# Misc 10 - Chapter 8 Class 12 Application of Integrals

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 10 Find the area of the region enclosed by the parabola 𝑥2=𝑦, the line 𝑦=𝑥+2 and the 𝑥−axis Step 1: Draw the Figure Parabola is 𝑥2=𝑦 Also, 𝑦=𝑥+2 is a straight line Step 2: Finding point of intersection A & B Equation of line is 𝑦=𝑥+2 Putting value of y in equation of parabola 𝑥2=𝑦 𝑥2=𝑥+2 𝑥2−𝑥−2=0 𝑥2−2𝑥+𝑥−2=0 𝑥(x−2) +1(𝑥−2)=0 (𝑥+1)(𝑥−2)=0 So, x = –1, x = 2 Required Area Area required = Area ADOEB – Area ADOEBC Area ADOEB Area ADOEB = −12𝑦 𝑑𝑥 y → Equation of line y = x + 2 Therefore, Area ADOEB = −12 𝑥+2 𝑑𝑥 = 𝑥22+2𝑥−12 = 222+2 2− −122+2 −1 = 2+ 4 – 12 + 2 = 152 Area ADOEBC Area ADOEBC = −12𝑦 𝑑𝑥 y → Equation of parabola 𝑥2=𝑦 𝑦= 𝑥2 Therefore, Area ADOEBC = −12 𝑥2 𝑑𝑥 = 𝑥33−12 = 13 23− −13 = 93 = 3 Area required = Area ADOEB – Area ADOEBC = 152 – 3 = 92

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.