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

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Question 8 Find the area bounded by the curve 𝑥2=4𝑦 and the line 𝑥=4𝑦 – 2
Here,
𝑥2=4𝑦 is a parabola
And,
x = 4y – 2 is a line
which intersects the parabola at points A and B
We need to find Area of shaded region
First we find Points A and B
Finding points A and B
Points A & B are the intersection of curve and line
We know that,
𝑥=4𝑦−2
Putting in equation of curve , we get
𝑥^2=4𝑦
(4𝑦−2)^2=4𝑦
16𝑦^2+4−16𝑦=4𝑦
16𝑦^2−16𝑦−4𝑦+4=0
16𝑦^2−20𝑦+4=0
4[4𝑦^2−5𝑦+1]=0
4𝑦^2−5𝑦+1=0
4𝑦^2−4𝑦−𝑦+1=0
4𝑦(𝑦−1)−1(𝑦−1)=0
(4𝑦−1)(𝑦−1)=0
So, y = 1/4 , y = 1
For 𝒚=𝟏/𝟒
𝑥=4𝑦−2
𝑥=4(1/4)−2
𝑥 =−1
So, point is (–1, 1/4)
For y = 1
𝑥=4𝑦−2
𝑥=4(1)−2
𝑥 =2
So, point is (2, 1)
As Point A is in 2nd Quadrant
∴ A = (−1 , 1/4)
& Point B is in 1st Quadrant
∴ B = (2 , 1)
Finding required area
Required Area = Area APBQ – Area APOQBA
Area APBQ
Area APBQ = ∫_(−1)^2▒𝑦𝑑𝑥
Here, 𝑦 → Equation of Line
𝑥=4𝑦−2
𝑥+2=4𝑦
𝑦=(𝑥 + 2)/4
Area APBQ = ∫_(−1)^2▒(𝑥 + 2)/4 𝑑𝑥
= 1/4 ∫1_(−1)^2▒(𝑥+2)𝑑𝑥
= 1/4 [𝑥^2/2+2𝑥]_(−1)^2
= 1/4 [(2^2/2+2(2))−(〖(−1)〗^2/2+2(−1))]
= 1/4 [(2+4)−(1/2−2))]
= 1/4 [6−1/2+2]
= 1/4×15/2
= 15/8
Area APOQBA
Area APOQBA = ∫_(−1)^2▒〖𝑦 𝑑𝑥〗
Here, 𝑦 → Equation of Parabola
𝑥^2=4𝑦
4𝑦=𝑥^2
𝑦=1/4 𝑥^2
Area APOQBA = 1/4 ∫1_(−1)^2▒〖𝑥^2 𝑑𝑥〗
= 1/4 [𝑥^3/3]_(−1)^2
= 1/4 [((2)^3 − (−1)^3)/3]
= 1/4 [(8 − (−1))/3]
= 1/4 [(8 + 1)/3]
= 3/4
Now,
Required Area = Area APBQ – Area APOQBA
= 15/8 – 3/4
" = " (15 − 6)/8
= 9/8
∴ Required Area = 𝟗/𝟖 Square units

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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