Misc 10 - Chapter 8 Class 12 Application of Integrals (Term 2)
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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