑2(πœ‹/6) ] =10[sin⁑(πœ‹/2)βˆ’sin⁑(πœ‹/6) ] =10[1βˆ’βˆš3/2] =10[(2 βˆ’ √3)/2] =5(2 βˆ’ √3) =10βˆ’5(2 βˆ’ √3) =10[(2 βˆ’ √3)/2] =5(2 βˆ’ √3) =πŸπŸŽβˆ’πŸ“βˆšπŸ‘ Area BEC Area BEC = ∫_(πœ‹/4)^(πœ‹/( 3))▒〖𝑦 𝑑π‘₯γ€— 𝑦→20 cos⁑2π‘₯ = ∫_(πœ‹/4)^(𝝅/( πŸ‘))β–’γ€–πŸπŸŽ π’„π’π’”β‘πŸπ’™ 𝒅𝒙〗 = 20[sin⁑2π‘₯/2]_(πœ‹/4)^(πœ‹/3) =10[sin⁑2(πœ‹/3)βˆ’sin⁑2(πœ‹/4) ] =10[sin⁑(2πœ‹/3)βˆ’sin⁑(πœ‹/2) ] =10[sin⁑(πœ‹βˆ’πœ‹/3)βˆ’sin⁑(πœ‹/2) ] =10[sin⁑(πœ‹/3)βˆ’sin⁑(πœ‹/2) ] =10[√3/2βˆ’1] =10 Γ—βˆš3/2βˆ’10 =5√3βˆ’10 Since √3 = 1.73, 5√3βˆ’10 is negative And, area cannot be negative ∴ Area BEC = πŸπŸŽβˆ’πŸ“βˆšπŸ‘ Therefore Area Required = Area ADB + Area BEC = (10βˆ’5√3)+(10βˆ’5√3) = 2 Γ—(10βˆ’5√3) = πŸπŸŽβˆ’πŸπŸŽβˆšπŸ‘ square unit

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo