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Area between curve and line
Ex 8.1, 6 Important
Ex 8.2, 6 (MCQ) Deleted for CBSE Board 2023 Exams
Example 8 Important Deleted for CBSE Board 2023 Exams
Misc 8
Misc 9 Important
Ex 8.2 , 7 (MCQ) Important Deleted for CBSE Board 2023 Exams
Misc 6 Important
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 10
Ex 8.1, 10 Important You are here
Misc 7
Ex 8.1, 9
Misc 12 Important Deleted for CBSE Board 2023 Exams
Ex 8.2, 3 Important Deleted for CBSE Board 2023 Exams
Example 14 Important Deleted for CBSE Board 2023 Exams
Area between curve and line
Last updated at Dec. 12, 2019 by Teachoo
Ex 8.1, 10 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