Question 5 - Chapter 8 Class 12 Application of Integrals (Important Question)

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Ex 8.2 , 5 Using integration find the area of the triangular region whose sides have the equations 𝑦=2𝑥+1, 𝑦=3𝑥+1 and 𝑥=4
Lets Draw the figure
& x = 4
Therefore,
Required Area = Area ABC
Finding point of Intersection B & C
For B
B is intersection of y = 3x + 1 & x = 4
Putting x = 4 in y = 3x + 1
y = 3(4) + 1 = 13
So, B(4, 13)
For C
C is intersection of y = 2x + 1 & x = 4
Putting x = 4 in y = 2x + 1
y = 2(4) + 1 = 9
So, C(4, 9)
Finding Area Required
Area ABC = Area OABD – Area OACD
Area OABD
Area OABD = ∫1_0^4▒〖𝑦 𝑑𝑥〗
Here, y = 3x + 1
Area OABD = ∫1_0^4▒〖(3𝑥+1) 𝑑𝑥〗
= [(3𝑥^2)/2+𝑥]_0^4
= [(3〖(4)〗^2)/2+4−[(3〖(0)〗^2)/2+0]]
= (3 × 16)/2 + 4 − 0
= 24 + 4
= 28
Area OACD
Area OACD = ∫1_0^4▒〖𝑦 𝑑𝑥〗
Here, y = 2x + 1
Area OACD = ∫1_0^4▒〖(2𝑥+1) 𝑑𝑥〗
= [(2𝑥^2)/2+𝑥]_0^4
= [(2〖×4〗^2)/2+4−[(2〖 × 0〗^2)/2+0]]
= 16 + 4 − 0
= 20
Area Required = Area ABDO − Area ACDO
= 28 − 20
= 8 square unit

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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