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.Now,
𝑦=20 cos 2𝑥;
Since 𝜋/6≤𝑥≤𝜋/3,
We find value of y at key points
At x = 𝝅/𝟔
𝑦=20 cos 2(𝜋/6)
= 20 cos 𝜋/3
= 20 ×1/2
= 𝟏𝟎
At x = 𝝅/𝟑
𝑦=20 cos 2(𝜋/3)
= 20 cos 2𝜋/3
= 20 cos(𝜋−𝜋/3)
= 20 × − cos 𝜋/3
= 20 ×(−1)/2
= −𝟏𝟎
At x = 𝝅/𝟒
𝑦=20 cos 2(𝜋/4)
= 20 cos 𝜋/2
= 20 ×0
= 𝟎
Thus, graph of 𝑦=20 cos 2𝑥 is
Now,
Area Required = Area ADB + Area BEC + Area DEF
Area ADB
Area ADB = ∫_(𝜋/6)^(𝜋/( 4))▒〖𝑦 𝑑𝑥〗
𝑦→20 cos2𝑥
= ∫_(𝜋/6)^(𝝅/( 𝟒))▒〖𝟐𝟎 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙〗
= 20[sin2𝑥/2]_(𝜋/6)^(𝜋/4)
=10[sin2(𝜋/4)−sin2(𝜋/6) ]
=10[sin(𝜋/2)−sin(𝜋/6) ]
=10[1−√3/2]
=10[(2 − √3)/2]
=5(2 − √3)
=10−5(2 − √3)
=10[(2 − √3)/2]
=5(2 − √3)
=𝟏𝟎−𝟓√𝟑
Area BEC
Area BEC = ∫_(𝜋/4)^(𝜋/( 3))▒〖𝑦 𝑑𝑥〗
𝑦→20 cos2𝑥
= ∫_(𝜋/4)^(𝝅/( 𝟑))▒〖𝟐𝟎 𝒄𝒐𝒔𝟐𝒙 𝒅𝒙〗
= 20[sin2𝑥/2]_(𝜋/4)^(𝜋/3)
=10[sin2(𝜋/3)−sin2(𝜋/4) ]
=10[sin(2𝜋/3)−sin(𝜋/2) ]
=10[sin(𝜋−𝜋/3)−sin(𝜋/2) ]
=10[sin(𝜋/3)−sin(𝜋/2) ]
=10[√3/2−1]
=10 ×√3/2−10
=5√3−10
Since √3 = 1.73, 5√3−10 is negative
And, area cannot be negative
∴ Area BEC = 𝟏𝟎−𝟓√𝟑
Therefore
Area Required = Area ADB + Area BEC
= (10−5√3)+(10−5√3)
= 2 ×(10−5√3)
= 𝟐𝟎−𝟏𝟎√𝟑 square unit

Made by

Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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