Misc 10 - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Dec. 8, 2016 by Teachoo
Miscellaneous
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Miscellaneous
Last updated at Dec. 8, 2016 by Teachoo
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